CHAPTER
14
Theor y of Equations
As per the past years CAT papers, on an average two-three questions are being asked.
The problems of this chapter are not so much difficult to solve. Since this chapter is
concerned with secondary level syllabus therefore it is important in JMET, XAT and
SNAP etc. In this chapter we study quadratic equation, inequation and higher degree
polynomial expressions.
14.1 Definition of the Standard
Quadratic Expression
For the real numbers a, b and c, the expression ax 2 + bx + c is called the quadratic
expression, if a ≠ 0. Here a is the coefficient of x 2 , b is the coefficient of x and c is a
constant term.
(a) If a = 0 , then quadratic expression ax 2 + bx + c will reduce to a linear expression
bx + c. That means if a (the coefficient of x 2 ) is zero, the given expression will no
longer be a quadratic expression.
(b) When we say that a, b and c are real numbers, it means a, b and c can be any sort of
numbers like rational numbers or irrational numbers or fractions or integers.
Nomenclature
Format
Restrictions
Quadratic Expression
ax + bx + c
—
Quadratic Function
y = ax + bx + c
Any value
Quadratic Function
f (x ) = ax + bx + c Any value
Quadratic Equation
ax 2 + bx + c = 0
Strictly equal to zero
Quadratic Inequation
ax + bx + c ≥ 0
Non-negative values
Quadratic Inequation
ax + bx + c > 0
Positive values
Quadratic Inequation
ax + bx + c ≤ 0
Non-positive values
Quadratic Inequation
ax + bx + c < 0
Negative values
2
2
2
2
2
2
2
Thus by substituting the arbitrary numbers in place of x in the function you can obtain
various values of y or f ( x ).
NOTE You can use y and f ( x ) interchangeably as per the requirement or convenience. It means,
conceptually, y and f ( x ) both represent the same thing. Technically, the relation y = f ( x ) represents a
function, which means by substituting the values of x you can get the corresponding values of y.
Chapter Checklist
Definition of the Standard
Quadratic Expression
Characteristics of the Graph
of the Quadratic Function
Different Ways to Express the
Quadratic Equation
Solutions or Roots of the Quadratic
Equations
Methods of Solving the Quadratic
Equation
Formation of a Quadratic Equation
From the Known Roots
Nature of the Roots of the Quadratic
Equation
Maximum or Minimum value of a
Quadratic Equation
Condition for Common Roots
Between two Quadratic
Equations
Basics of Inequality
Position of Roots of a Quadratic
Equation with Respect to One
or Two Real Numbers
Relation Between the Roots of
Two Quadratic Equations
Polynomial Equations or Functions
of Higher Degree
End (or Long Term) Behaviour
of a Polynomial Function
Solutions (or Roots) of a
Polynomial Equation
Polynomial Inequality
Relation between Roots and
Coefficients in a Polynomial
Rational Polynomials
Rational Polynomial Inequalities
Maximum and Minimum value
of a Rational Expression
CAT Test
756
QUANTUM
Variants of the Quadratic Function
All of the following functions are quadratic functions since
the degree of each equation is 2. That means in each function
there is always a term x 2 or ax 2 for every non-zero value of a.
Here x is a variable and a, b, c are the constant numbers.
y =x2
y = ax 2
y = ax 2 + x
y = ax 2 + bx
y = ax 2 + c
y = ax + x + c
2
y = ax 2 + bx + c
Graphical Illustration
Let us consider a quadratic function y = x 2 − 5x + 6 . By
substituting the arbitrary numerical values of x in the above
function we get the corresponding values of y as shown in
the following table
x
y
−2
−1
0
1
20
12
6
2
2
0
3
4
5
6
7
0
2
6
12
20
Now if we connect all the points ( x, y) on the coordinate
plane, we will get a quadratic graph as shown below.
Please take note of the fact that you can consider any real
number for x to draw your graph.
Y-axis
20
18
16
12
8
6
4
–1
0
NOTE Even though there CANNOT be more than two roots of
any given quadratic equation, however, from any two given roots
you can have an infinite number of quadratic equations. To get all
the possible quadratic equations with both the roots common, just
multiply or divide the original (or basic) quadratic equation that you
have. So it follows that ax 2 + bx + c ⇒ k ( x − α )( x −β) = 0; Wherek is
any real number.
Therefore, you have to use your own discretion with regard to the
problem that k should be 1 or other values of k are
possible/valid/required.
8. If ( x −α ) and ( x − β) are the two factors of the quadratic
equation ax 2 + bx + c = 0, then ( x − α )( x − β) = 0
( x − α ) = 0 ⇒ x = α or ( x −β) =0 ⇒ x =β
Whereα andβ are the roots of the equation ax 2 + bx + c = 0
9. If quadratic equation is satisfied by more than two
distinct numbers (real or imaginary) then it becomes an
identity i.e., a = b = c = 0.
Becoming an identity implies that whatever value you
substitute for x, the equation will be satisfied for every
value of x.
2. A quadratic equation ax2 + bx + c = 0 may have
2
–2
7. If α and β are the two roots of the quadratic equation
ax 2 + bx + c = 0, then ax 2 + bx + c = 0 can be expressed
as the product of two factors ( x −α ) and ( x − β) as
following ax 2 + bx + c = ( x − α )( x − β) = 0
1. Which of the following is/are the solution(s) of
x2 − 7 x + 12 = 0?
(a) 2
(b) 3
(c) 4
(d) both 3 and 4
10
–3
4. A quadratic equation cannot have more than two
different roots.
5. A quadratic equation can have either two or zero REAL
roots.
6. If α is a root of the quadratic equation ax 2 + bx + c = 0,
then ( x − α ) is a factor of ax 2 + bx + c = 0.
Practice Exercise
14
–4
CAT
1
2
3
4
5
6
7
8
X-axis
–2
Properties of the Quadratic
Equations
1. The degree of any quadratic equation is always 2.
2. The value of x that satisfies the relation ax 2 + bx + c = 0
is called the root or zero or solution of this equation.
3. A quadratic equation has exactly two roots
(or solutions or zeros).
(a) two real roots
(b) two non-real roots
(c) one real and one non-real root
(d) Either (a) or (b)
3. If 5 and −6 are the roots of the quadratic equation,
then what will be the factors of the quadratic equation?
(a) (x − 5 ) and (x − 6 )
(b) (x − 5 ) and (x + 6 )
(c) (x + 5 ) and (x − 6 )
(d) (x + 5 ) and (x + 6 )
4. If 3/2 and 4 are the two roots of a quadratic equation,
then which one of the following is not the correct
quadratic equation?
(a) 2 x2 − 11x + 6 = 0
(b) 6 x2 − 33 x + 18 = 0
2
(c) −10 x − 55 x − 30 = 0 (d) 4 x2 − 41x + 24 = 0
Theory of Equations
757
5. If (x − 7 ) and (x − 17 ) are the factors of a quadratic
equation, then which one of the following is the correct
equation?
(a) x2 − 10 x + 24 = 0
(b) x2 − 33 x + 119 = 0
(c) x2 + 24 x − 119 = 0
(d) x2 − 24 x + 119 = 0
6. For what value of p, the equation ( p − 1)(2 p + 1)x2
+ ( p2 − 1)x + ( p − 1)( p − 3 ) = 0 is an identity (or has
more than two solutions)?
(a) −1
(b) −1 / 2
(c) 1
(d) 1/3
It implies that how far the vertex is from the X-axis. The
positive (or negative) value of vertex y indicates that the
vertex is above (or below) the X-axis.
11. The parabolic graph must intersect Y-axis but only once;
except in rare cases where the values of x = 0 is not
allowed, it may not intercept. It implies that the graph of
quadratic function, in general, may exist anywhere across
the whole X-axis, depending on the quadratic function.
NOTE For more info about the axis and Cartesian plane you can
refer Coordinate Geometry.
Parabola
7. In the given equation ax2 + bx + c, if a = 0 and b ≠ 0,
then
(a) The equation represents a linear equation
(b) The equation represents a quadratic equation
(c) The equation represents a bi-quadratic equation
(d) none of the above
Fig (i)
2. (d)
7. (a)
3. (d)
Vertex (Highest
point of the graph)
Axis of
Symmetry
Answers
1. (d)
6. (c)
Vertex (Lowest
point of the graph)
Axis of
Symmetry
4. (d)
5. (d)
Parabola
14.2 Characteristics of the Graph
of the Quadratic Function
1. A graph of a quadratic equation is exactly a parabola as
shown below.
2. The parabola is always symmetric.
3. The line that splits the parabola through the middle is
called the "axis of symmetry".
4. Axis of symmetry is obtained by x = − b/2a. It implies
that how far the axis of symmetry is from the Y-axis.
5. The axis of symmetry may or may not overlap the
Y-axis, which depends on the quadratic function. But
axis of symmetry is always parallel to Y-axis.
6. The point on the axis of symmetry that bisects the
parabola is called the "vertex", and it is the point where
the curvature is greatest.
7. The parabola can open UP or DOWN only.
8. If the parabola opens up, the vertex will be the lowest
point and if it opens down, the vertex will be the highest
point of the graph.
9. The lowest point of the graph is called the minimum
(plural is minima) and the highest point of the graph is
called the maximum (plural is maxima).
10. The vertex of the quadratic graph is obtained by
( 4ac − b 2 )
.
y=
4a
Fig (ii)
Practice Exercise
1. Which of the following is not a graph of a quadratic
equation?
(i)
(ii)
(iii)
(iv)
(a) only (iii)
(b) (i) and (iii) (c) (ii) and (iv) (d) only (iv)
2. Which of the following is a correct graph of a quadratic
equation, where the vertical line represents the axis of
symmetry?
(i)
(ii)
(iii)
(iv)
(a) Only (iii)
(c) (i) and (iv)
(b) (i) and (iii)
(d) none of the graphs
758
QUANTUM
3. Consider the following graph of a quadratic function.
Then which of the following facts are true regarding
this quadratic graph?
Y
12
8
4
–8 –7 –6 –5 –4 –3 –2 –1 0
1
2 3
4
5
X
–4
–8
–12
y′
(i) Y−intercept is −8
(iii) a>0
(v) D>32a
(ii) X−intercepts are −4 and 2.
(iv) D>0
(vi) b = 2a
(vii) Vertex of the graph is less than −8.
(viii) The difference between the Y-axis and axis of
symmetry is 1 unit.
(ix) The equation of axis of symmetry is x = − 1.
(x) The value of y is same, if it is equidistant from x = − 1.
(a) (i), (ii), (iii) and (vi)
(b) (ii), (iii), (iv) and (x)
(c) Except (v), (vi) and (x) (d) all of these
4. For the quadratic equation 3 x2 + 12 x + 10 = 0, what is
the distance between the Y-axis and the axis of
symmetry of the quadratic graph?
(a) 2 units (b) 4 units (c) 3 units
(d) 6 units
5. The two roots of a quadratic equation are −6 and 12,
what is the distance between the axis of symmetry of
the quadratic graph and the Y-axis?
(a) 6 units (b) 1unit
(c) 2 units
(d) 18 units
6. For the quadratic equation x2 + 6 x − 16 = 0, what will
be the position of the axis of symmetry of the
quadratic graph with reference to the Y-axis?
(a) 2 units left
(b) 3 units right
(c) 3 units left
(d) can’t be determined
7. The axis of symmetry, of a quadratic graph, is always
parallel to
(a) X-axis
(b) Y-axis
(c) Z-axis
(d) none of these
8. For a quadratic graph, the axis of symmetry and Y-axis
coincide (or overlap) then,
(a) a − b = c
(b) 2 a − b = 0
(c) b = 0
(d) c = 0
9. For the quadratic equation 3 x2 + 12 x + 10 = 0, what is
the vertex of the graph?
(a) y = − 2
(b) y = − 4 (c) y = 3
(d) 6
10. For the quadratic equation x2 + 12 x + 20 = 0, how far
is the vertex from the X-axis?
(a) 8 unit
(b) 16/3 unit (c) 16 unit (d) 8/3 unit
CAT
11. For the quadratic equation x2 − 11x + 24 = 0, what will
be the position of the vertex of the quadratic graph
with reference to the X-axis?
(a) 25/4 units below
(b) 25/4 units above
(c) 21/4 units below
(d) can’t be determined
12. For the quadratic equation x2 + 9 x + 18 = 0, what will
be the position of the vertex of the quadratic graph
with reference to the origin (or intersection of both the
axes)?
(a) 9/2 units left, 113/4 units below
(b) 9/2 units left, 113/4 units above
(c) 9 units right, 81 units below
(d) 9 units right, 9 units below
13. Find the value of y where the
3 x2 + 19 x − 18 = 0 intersects the Y-axis.
(a) −18
(b) 0
(c) 4
(d) 16
graph
of
14. Which of the following is false?
(i) If only the positive values of x are taken, the
quadratic graph does not intersect the Y-axis.
(ii) If only negative values are taken, the quadratic
graph does not intersect the Y-axis.
(iii) The interception of Y-axis can be known when we
substitute x = 0, in the quadratic function.
(iv) There can be minimum zero and maximum two
x-intercepts.
(a) (i) and (iv)
(b) (i), (iii), (iv)
(c) (ii), (iii), (iv)
(d) none of these
15. If D be the discriminant of the quadratic equation
ax2 + bx + c = 0, what will be the coordinates of its
vertex?
(a) −b /2 a, −D / 4 a
(b) −b /2 a, D / 4 a
(c) b /2 a, −D /2 a
(d) −b /2 a, D / 4 a
Answers
1. (d)
6. (c)
11. (a)
2. (d)
7. (b)
12. (a)
3. (d)
8. (c)
13. (a)
4. (a)
9. (a)
14. (d)
5. (b)
10. (b)
15. (a)
14.3 Different Ways to Express
the Quadratic Equation
A quadratic equation can be expressed in the following forms
(1) ax 2 + bx + c = 0
 b
 c
(2) x 2 +   x +   = 0;
a
a
where a, b and c are the standard notations.
(3) x 2 − (α + β) x + (αβ) = 0; where α and β are the roots of
the equation.
Theory of Equations
759
Exp.) We can express a quadratic equation in three
different forms, whose roots are (−2) and (1/2).
(1) 2x 2 + 3 x − 1 = 0
 3
 1
(2) x 2 +   x +  −  = 0; where a = 2, b = 3 , c = − 1
 2
 2
3
1
x − =0
2
2
1
 3
 1
2
(3) x −  −  x +  −  = 0; where α = − 2, β =
 2
 2
2
3
1
x2 + x − = 0
⇒
2
2
x2 +
⇒
Discriminant of the Quadratic Equations
For the given quadratic equation ax 2 + bx + c = 0, the
discriminant is denoted by D and expressed as D = b 2 − 4ac.
Exp.) Determine the discriminant of the following
quadratic equations.
(i) x 2 + 10 x + 6
(iii) − x − 8 x + 15
2
(v) x − 16
2
(ii) 8 x + 17 + 5 x 2
(iv) x + 6 x
2
(vi) 9 x 2
Solution (i) Discriminant of the equation x 2 + 10x + 6 is
102 − 4 × 1 × 6 = 76
(ii) Discriminant of the equation 5 x 2 + 8x + 17 is
82 − 4 × 5 × 17 = − 276
(iii) Discriminant of the equation − x 2 − 8x + 15 is
(ii) Discriminant of the equation 5x2 + 8x + 17 is −276 < 0, so
it cannot be factorised into two linear factors.
(iii) Discriminant of the equation x2 − 16 is 64 > 0, so it is
factorisable into two linear factors as x2 − 16 = (x + 4)(x − 4)
(iv) Discriminant of the equation 9x2 is 0, so it is factorisable
into two linear factors as 9x2 = (3x )(3x ) or 9x2 = x (9x ).
How to Factorize a Given
Quadratic Equation
Step 1 First of all write down the given quadratic equation
in its standard form of ax 2 + bx + c = 0
Step 2 If any of the coefficients a, b and c is a rational
number, make it integer by multiplying whole
equation by the denominator of that rational number.
Step 3 Find the product ac.
Step 4 Find all the factor pairs of ac, such that the product of
any pair is equal to ac.
Step 5 Choose one such pair, say ( m, n), such that if ac is
positive then m + n = b and if ac is negative then
m − n = − b, provided m ≤ n.
Step 6 Write down the expression as ax 2 + ( m + n) + c = 0
or ax 2 + ( m − n) + c = 0, as per the nature of b.
Step 7 Now make two pairs out of four terms considering
there will be something common in the pairs as shown
below. If ax 2 + ( m + n) + c = 0
⇒
( ax 2 + m) + ( n + c) = 0
(iv) Discriminant of the equation x 2 + 6x is 62 − 4 × 1 × 0 = 36
or
( ax 2 + n) + ( m + c) = 0
(v) Discriminant of the equation x 2 − 16 is 02 − 4 × 1 × −(16) = 64
If
ax 2 + ( m − n) + c = 0
(vi) Discriminant of the equation 9x 2 is 02 − 4 × 9 × 0 = 0
⇒
( ax 2 + m) + ( −n + c) = 0
Condition that the Quadratic Equation is
Factorisable into two Linear Factors
When the discriminant D ≥ 0, then the equation
ax 2 + bx + c = 0 is factorisable into two linear factors. That is
ax 2 + bx + c = 0
⇒ ( x − α )( x − β) =0; where α and β are the roots of the
quadratic equation.
or
( ax 2 − n) + ( m + c) = 0
( −8) 2 − 4 × ( −1) × 15 = 124
Exp.) Which of the following quadratic expressions are
factorisable into two linear factors?
(i) 2 x 2 + 10 x + 6
(iii) x 2 − 16
(ii) 5 x 2 + 8 x + 17
(iv) 9 x 2
Solution
(i) Discriminant of the equation x2 + 10x + 6 is 76 > 0, so it
is factorisable into two linear factors as x2 + 10x + 6
= (x + 19 ) (x − 19 ).
Step 8 Take out common if any, and then simplify.
NOTE This technique has some limitations that you will come
across when you try to find the factors of an equation whose roots
are irrational. Therefore, you may use other techniques to find the
factors, like using sum and product of roots method or
Sridharacharya’s method.
Exp. 1) Factorize the given quadratic equation
2x 2 + 3x − 9 = 0.
Solution Product a ⋅ c = 2 × −9 = − 18
Now,
18 = 1 × 18 or 2 × 9 or 3 × 6
Since a ⋅ c is negative, therefore we will take the difference of 3
and 6 and which is 3 = b. Therefore we choose 3 and 6.
Now we write the equation, as following 2x 2 + 3 x − 9 = 0
⇒
⇒
2x 2 + ( 6x − 3 x) − 9 = 0 ⇒( 2x 2 + 6x) − ( 3 x + 9) = 0
2x( x + 3) − 3( x + 3) = 0 ⇒( x + 3)( 2x − 3) = 0
760
QUANTUM
Exp. 2) Factorize the given quadratic equation
5
3 − x − 3x 2 = 0.
2
Solution The given quadratic equation can be expressed in the
5
standard form as −3 x 2 − x + 3 = 0
2
Since the value of b is a rational number therefore we will
multiply the whole equation by the denominator of b that is 2
and then we will get the whole equation with integer
coefficients −6x 2 − 5 x + 6 = 0. Since product a ⋅ c = − 6 × 6 = −36
Therefore, 36 = 1 × 36 or 2 × 18 or 3 × 12 or 4 × 9 or 6 × 6
Now, since a ⋅ c is negative, therefore we will take the
difference of 4 and 9 which is 5 = b. Therefore, we choose 4 and
9. Now, we write the equation as following
−6x 2 − 5 x + 6 = 0 ⇒ −6x 2 + ( 4x − 9x) + 6 = 0
⇒
( 4x − 6x 2 ) + ( 6 − 9x) = 0
⇒ 2x( 2 − 3 x) + 3( 2 − 3 x) = 0 ⇒( 2 − 3 x)( 2x + 3) = 0
Exp. 3) Factorize the given biquadratic equation
x 4 + 2x 2 − 8 = 0.
Solution The given biquadratic equation can be expressed in the
standard form as ( x 2 ) 2 + 2x 2 − 8 = 0
Product
a ⋅ c = 1 × −8 = − 8
Now,
8 = 1 × 8 or 2 × 4
Since a ⋅ c is negative, therefore we will take the difference
of 2 and 4 and which is 2 = b. Therefore, we choose 2 and 4.
Now we write the equation as following ( x 2 ) 2 + 2x 2 − 8 = 0
⇒
( x 2 ) 2 + ( 2x 2 − 4x 2 ) − 8 = 0
⇒
[( x 2 ) 2 + 2x 2 ] + ( −4x 2 − 8) = 0
⇒
⇒
2
2
2
2
( x 2 + 2)( x + 2)( x − 2) = 0
Exp. 4) Factorize the given quadratic equation
x 6 + 6x 3 + 5 = 0.
Solution The given quadratic equation can be expressed in the
standard form as ( x 3 ) 2 + 6x 3 + 5 = 0. Product a ⋅ c = 1 × 5 = 5
Now, 5 = 1 × 5 is the only pair, so it has got to be easy.
Since a ⋅ c is positive, therefore we will take the addition of 1 and
5 and which is 6 = b. Now we write the equation as following
( x 3 ) 2 + 6x 3 + 5 = 0
⇒
(x3) 2 + (x3 + 5x3) + 5 = 0
⇒
[( x 3 ) 2 + x 3 ] + (5 x 3 + 5) = 0
⇒
[( x 3 ) 2 + x 3 ] + (5 x 3 + 5) = 0
⇒
x 3 ( x 3 + 1) + 5( x 3 + 1) = 0
⇒
( x 3 + 5)( x 3 + 1) = 0
⇒
( x 3 + 5)( x 3 + 13 ) = 0
⇒ ( x + 5)( x + 1)( x − x + 1) = 0
3
2
Practice Exercise
Find the facrtors of following equations.
1. x2/3 − x1/3 − 56
2. x + 9 x1/2 + 8
3. 4 x 4 − 81
4. (x − 3 )4 + 2 (x − 3 )2 − 8
5. 6 x2 − xy − 12 y2
6. 20 x2 + 17 x − 91
7. 35 x + 84 − 21x2
Answers
1. Assume
1
x3
= p, then the given equation will become like
p2 − p − 56
p2 − p − 56 = p2 − 8p + 7p − 56
Then,
= (p + 7)(p − 8)
Therefore,x2/3 − x1/3 − 56 = (x1/3 + 7)(x1/3 − 8)
1
2. Assume x2 = p, then the given equation will become like
p2 + 9p + 8
Then, p2 + 9p + 8 = p2 + p + 8p + 8 = (p + 8)(p + 1).
Therefore, x + 9x1/2 + 8 = (x1/2 + 8)(x1/2 + 1)
3. We can apply (a2 − b2 ) = (a + b)(a − b)
4x 4 − 81 ⇒ (2x )2 − 92 = (2x2 + 9)(2x2 − 9)
4. Assume (x − 3)2 = p, then the given equation will become
like p2 + 2p − 8.
Then, p2 + 2p − 8 = p2 + 4p − 2p − 8 = (p + 4)(p − 2).
5. 6x2 − xy − 12 y2 = 6x2 − 9xy + 8xy − 12 y2
x ( x + 2) − 4( x + 2) = 0 ⇒( x + 2)( x − 4) = 0
2
CAT
Then find a pair of factors of 6 × 12, such that their
difference is 1. That is 6 × 12 = 9 × 8
= 3x (2x − 3 y ) + 4 y (2x − 3 y )
= (2x − 3 y )(3x + 4 y ).
6. Multiply 20 and 91. That is 2 × 2 × 5 × 7 × 13. Then split
these prime factors in two groups such that the difference
of these two factors is 17. That is,
(2 × 2 × 13)(5 × 7) = 52 × 35
Now, we have, 20x2 + 17x − 91
= 20x2 + (52x − 35x ) − 91
= (4x − 7)(5x + 13)
7. 35x + 84 − 21x = − 21x2 + 35x + 84
2
= − 7(3x2 − 5x − 12)
Now multiply 3 and 12. That is 3 × 2 × 2 × 3
Then find the pair of factors of 3 × 2 × 2 × 3 such that the
difference of these two factors is 5.
That is
3 × 2 × 2 × 3 = 4 × 9.
Now,
−21x2 + 35x + 84
= − 7(3x2 − 5x − 12)
= − 7(3x2 − 9x + 4x − 12)
= − 7[(x − 3)(3x + 4)]
Theory of Equations
14.4 Solutions or Roots of the
Quadratic Equations
The value of x that satisfies the relation ax 2 + bx + c = 0 is
called the root or solution of this equation. Simply, the root of
the quadratic function y = ax 2 + bx + c = 0 is the value of x for
which y becomes 0.
NOTE It is preferable to use the term zero while referring a
quadratic function and use the term root while referring to a
quadratic equation.
Exp. 1) The roots of the quadratic equation
x 2 − 8x + 15 = 0 are 3 and 5. As, by substituting 3 or 5 in
place of x, both sides of the equation become 0.
Exp. 2) The roots of the quadratic function
y = x 2 − 8x + 15 are 3 and 5. As by substituting 3 or 5 in
place of x, the value of y becomes 0.
Exp. 3) The roots of the quadratic equation
x 2 − 8x + 22 = 7 are 3 and 5. The given equation can be
simplified as x 2 − 8x + 15 = 0. As, by substituting 3 or 5 in
place of x, both sides of the equation become 0.
Exp. 4) The roots of the quadratic equation x 2 + 9x − y
= 17 x − 15 are 3 and 5. The given equation can be
simplified as y = x 2 − 8x + 15.
As, by substituting 3 or 5 in place of x, the value of y becomes 0.
Sum and Product of the Roots of a
Quadratic Equation
If the two roots of the quadratic equation ax 2 + bx + c = 0 are
α and β,
−b
(i) Sum of the roots = (α + β) =
a
c
(ii) Product of the roots = (αβ) =
a
Exp. 1) Find the sum and product of the roots of the
quadratic equation x 2 − 11x + 28 = 0.
Solution Comparing the given equation with the standard
form of the quadratic equation ax 2 + bx + c = 0, we get
a = 1, b = − 11, c = 28. Therefore, sum of roots
− b −( −11)
c 28
= =
= 11 and product of the roots = =
= 28.
a
1
a
1
Exp. 2) Find the sum and product of the roots of the
quadratic equation 3x 2 + 15x + 12 = 0.
Solution Comparing the given equation with the standard form
of the quadratic equation ax 2 + bx + c = 0, we get
− b −15
=
= −5
a = 3 , b = 15 , c = 12. Therefore, sum of roots =
a
3
c 12
and product of the roots =
= 4.
a
3
761
Exp. 3) If the sum of the roots of the quadratic equation
ax 2 + bx + c = 0 is equal to the sum of the squares of their
reciprocals, show that bc 2 , ca 2 , ab 2 are in Arithmetic
Progression (AP). It is well known that if any three
consecutive terms, form a sequence, are such that the
difference between any two consecutive terms is same,
then they are in AP.
For Example if x , y , z are in AP, then z − y = y − x = d (say).
Solution If α and β be the roots of the quadratic equation
−b
...(i)
α +β =
ax 2 + bx + c = 0, then
a
c
...(ii)
αβ =
a
1
1
Given that, α + β = 2 + 2
α
β
⇒
α+β=
α 2 + β2
(αβ) 2
⇒α+β=
(α + β) 2 − 2αβ
(αβ) 2
2
−b
=
a
⇒
⇒
 −b
 c
  − 2 
 a
 a
 c
 
 a
bc2 + ab 2 = 2ca 2
2
2
2
⇒
− b b 2 − 2ac
=
a
c2
⇒ bc2 − ca 2 = ca 2 − ab 2
2
Therefore bc , ca , ab are in Arithmetic Progression.
NOTE For better understanding of Arithmetic Progression you
may refer the chapter Sequence, Series and Progression.
Practice Exercise
1. If one root of a quadratic equation is zero, which one of
the following is correct?
(a) ax2 + bx + c = 0
(b) ax2 + c = 0
2
(c) ax + bx = 0
(d) bx + c = 0
2. If both the roots of a quadratic equation are equal in
magnitude but opposite in nature, which one of the
following is correct?
c
 b
(a) x2 +   x −   = 0
(b) ax2 + bx = 0
 a
 a
(c) x2 = 0
(d) ax2 + c = 0
3. If both the roots of a quadratic equation are
coincident, which one of the following is not correct?
(b) x2 + 2 kx + k2 = 0
(a) x2 − 2 cx + c = 0
2
2
(c) x − 2 bx − b = 0
(d) x2 − 2 kx + k2 = 0
4. If one root of the equation is reciprocal to the other
root, which of the following is correct?
(a) ax2 + bx = 0
(b) ax2 + bx + 1 = 0
2
(c) ax − bx = 0
(d) x2 + x − c = 0
5. If one root of the quadratic equation x2 − x − 1 = 0 is α,
the other root is
(a) α 3 + 3 α (b) α 3 − 3 α (c) α 2 + 3
(d) α 2 − 3
Answers
1. (c)
2. (d)
3. (c)
4. (b)
5. (b)
762
QUANTUM
14.5 Methods of Solving the
Quadratic Equation
(1)
(2)
(3)
(4)
⇒
⇒
Factorization Method
Step 1 If the equation is not given in the standard form then
express the given equation in the standard form of
ax 2 + bx + c = 0.
2
Step 4 Simplify each linear factor.
55
Exp.) Find the roots of
x + x 2 − 1 = 0.
24
Step 5 Find the square root of both the sides.
55
x − 1 =0
24
24x 2 + 55 x − 24 = 0 ⇒( 3 x + 8)( 8x − 3) = 0
Solution The given expression can be written as x 2 +
8 
3
8
3

 x +   x −  = 0 ⇒ x + = 0 or x − = 0

3 
8
3
8
−8
3
or x =
⇒
x=
3
8
8
3
Therefore, there are two roots − and .
3
8
⇒
−b + b 2 − 4ac
2a
and
−b − b − 4ac
2
55
Exp.) Find the roots of
x + x 2 − 1 = 0.
24
2a
Solution The given expression can be written as x 2 +
⇒
⇒
⇒
2
2
2
2
55 
 73 

x +  =  
 48

48
x+
55
x − 1= 0
24
24x 2 + 55 x − 24 = 0
− b ± b 2 − 4ac −55 ± 55 2 − 4 × 24 × ( −24)
=
2a
2 × 24
55
x = 1.
24
2
55
73
55 73
=±
±
⇒x = −
48
48
48 48
128
8
18
3
and
x=−
⇒ x = − and
48
3
48
8
Exp. 2) Find the roots of 2x 2 − 3x − 1 = 0.
⇒
⇒
⇒
Here, a = 24, b = 55 and c = − 24.
Now,
2
55
 55 
 55 
x +   =1 +  
 48
 48
24
Solutions The given expression can be written as x 2 −
.
Step 3 Then one root (say α) is equal to
⇒
Solution The given expression can be written as x 2 +
 55 
 55 
 73 
⇒ x 2 + 2  x +   =  
 48
 48
 48
Step 2 Substitute the values of a, b and c in the formula
another root (say β) is equal to
Step 6 Simplify the resulting equation.
55
Exp. 1) Find the roots of
x + x 2 − 1 = 0.
24
Then, x 2 +
Sridharacharya Method
Step 1 If the equation is not given in the standard form then
express the given equation in the standard form of
ax 2 + bx + c = 0.
2a
Step 2 Modify the equation as ax 2 + bx = − c
 b
Step 4 Add   to both the sides.
 2
Step 3 Equate each linear factor with zero.
−b ± b 2 − 4ac
Square Completion Method
Step 1 If the equation is not given in the standard form then
express the given equation in the standard form of
ax 2 + bx + c = 0.
Step 3 If a ≠1, divide the whole equation by a
Step 2 Factorize the equation ax 2 + bx + c = 0 into two
linear factors.
⇒
−55 ± 5329 −55 ± 73
=
48
48
−55 + 73
−55 − 73
and
48
48
8
3
− and .
3
8
⇒
Factorization Method
Sridharacharya Method
Square Completion Method
Sum and Product of Roots Method
CAT
⇒
⇒
1
1
 3
 3
x 2 −   x − = 0 ⇒ x 2 − 2  x =
 2
 4
2
2
2
1  3
 3
 3
x 2 − 2  x +   = +  
 4
 4
2  4
2
2
3
17

⇒ x −  =±

4
4
3
±
4
3
x= +
4
3+
x=
4
17
4
3
17
17
and −
4
4
4
3 − 17
17
and
4
3
17

x −  =

4
16
x=
3
1
x − = 0.
2
2
Theory of Equations
763
Exp. 3) Find the roots of x 2 + 2x + 5 = 0.
p( x − a ) +
and
Solution The given expression can be written as x 2 + 2x = − 5.
⇒
⇒
x 2 + 2x + 1 = − 5 + 1
⇒
( x + 1) = − 4 ⇒( x + 1) = ± −4
⇒
⇒
( x + 1) = ± 2i ⇒ x = − 1 ± 2i
x = − 1 + 2i and −1 − 2i
2
Type 3.
Sum and Product of Roots Method
Step 1 If the equation is not given in the standard form then
express the given equation in the standard form of
ax 2 + bx + c = 0.
b
Step 2 Find sum of the roots (α + β) = − and product of
a
c
the roots (αβ) = .
a
Step 3 Substitute the values of α + β and αβ in the formula
(α − β) 2 = (α +β) 2 − 4αβ.
Step 4 Find the value of α − β.
Step 5 Solve the equations α +β and α − β to get the roots
α and β.
Exp.) Find the roots of 55 x + x 2 − 1 = 0.
24
Solution The given expression can be written as x 2 +
The sum of the roots = α + β = −
55
x − 1 = 0.
24
55
and the product of the roots
24
= αβ = − 1.
Therefore using (α −β) 2 = (α + β) 2 − 4αβ, we get
Type 4.
p( x − a )( x − b) − r ( x − b) + q = 0
a − x = bx + c. Square both sides and simplify.
2
i. e.,
a − x 2 = b 2 x 2 + 2bcx + c 2
⇒
(1 + b 2 ) x 2 + 2bcx + ( c 2 − a ) = 0
ax + b + cx + d = e Transform one of the
radicals to RHS and square ax + b = e − cx + d ,
such equations may require squaring and your
solution must satisfy ax + b ≥ 0 and cx + d ≥ 0.
Type 5.
1 
1


a  x 2 + 2  + b  x +  + c = 0 For this type of



x
x
equations we use the following identity
1
2


1
1

Thus a   x +  − 2 + b  x +  + c = 0

x
x
 

1
Now put x + = y to get a quadratic equation
x
i. e.,
ay 2 + by + ( c − 2a ) = 0
1 
1


Type 6. a  x 2 + 2  + b  x −  + c = 0



x
x
2
 8
Solving α + β and α − β, we will get (α , β) =  − ,
 3
3
.
8
Solutions of Equations Reducible
to Quadratic Form
Equation which are not quadratic at a glance but can be
reduced to quadratic equations by suitable transformations.
Some of the common types are :
Type 1. ax 4 + bx 2 + c = 0. This can be reduced to a
quadratic equation by substituting x = y.
2
ay 2 + by + c = 0
q
q
Tpye 2. px + = r, p( x − a ) +
= r. Multiply both
x
( x − b)
sides by the LCD of LHS to get a quadratic
q
equation px + = r ⇒ px 2 − rx + q = 0
x
i. e.,
2
1

x + 2 = x +  − 2

x
x
2
Use the following identity x 2 +
73
 −55 
(α − β) 2 = 
 − 4( −1) ⇒α −β = ±
 24 
24
q
=r
( x − b)
2
1

= x −  + 2
2

x
x
1
2


1
1

Thus, a   x −  + 2 + b  x −  + c = 0, put



x
x


1
x − = y to get a quadratic equation in y.
x
i.e.,
ay 2 + by + ( c − 2a ) = 0
Type 7. (i) x 2a + x a + b = 0 (ii) x a + x − a = b
Put x a = y to get a quadratic equation in y,
i.e., (i) y 2 + y + b = 0 (ii) y +
1
=b
y
NOTE In all the equations involving radical signs, the answer
must be checked by substituting in the original equation.
Type 8. ( x + a )( x + b)( x + c)( x + d ) + k = 0. When sum of
the quantities a, b, c, d is equal to the sum of the
other two, can be solved as shown in example 16.
764
QUANTUM
Exp. 1) Solve for y : 9y 4 − 29y 2 + 20 = 0.
9y 4 − 29y 2 + 20 = 0
Solution
y = x.
Put
9x − 29x + 20 = 0 ⇒ 9x 2 − 20x − 9x + 20 = 0
2
20
( x − 1)( 9x − 20) = 0 ⇒ x = 1 or x =
9
20
2 5
2
2
⇒ y = ± 1 and y = ±
y = 1 or y =
3
9
⇒
⇒
Exp. 2) Solve for x : x 6 − 26x 3 − 27 = 0.
x 6 − 26x 3 − 27 = 0
Solution
x3 = y
Let
y 2 − 26y − 27 = 0
then,
⇒
1
1
Exp. 6) Solve for x : 2  x 2 + 2  − 9  x +  + 14 = 0.


x
x 2
Solution Put x 2 +
2
y − 27 y + y − 27 = 0 ⇒( y + 1)( y − 27) = 0
∴
y = − 1 or y = 27 ⇒ x 3 = − 1 or x 3 = 27
⇒
x = − 1 or x = 3
Exp. 3) Solve : 2x −
2x −
Solution
⇒
3
= 5.
x
2


1
1

2  x +  − 2 − 9  x +  + 14 = 0



x
x


1
Substitute
x+ =y
x
∴
2( y 2 − 2) − 9y + 14 = 0 ⇒ 2y 2 − 4 − 9y + 14 = 0
⇒
( y − 2)( 2y − 5) = 0 ⇒ y = 2 or y =
Since
Exp. 4) Solve 2x + 9 + x = 13.
2x + 9 + x = 13 ⇒ 2x + 9 = 13 − x
Solution
⇒
Also,
⇒
3
= 5 ⇒ 2x 2 − 3 = 5 x ⇒( 2x + 1)( x − 3) = 0
x
1
or x = 3.
x=−
2
∴
( x − 8)( x − 20) = 0 ⇒ x = 8, x = 20
Exp. 5) Solve 2x + 9 − x − 4 = 3.
Solution
⇒
∴
Let
⇒
2x + 9 − x − 4 = 3
Since
2x + 9 = 3 +
∴
x−4
Squaring both sides and simplifying, we get
x + 4=6 x −4
⇒
∴
Again squaring both sides,
( x + 4) 2 = 36( x − 4)
⇒
⇒
Verification:
and
and
x 2 − 28x + 160 = 0 ⇒( x − 8)( x − 20) = 0
⇒
x =1
x+
⇒
2
1 
1
= x −  + 2
x
x2 
2


1
1

6  x −  + 2 − 25  x −  + 12 = 0



x
x


1
x − = y ⇒ 6( y 2 + 2) − 25 y + 12 = 0
x
3
8
y= , y=
2
3
1
1 8
and x − =
y=x−
x
x 3
1 3
1
and
x− =
x=− , x=3
x 2
3
1
and
x=−
x=2
2
1 1
x = − , − , 2, 3
3 2
Exp. 8) Solve for x : x 2 + x − 6 − x + 2.
x = 8, x = 20
2x + 9 ≥ 0
x − 4 ≥ 0 ⇒x ≥
5
2
⇒ x 2 − 2x + 1 = 0
1 5
=
x 2
1
2x 2 − 5 x + 2 = 0 ⇒ x = 2 or x =
2
1
x = , 1, 2
2
y=
Solution Put x 2 +
NOTE Here we must have 2x + 9 ≥ 0 thus for x = 8, 2x + 9 > 0
Also for x = 20, 2x + 9 > 0
∴ Both the values satisfy the condition.
1
=2
x
( x − 1) 2 = 0
x+
= x 2 − 7 x + 10, x ∈ R
−9
2
Solution
x 2 + x − 6 − x + 2 = x 2 − 7 x + 10
( x + 3)( x − 2) − ( x − 2) = ( x − 5)( x − 2)
x≥4
Since the values x = 8 and 20 satisfy both these conditions
∴
x = 8, x = 20
5
2
1
1
Exp. 7) Solve 6  x 2 + 2  − 25  x −  + 12 = 0.


x
x 
Squaring both sides 2x + 9 = (13 − x) 2 ⇒ x 2 − 28x + 160 = 0
⇒
1 
1
= x +  − 2
2

x
x
1
1


2  x 2 + 2  − 9  x +  + 14 = 0


x
x 
⇒
2
⇒
CAT
⇒
( x − 2) [ ( x + 3) − ( x − 2) − ( x − 5)] = 0
Theory of Equations
( x − 2) = 0 ⇒ x = 2
Either
or
765
( x + 3) − ( x − 2) − ( x − 5) = 0
Practice Exercise
Find the solutions of the following equations.
2
⇒
( x + 3) − ( x − 2) = x − 5
Squaring both sides
x 2 + 12x + 36 = 4( x 2 + x − 6)
10
3
Since the equation involves radical therefore substituting
10
10
x = 2, 6 and − in the original equation, we find that x = −
3
3
does not satisfy the equation.
∴
x = 2, 6
⇒
x = 6, x = −
Solution
3 x + 2 + 3 − x − 10 = 0 ⇒ 3 x ⋅ 3 2 +
3x = y ⇒
⇒
9y 2 − 10y + 1 = 0
⇒
( 9y − 1)( y − 1) = 0
1
or y = 1
y=
9
1
1
y=
⇒ 3x = 2
9
3
⇒
When
when
y =1
⇒
⇒
∴
x=0
x = − 2, 0
1
− 10 = 0
3x
1
9y + − 10 = 0
y
Let
⇒
⇒
⇒ x = −2
3x = 1
( x + 1)( x + 2)( x + 3)( x + 4) = 24
[( x + 1)( x + 4)][( x + 2)( x + 3)] = 24
( x 2 + 5 x + 4)( x 2 + 5 x + 6) = 24
x2 + 5x = y
Let
∴
( y + 4)( y + 6) = 24 ⇒ y 2 + 10y = 0
⇒
Now,
When
y = 0 and y = − 10
y = x2 + 5x
y = 0, x 2 + 5 x = 0
⇒
x = 0, x = − 5
Again when y = − 10, x + 5 x = − 10
2
⇒
3. p−6 − 9 p−3 + 8 = 0
4. t − 9 t + 14 = 0
5. y
−4
−4=0
7. x =
9. m +
11.
6. 2 x10 − x5 − 4 = 0
x+ 6
8. x +
2m − 3 = 1
2h − 1 −
x −4 =4
10. y = 5 y + 6 − 2
h − 4 =2
12. 3 − r = r + 7 + 2
Answers
3. 1 and
1
2
2. −27 and 125
4. 4 and 49
5. ± 2i
7. 3
9. There is no solution.
11. 5 and 13
6. 1.11014 and −103473
.
8. 4
10. −1 and 2
12. −6
14.6 Formation of a Quadratic
Equation
Exp. 10) Solve for x : ( x + 1)( x + 2)( x + 3)( x + 4) = 24.
Solution
2. x3 − 2 x3 − 15 = 0
1. ±2 and ± 3
Exp. 9) Solve for x : 3 x + 2 + 3 − x = 10.
1
1. x 4 − 7 x2 + 12 = 0
x 2 + 5 x + 10 = 0
Since LHS expression cannot be factorized, therefore we
should use the formula for finding the value of x.
Here
D = b 2 − 4ac = 25 − 4 × 1 × 10 = − 15
If you know the two roots, say α and β, then you can form the
possible quadratic equation using these two roots. Why I’m
saying ‘‘possible’’ quadratic equation because there can be
innumerable (or infinite) number of quadratic equations
having the same two roots. The underlying logic is that if you
multiply or divide a given quadratic equation by any
non-zero real number, the equation gets changed (i.e., the
coefficients/constants get multiplied or divided), however,
the roots remain the same for every such quadratic equation.
So it implies that the roots of ax 2 + bx + c = 0 and
k [ ax 2 + bx + c] = 0 are same.
Now to form a quadratic equation from the known roots you
can use the following approach. If α and β are the two roots
then possible quadratic equation will be ( x − α )( x − β) = 0.
But since you are working backward, that is from roots to
equation, therefore you never know about the exact
quadratic equation these roots belong to. Thus to include all
the possibilities you can multiply the above equation by any
non-zero real number as it’s given below
⇒ k ( x − α )( x − β) =0; Where k is a non-zero real number.
k [ x 2 − (α +β) x +αβ] = 0
Since D < 0, the equation x 2 + 5 x + 10 = 0 has no real solution.
⇒
∴
⇒ k [ x 2 − (sum of the roots) x + (product of the roots) ] = 0
x = 0, − 5
766
QUANTUM
= [(α + β) 2 − 2αβ](α + β) (α + β) 2 − 4αβ
Exp. 1) Form the quadratic equation whose roots are −7
and 3/2.
 3  −11
and product of the
Solution Sum of the roots = − 7 +   =
 2
2
−21
. Therefore, the possible required equation is
roots =
2

 −21 
 11
k x 2 −  −  x + 
 =0
 2  
 2

⇒
k[2x 2 + 11x − 21] = 0
Here k is a non-zero real number.
Exp. 2) Form the quadratic equation whose roots are
5 + 2 7 and 5 − 2 7 .
Solution Sum of the roots = 5 + 2 7 + 5 − 2 7 = 10 and product of
 b 2 − 2ac  b 
=
 − 
 a2   a
=−
K[x 2 − 10x − 3] = 0
Here, k is a non-zero real number.
Exp. 3) Form the quadratic equation whose roots are
5 + 2 7 and 5 − 2 7 and the constant term is 45.
Solution Sum of the roots = 5 + 2 7 + 5 − 2 7 = 10 and product
of the roots = 5 2 − ( 2 7 ) 2 = − 3.
Therefore, the required equation is k [x 2 − (10) x + ( −3)] = 0
⇒ k [x 2 − 10x − 3] = 0
⇒15 x 2 − 150x − 45 = 0; (since the constant term is 45)
Exp. 4) If α and β are the roots of the equation
ax 2 + bx + c = 0, find the values of the following
expressions in terms of a, b and c.
(a) α 2 + β 2
(b)
1 1
+
α β
(c) α 4 − β 4
Solution
(a) Let ax 2 + bx + c = 0 and α , β be the roots of the equation,
b
c
then
α + β = − , αβ =
a
a
∴
α 2 + β 2 = (α + β) 2 − 2αβ
2
2c b 2 2c
 b
= 2 −
= −  −
 a
a
a
a
b 2 − 2ac
.
=
a2
1 1 α + β −b / a
b
(b)
+ =
=
=−
α β
αβ
c/ a
c
(c) α − β = (α + β )(α + β)(α − β)
4
4
2
2
b2
c
−4
2
a
a
b 2
( b − 2ac)
a4
b 2 − 4ac
Exp. 5) If α , β are the roots of the equation
2x 2 − 3x + 2 = 0, form the equation whose roots are
α2 , β2.
Solution 2x 2 − 3 x + 2 = 0, andα,β are the rootsα + β =
3
,αβ = 1
2
For the new equation, roots are α 2 and β 2
∴ Sum of the roots α 2 + β 2 = (α + β) 2 − 2αβ
2
 3
=   − 2(1)
 2
the roots = 5 2 − ( 2 7 ) 2 = −3.
Therefore, the required equation is
K[x 2 − (10) x + ( −3)] = 0
CAT
9
1
−2=
4
4
and product of the roots = α 2β 2 = (αβ) 2 = (1) 2 = 1
=
∴ the required equation is
x 2 − (sum of roots) x + product of roots = 0
1
x +1=0
4
4x 2 − x + 4 = 0
⇒
x2 −
⇒
Exp. 6) Two candidates attempt to solve a quadratic
equation of the form x 2 + px + q = 0. One starts with a
wrong value of p and finds the roots to be 2 and 6. The
other starts with a wrong value of q and finds the roots to
be 2 and –9. Find the correct roots and the equation.
Solution When p is wrong i.e.,
−b
( = α + β) is wrong but
a
c
( = α ⋅ β) is correct.
a
c
= 2 × 6 = 12.
a
c
Again when q is wrong i.e., = (αβ) is wrong
a
−b
but
= α + β is correct.
a
−b
∴
= α + β = 2 + ( −9) = − 7
a
∴the required correct quadratic equation is
x 2 − (α + β) x + αβ = 0
Hence
αβ =
x 2 − ( −7) x + 12 = 0
⇒
x 2 + 7 x + 12 = 0
and the correct roots of this equation are − 3 , − 4.
Theory of Equations
767
Practice Exercise
−3
−5
and
,
2
3
which one of the following is the concerned quadratic
equation?
(a) 6 x2 + 19 x + 15
(b) 19 x + 6 x2 + 15
2
(c) 30 x + 95 x + 75
(d) both (a) and (c)
1. If the two roots of a quadratic equation are
2. If the two roots of any quadratic equation are
− 3, how many such equations are possible?
(a) 0
(b) 1
(c) 2
(d) infinite
3 and
3. If the two factors of any quadratic equation are
(x + 3 ) and (x − 3 ), how many such equations are
possible?
(a) 0
(b) 1
(c) 2
(d) infinite
4. If a quadratic equation is multiplied by 4, the roots of
the new equation will be multiplied by
(a) 16
(b) 4
(c) 2
(d) 1
5. If one root is 3 − 5 and the other root is 3 +
5 , find
the possible sum of coefficients of x2 and x.
(a) −20
(b) −6
(c) 14
(d) 11
Answers
1. (d)
2. (d)
3. (d)
4. (d)
5. (a)
14.7 Nature of the Roots of the
Quadratic Equation
For a quadratic equation ax 2 + bx + c = 0 ; where a, b, c are
real numbers and a ≠ 0, the nature of the roots can be easily
determined by knowing the value of the discriminant
(D = b 2 − 4ac) of the above quadratic equation.
D<0
D =0
Complex
Real
Non-zero
Rational
imaginary parts
Unequal
Equal
(Conjugate pairs)
D > 0 (D is not
D > 0 (D is a
perfect square) a perfect square)
Real
Rational
Real
Irrational
Unequal
Unequal
(Conjugate Pairs)
1. Roots are real only when D is non-negative.
2. Roots are complex (or imaginary) when D is negative.
3. Roots are rational only when D is a perfect square
number like 0, 1, 4, 9, 16, …… etc.
4. Roots are equal only when D =0.
5. The equal roots are called Double Root.
6. When the roots are irrational they are in conjugate pairs
as if one root is p + q , the other root will be p − q .
Here, q is the irrational part of the root.
7. When the roots are complex they are in conjugate pairs
as if one root is p + iq, the other root will be p − iq. Here
iq is the imaginary part of the root.
8. If D > 0; a =1; b, c ∈ Z (integer numbers) and roots are
rational, the roots are integers.
9. If a quadratic equation has one real root and a, b, c ∈ R ,
other root is also real.
10. If the roots are real and equal, the graph will touch the
X -axis.
11. If the roots are real and unequal, the graph will intercept
the X -axis.
12. If the roots are non-real, the graph does not touch
the X -axis.
13. The points on the X -axis, where the quadratic graph
touches or intercepts are called the x-intercepts.
Graphical Illustration
Following graphs are depicted on the Cartesian plane, where
horizontal arrow and vertical arrow represent X -axis and
Y-axis, respectively. In these illustrative graphs my main
consideration is the position of parabolic graph with relation
to X -axis. In order to avoid complications I have not
discussed about position of Y -axis and which is out of
context as well.
D<0
D =0
D>0
When a > 0,
graph will open
upward and its
vertex will be the
lowest point of
the graph.
When a < 0,
graph will open
downward and
its vertex will be
the highest point
of the graph.
(i) The points, where parabola (quadratic graph) meets
the X-axis, are called the real roots of the equation.
(ii) When roots are real, the parabolic graph must touch or
intersect the X-axis.
(iii) When roots are non-real, the parabolic graph will not
touch or pass through the X-axis.
(iv) Either both the roots will be real or both the roots will
be non-real.
768
QUANTUM
(v) When both the roots are equal, the vertex of parabolic
graph tangentially passes through the X-axis.
(vi) Double root has a multiplicity of 2. That means two
roots are same and this is the reason why the quadratic
graph just kisses the X -axis and bounces back to the
same side of X -axis.
(vii) In case of real roots when the roots are unequal the
parabolic graph intersect the X-axis at two different
points. These points are called the X-intercepts.
(viii) The quadratic graph does not intersect the Y-axis more
than once. Also the quadratic graph does not intersect
the X-axis more than two times.
CAT
Exp. 3) Determine k for which the roots of the equation
9x 2 + 2kx + 4 = 0 are equal.
Solution Let the roots be α and α.
∴
Sum of roots = α + α = 2α = −
α=−
⇒
and
k
9
product of the roots = α 2 =

−

∴
⇒
2k
9
4
9
2
k
4
⇒
 =

9
9
k 2 = 36 ⇒
k2 4
=
81 9
k = ± 6.
Exp. 1) Find the nature of the roots of the quadratic
equation ( a − b) x 2 + (b − c) x + ( c − a) = 0 without really
knowing the exact values of the roots.
Alternatively In order that roots of a quadratic equation
are equal, its discriminant must be zero.
i.e.,
b 2 − 4ac = 0
(a) Rational
(b) Irrational
(c) Non-real
(d) Both (a) and (b)
Solution Let us consider x = 1, therefore
∴
( a − b) + ( b − c) + ( c − a) = 0
It implies that x =1 is a root of the given equation. Since one
root of this equation is real and rational, so the other root
definitely has to be real and rational. As the non-real roots and
irrational roots occur in pairs only so there is no scope for the
other root to be non-real or irrational.
Hence choice (a) is the answer.
Alternatively Let us consider a = 3 , b = 2, c = 1 keeping in
mind that a, b, c are rational numbers.
Then
( a − b) x 2 + ( b − c) x + ( c − a) = 0 ⇒ x 2 + x − 2 = 0
The discriminant (D) of the above quadratic equation
= (1) 2 − 4(1 × −2) = 9
We know that when D is a perfect square number the roots are
real and rational.
Exp. 2) If ( a + b + c = 0) , find the nature of the roots of
the equation ( c 2 − ab) x 2 − 2( a 2 − bc) x + (b 2 − ac) = 0.
(a) Imaginary
(b) Real and equal
(c) Non-real
(d) Irrational
Solution Since it is given that a + b + c = 0, we can consider
a = − 1, b = 0, c = 1
Now, we have ( c2 − ab) x 2 − 2( a 2 − bc) x + ( b 2 − ac) = 0
⇒
x − 2x + 1 = 0
2
The discriminant (D) of the above equation = ( −2) − 4(1 × 1) = 0
2
We know that when D = 0, the roots of the quadratic equation
are real, rational and equal. Hence, choice (b) is the correct one.
( 2k) 2 − 4 × 9 × 4 = 0
⇒
k = ± 6.
Exp. 4) Which of the following quadratic expression
can be expressed as a product of real linear factors?
(a) x 2 − 2x + 3
(b) 3 x 2 − 2x −
3
(c) 2x 2 + 3 x − 4
(d) 2x 2 − 5 x + 3
Solution (a) D = b 2 − 4ac < 0 cannot be expressed
(b) D = b 2 − 4ac > 0 can be expressed
(c) D = b 2 − 4ac > 0 can be expressed
(d) D = b 2 − 4ac < 0, can’t be expressed.
NOTE If D < 0, then the quadratic equation cannot be expressed
into two linear factors.
Exp. 5) Find the set of values of p for which the
quadratic equation has real linear factors.
9x 2 − px + 4
Solution For any quadratic polynomial to have real linear
factors, we must have D ≥ 0
∴
b 2 − 4ac ≥ 0
p2 − 4 × 9 × 4 ≥ 0
⇒
⇒
p 2 − 144 ≥ 0
p 2 ≥ 144
⇒
p ≥ ± 12
Therefore, either p ≤ − 12 or p ≥ 12.
Theory of Equations
769
Practice Exercise
1. If the roots of a quadratic equation are real and
rational, which one of the following best describes the
value of the discriminant D ?
(a) D > 0
(b) D = 0
(c) D < 0
(d) D ≥ 0
2. If the roots of a quadratic equation are real and
unequal, which one of the following is necessarily true?
(a) D < 0
(b) D = 0
(c) D > 0
(d) D ≥ 0
3. If the roots of a quadratic equation are unequal, which
one of the following cannot be true?
(a) D > 0
(b) D < 0
(c) D = 0
(d) Roots are conjugate also
4. Which one of the following is a correct statement?
(a) When roots are real, these are definitely equal.
(b) When roots are rational, these are definitely real.
(c) When roots are unequal, these are definitely
complex roots.
(d) When roots make the conjugate pair, these are
either real or complex ones.
5. If one root of the quadratic equation is 2 + 5 7 , find
the other root of this equation.
(a) 5 + 2 7
(b) −2 + 5 7
(c) −2 − 5 7
(d) none of these
6. If one root of the quadratic equation is 3 − 2 i, find the
sum of the roots.
(a) 0
(b) 9
(c) 6
(d) −4 i
7. Given the two roots 3 and −3 3 , how many
quadratic equations can be formed?
(a) 2
(b) 1
(c) 0
(d) none of these
8. If one root of a quadratic equation is 8, then what
would, definitely, be the other root?
(a) −8
(b) 2 2
(c) i 8
(d) none of these
9. Roots can be conjugate except when
(a) D < 0
(b) a < 0
(c) D > 0
(d) D = 0
10. If the roots are complex conjugate, then the sum of the
roots is
(a) An integer
(b) An irrational
(c) An imaginary
(d) none of these
11. If the roots are real and equal,then which of the
following is always valid?
(a) D > 0
(b) D < 0
(c) D = 0
(d) D is a perfect square
12. If the roots are real and rational, then which of the
following is the most appropriate one?
(a) D > 0
(b) D < 0
(c) D = 0
(d) D is a perfect square
13. If the two roots are real, equal and irrational, then
which of the following is the most appropriate one?
(a) D > 0
(b) D < 0
(c) D ≤ 0
(d) doesn’t exist
14. If the roots of a quadratic equation are rational, which
of the following cannot be the value of discriminant D?
(a) 0.01
(b) 6.25
(c) 0.036
(d) 1.44
15. A quadratic equation has double root, then
(a) the two roots lie on the different sides of the Y-axis.
(b) the quadratic equation cannot be expressed in the
form of a perfect square.
(c) the product of the roots is always negative.
(d) the sum of the roots may be positive or negative.
16. If roots are real and irrational, then which of the
following is the most appropriate one?
(a) D > 0, except rational numbers
(b) D > 0, except perfect squares
(c) D ≤ 0, except rational numbers
(d) none of the above
17. If the roots are unequal, then the possible values of D
is/are
(a) D = 0
(b) D − {0 }
(c) D > 0
(d) D < 0
18. If D = 7744, then roots are
(a) equal and real
(c) conjugate pairs
(b) irrational
(d) rational and unequal
19. If D = 4477, then the roots are
(a) equal and real
(c) complex conjugate
(b) irrational
(d) rational and unequal
20. If a quadratic equation has one real root then the other
root will necessarily be
(a) real
(b) equal
(c) complex conjugate
(d) rational and unequal
21. If the roots are same, then the roots must be
(a) negative
(b) rational
(c) irrational
(d) complex conjugate
22. Which one of the following is false about the equation
4 x2 − 12 x + 9 = 0?
(a) It can be expressed as a perfect square.
(b) The solution of this equation is a double root.
(c) The multiplicity of this equation is 2.
(d) The roots are 1.5 units away from the Y-axis in
the left side.
770
23. When the parabola opens upward, then which of the
following condition must be satisfied?
(a) D > 0
(b) a < 0
(c) a > 0
(d) a > b > c
24. When the parabola kisses the X-axis then which of the
following condition must be satisfied?
(a) D > 0
(b) a > 0
(c) D = 0
(d) b2 < 4 ac
25. When the parabola never intersects the X-axis, then
which of the following is certainly true?
(a) D > 0
(b) a < 0
(c) D = 0
(d) b2 ≤ 4 ac
26. The intersection of X-axis and parabola primarily
depends on the value of
(a) D
(b) a
(c) b
(d) none of these
27. The parabola intersects the Y-axis at most
(a) 4 times
(b) 3 times
(c) 2 times
(d) 1 time
28. For a standard quadratic equation the parabola
intersects the Y-axis at most
(a) 4 times (b) 3 times (c) 2 times (d) 1 time
29. If the quadratic graph opens downward, then the roots
are always
(a) imaginary
(b) real
(c) equal
(d) cannot be determined
30. If the graph intersects X-axis at two distinct points,
then the roots are necessarily
(a) complex
(b) irrational
(c) rational
(d) real
31. If a given quadratic equation is multiplied by −2, now
the parabola will open
(a) upward
(b) downward
(c) in the original direction only
(d) the direction will be reversed
32. If the comprehensive graph of quadratic function lies
in all the four quadrants, then roots are necessarily
(a) real
(b) non-real
(c) real and equal
(d) equal and irrational
33. If the graph intersects Y-axis below the X-axis then
which of the following can be valid?
(a) Roots are real and parabola opens downward
(b) Roots are non-real and parabola opens upward
(c) Roots are real only
(d) If the roots are real then parabola opens upward
and if the roots are non-real then parabola opens
downward
34. If the quadratic graph intersects both the X and Y axes
at the same point, which of the following is not correct?
(a) Roots are imaginary
(b) Roots are neither negative nor positive
(c) b = c = 0
(d) Graph may open in either side–up and down
QUANTUM
CAT
35. If the two real roots are distinct, then which of the
following may be correct?
(a) Both are negative
(b) One is positive and another one is negative
(c) Both are positive
(d) all of the above
36. The graph of a quadratic equation is symmetrical
about Y-axis, which ones of the following can’t be the
roots of a quadratic equation?
 −2 2 
(a) (0, 0)
(b) 
, 
 3 3
(c) (2 3 ,−2 3 )
(d) (4 + 2 , 4 − 2 )
37. If the roots of a quadratic equation are (4, 10), then
what is the distance between Y-axis and the axis of
symmetry of the pertinent graph?
(a) 7 unit
(b) 6 unit
(c) 5/2 units
(d) cannot be determined
38. The distance between Y-axis and the axis of symmetry
is 4 and one of the two roots is −9, then what can be the
other root?
(a) 9 or 19
(b) 4 or 14
(c) 1 or 17
(d) −5 or 13
39. In a set of 200 quadratic equations, the discriminant D
for
different
equations
is
such
that
D = − 99 , − 98... ,99 ,100. Then which one of the
following is not true?
(a) There are exactly 45% equations that have
irrational roots.
(b) There are at most 21 distinct rational roots.
(c) The roots of 66% equations are in conjugate pairs.
(d) Number of non-real roots is less than that of real
roots.
40. Which one(s) of the following graphs is/are correctly
sketched for the given quadratic equation which has
double root (i.e., both the roots are equal),
considering that all of these graphs appear
considerably thick when seen with a magnifying lens?
The horizontal lines, in these graphs, represent the
X-axis.
(i)
(ii)
(iii)
(iv)
(a) (i) and (ii)
(c) (i) and (iii)
(b) (ii) and (iv)
(d) (ii) and (iii)
Theory of Equations
771
41. Graph of the equation ax2 + bx + c = 0 is shown below.
If D denotes the discriminant of this equation; S and P
denote the sum and product of the roots of this
equation, respectively. Then which of the following
facts is/are true pertaining to this graph?
Y
(iii) S > 0
b
(v) − > 0
a
(a) (ii), (iii)and (iv)
(b) (ii), (iii) and (vi)
(c) Except (i) and (vi)
(d) none of the above
Case
αβ
α+β
α
β
1
+
+
+
+
(ii) a < 0
2
+
−
−
−
(iv) P < 0
c
(vi) > 0
a
3
−
+
−
+
4
−
−
+
−
Exp. 1) If α and β are the roots of the quadratic equation
x 2 + 5x + 6 = 0, what are the signs of the roots?
Solution Since αβ is positive and α + β is negative, therefore
both the roots are negative.
42. Some facts are given below for a quadratic equation
ax2 + bx + c = 0, and then four graphs are also drawn.
If D denotes the discriminant of this equation; S and P
denote the sum and product of the roots of this
equation, respectively, which one of the following
graphs is the best representative of all these facts?
(i) D > 0
(ii) a < 0
(iii) S < 0
b
(v) − < 0
a
(iv) P < 0
c
(vi) < 0
a
Y
X
(b)
Y
(c)
Solution Since (αβ) is negative and (α + β) is also negative,
therefore one root is positive and another one is negative.
More precisely, as we know|α| < |β|therefore α is positive and
β is negative.
1. If sum of the roots is −2 and product of the roots is −15,
(a) both the roots lie on the right-side of the Y-axis.
(b) both the roots lie on the left-side of the Y-axis.
(c) roots lie on both the sides of the Y-axis.
(d) one root is zero and another root is imaginary
number.
X
2. If sum of the roots is −11and product of the roots is 24,
(a) both the roots lie right-side of the Y-axis
(b) both the roots lie left-side of the Y-axis.
(c) roots lie on both the sides of the Y-axis.
Y
X (d)
Exp. 2) If α and β are the roots of the quadratic equation
24x 2 + 55x − 1 = 0, what are the signs of the roots?
Practice Exercise
Y
(a)
A quadratic equation ax 2 + bx + c = 0 can be expressed using
the roots (α, β) as following
 b
 c
ax 2 + bx + c = 0 ⇒ x 2 +   x +   = 0
a
a
 b
 c
2
2
⇒ x −  −  x +   = 0 ⇒ x − (α + β) x + (αβ) = 0
 a
a
Assume | α| < | β,| then we have the four conditions as shown
below, provided a > 0.
X
(i) D < 0
Sign of the Roots of a Quadratic Equation
X
(d) one root is zero and another root is imaginary number.
3. If the quadratic equation is x2 − 34 x + 93 = 0.
Answers
1. (d)
2. (c)
3. (c)
4. (d)
5. (d)
6. (c)
7. (c)
8. (d)
9. (d)
10. (a)
11. (c)
12. (d)
13. (d)
14. (c)
15. (d)
16. (b)
17. (b)
18. (d)
19. (b)
20. (a)
25. (d)
21. (b)
22. (d)
23. (c)
24. (c)
26. (a)
27. (d)
28. (c)
29. (d)
30. (d)
31. (d)
32. (a)
33. (d)
34. (a)
35. (d)
36. (d)
37. (a)
38. (c)
39. (c)
40. (c)
41. (c)
42. (b)
(a) both the roots are positive.
(b) both the roots are negative.
(c) one root is positive and another one is negative.
(d) one root is rational and another root is irrational.
4. If the quadratic equation is − x2 + 28 x + 93 = 0,
(a) both the roots are positive.
(b) both the roots are negative.
(c) one root is positive and another one is negative.
(d) one root is rational and another root is irrational.
Answers
1. (c)
2. (b)
3. (a)
4. (c)
772
QUANTUM
14.8 Maximum or Minimum Value
of a Quadratic Equation
A quadratic equation, as you know, when drawn on a plane
takes the shape of a parabola. This parabolic graph opens
only in two directions –
(a) Upward (like hands-up) when a person is happy.
(b) Downward (like hands-down) when a person is sad.
CAT
NOTE Essentially a quadratic equation y = ax 2 + bx + c either
gives minimum or maximum value, but not both.
Both the minimum and maximum values can be obtained in a
particular case only when there is a restriction on the values of x and
that tells nothing but it tells the range of the quadratic function.
Techniques for Obtaining the
Minimum/Maximum Value of a
Quadratic Equation
b
, we get the minimum/maximum value of y,
2a
where y = ax 2 + bx + c.
At x = −
b
, in the
2a
quadratic equation y = ax 2 + bx + c we can get the
minimum/maximum value of y.
Approach 1 By substituting the value of x = −
Positive attitude;
happy posture fig (i)
Negative attitude;
sad posture fig (ii)
These graphs can be related with the situations as in a person
is happy when his attitude a is positive and he is sad when his
attitude a is negative. So when in the quadratic equation
y = ax 2 + bx + c, the coefficient of x 2 i.e., a is positive the
parabolic graph opens upward like the fig (i) and it gives the
minimum value of y. Similarly, when in the quadratic
equation y = ax 2 + bx + c, the coefficient of x 2 i.e., a is
negative the parabolic graph opens downward like the fig (ii)
and it gives the maximum value of y.
Y
Y
Approach 2 By substituting the values of a, b, c in the
following formula we can get minimum/maximum value of y.
The minimum/maximum value of
y=
4ac − b 2
4a
Approach 3 Differentiation method of Calculus.
NOTE You don’t have to use the calculus at all in order to avoid
the unnecessary complications. Your objective is to get the answer
quickly and simply. That’s it.
Maximum
Exp.) Consider the following quadratic expressions and
determine
X′
X
Y′
Minimum
X′
X
Y′
Graphically, the maxima (or minima) denote the maximum (or
minimum) distance between X -axis and the vertex of the graph.
In the equation Parabolic
Graph may continue
y = ax 2 + bx + c, graph opens upwards the
when a > 0
upward
infinitely, so you
cannot determine
the maximum value
of y, unless the
value of x is
restricted.
Graph attains
the minimum
value of y and
it cannot go
below this
minimum
value.
In the equation Parabolic
Graph may continue
y = ax 2 + bx + c, graph opens downwards the
when a < 0
downward infinitely, so you
cannot determine
the minimum value
of y, unless the
value of x is
restricted
Graph attains
the maximum
value of y and
it cannot go
above this
maximum
value.
(i) Which of these quadratic expressions give minima and
which of these give maxima?
(ii) The value of x at which these expressions will yield
minimum or maximum value.
(iii) The minimum or maximum value of these quadratic
expressions.
(1) 9 x 2 + 28 x − 4
(2)−7 x 2 + 10 x
(3) x 2 − 9 x
(4) 17 x + 1 − 4 x 2
(5) −2 x 2 − 11 x + 3
(6) 10 + 9 x − x 2
(7) x 2 − x
(8) x 2 + 16
(9) −
3 2
x +7 x − 1
2
(10) 6 x − 9 x 2
(11) − x 2 + 8 x + 10
(12) x 2
Theory of Equations
773
Solution Consider the following table for the answers of all the 12
problems.
(i)
(ii)
Answer Value of Minima / Value of x for which
equation gives
Number coefficient Maxima
minimum /
maximum value of y
b
14
9
Minima
1.
=−
x=−
2a
9
2.
−7
Maxima
3.
1
Minima
4.
−4
Maxima
5.
−2
Maxima
6.
−1
Maxima
7.
1
Minima
8.
1
Minima
9.
−3
2
Maxima
10.
9
Minima
11.
−1
Maxima
12.
9
Minima
b
5
=
2a 7
b
9
=
x=−
2a 2
b 17
=
x=−
2a
8
b 11
=
x=−
2a −4
b
9
=
x=−
2a 2
b 1
=
x=−
2a 2
b
=0
x=−
2a
b 7
x=−
=
2a 3
b 1
x=−
=
2a 3
b
x=−
=4
2a
b
=0
x=−
2a
x=−
(iii)
Minimum/
Maximum
value of y
y min
−232
=
9
25
7
−81
y min =
4
305
y max =
16
145
y max =
8
121
y max =
4
−1
y min =
4
y max =
y min = 16
y max =
43
6
y min = 1
y max = 26
y min = 0
NOTE Practice to get the min/max value of a quadratic function
through both the techniques for superior confidence and better
speed of calculation.
Practice Exercise
1. What is the minimum distance between the vertex of
the quadratic graph and X-axis when the roots are
equal?
(a) 0
(b) 1
(c) ∞
(d) data insufficient
2. What is the minimum distance between the vertex of
the quadratic graph and X-axis when the roots are real
and irrational?
(a) 1
(b) 0
(c) −1
(d) data is sufficient
3. The axis of symmetry is 3 units away from Y-axis and
the minimum is 4 units, then what is the minimum
distance between minimum (extremum) of the graph
and origin?
(a) 1
(b) −1
(c) 5
(d) none
4. If the quadratic equation has negative real roots and
its maximum (extremum) of the graph is 10 unit, then
the distance between the X-axis and maximum of the
graph is
(a) 10
(b) −10
(c) 10 and −10 both
(d) data is insufficient
5. If the minimum value of the quadratic function is 0,
then which of the following is not definitely true about
the roots?
(a) Real
(b) Non-Real
(c) Equal
(d) Rational
6. If one of the roots is 0 and another one is real positive,
then which of the following is certainly incorrect?
(a) minimum value of the function is −1
(b) maximum value of the function is 1
(c) minimum of the graph is at X-axis
(d) the graph intersects Y-axis below the X-axis
7. For a quadratic equation ax2 + bx + c = 0, if you have
double roots and positive coefficient of x2, then the
minimum value of y is
(a) any negative real number
(b) the lowest negative number
(c) the lowest whole number
(d) cannot be determined
8. For a quadratic equation ax2 + bx + c = 0 , if you have
equal roots and negative coefficient of x2, then the
maximum value of y is
(a) the highest non-positive integer
(b) the lowest negative number
(c) depends on roots
(d) none of the above
9. There are two quadratic equations; none of them has
integer or complex roots. If each root of a quadratic
equation is three times the root of another quadratic
equation, then the maximum of this equation will be
m times to that of the other equation. What’s the
value of m?
(a) 1
(b) 3
(d) 9
(c) 3
10. There are two quadratic equations E1 and E2. Each
root of E1 is four times to that of E2, and the minimum
of E1 will be m times to that of E2. Then m is
(a) 0
(b) 2
(c) 16
(d) Cannot be determined uniquely
11. The two roots of a quadratic equation are −7 and −3,
what will be the maximum value of this equation?
(a) 1
(b) −1
(c) 4
(d) Cannot be determined uniquely
774
12. Given the two certain roots of any quadratic equation,
what is the locus of the vertices of all such quadratic
graphs?
(a) Circle
(b) Parabola
(c) Axis of symmetry
(d) A line parallel to X-axis
13. What is the locus of the vertices of all the quadratic
equations whose roots are m times the certain given
roots, say α and β?
(a) Polygon
(b) Parabola
(c) A line parallel to X-axis (d) Nothing can be said
14. The factors of a quadratic equation are (x − 4) and
(x − 8 ), what will be the vertex of the quadratic graph?
(a) −4
(b) 4
(c) 6
(d) −6
15. For a quadratic equation, the roots are (−1, 3) and the
minimum is 4 units, then the triangle formed by
connecting the roots and the minimum value of y is
(a) right angle triangle
(b) scalene triangle
(c) equilateral triangle
(d) isosceles triangle
16. For a quadratic equation, the roots are (−3, 3) and the
maximum of y is 3 3 units, then the triangle formed
by connecting the roots and the maximum value of y is
(a) right angle triangle
(b) scalene triangle
(c) equilateral triangle
(d) isosceles triangle
17. If the roots are (−3, 9), then the highest/lowest value of
y will be attained at x equal to
(a) −27
(b) 6
(c) −3
(d) 3
18. For all the non-negative real x, what is the minimum
and maximum non-negative value of y, where
y = − x2 + 8 x + 20 ?
(a) 0, 36
(b) 18, 36
(c) 0, 20
(d) none of these
19. For the non-positive real x, such that x ≥ − 11, what is
the minimum and maximum value of y, where
y = x2 + 12 x + 27?
(a) −9, 16
(b) −9, 27
(c) 0, 27
(d) none of these
QUANTUM
22. For −3 < x < 2, find the range of
y = − 3 x2 + 10 x − 21.
38 

(a)  −78 , −


3
(c) {−78 , − 13 }
(b) [ −78 , − 13 ]
(d) none of these
23. The minimum and maximum values of
(7 − x)2 − 8 are
(a) −8, 15
(b) −15, 15
(c) −8, 41
(d) none of these
24. The maximum value of 49 − (7 − x)2 is
(a) 49
(c) 0
(b) 42
(d) none of these
25. The equation x2 − 3 (x − 3 )2 − 2 = 0 has
(a) one minimum point
(b) one maximum point
(c) Both maximum and minimum points
(d) None of the above
26. A quadratic equation which attains its minimum −4 at
x = 5 and its y-intersect is 21, the largest root of this
equation must be less than
(a) 10
(b) 5
(c) 6
(d) 3
27. A quadratic equation which attains its maximum 4 at
x = − 4, and y = 3 at x = − 3 what will be the value of
equation at x = − 5?
(a) 5
(b) 2
(c) 3
(d) none of these
3
28. A quadratic equation attains minimum at x = . Also,
2
3
1
y = at x = , what’s the value of the equation when
4
2
5
x= ?
2
(a) 3/4
(b) 5/4
(c) 1/4
(d) cannot be determined
29. If one root of a quadratic equation is −2. At
x = − 5 , y = 27 and at x =`7 , y = 27. Then what’s the
minimum value of the equation?
(a) 5
(b) 0
(c) −9
(d) none of these
20. If y = 2 x2 + 12 x + 35 , the distance between the Y-axis
and the minimum of y is
(a) 1
(b) −1
(c) 0.5
(d) 1.5
−8
8
21. For
≤ x ≤ , the minimum and maximum values of
3
3
2
f (x) = x − 6 x + 18 are
370
11 26
82 370
(b)
(c) 83, 173 (d)
(a) 81,
,
,
9
9 9
9
9
CAT
Answers
1.
6.
11.
16.
21.
26.
(a)
(c)
(d)
(c)
(d)
(a)
2.
7.
12.
17.
22.
27.
(d)
(c)
(c)
(d)
(a)
(c)
3.
8.
13.
18.
23.
28.
(c)
(a)
(c)
(a)
(d)
(a)
4.
9.
14.
19.
24.
29.
(a)
(d)
(a)
(b)
(a)
(c)
5.
10.
15.
20.
25.
(b)
(d)
(d)
(c)
(b)
Theory of Equations
775
14.9 Condition for Common Roots
Between Two Quadratic Equations
18
Consider two quadratic equations,
ax 2 + bx + c = 0 ; (where a ≠ 0)
…(i)
14
And a ′ x 2 + b′ x + c′ = 0 ; (where a′ ≠ 0)
…(ii)
12
(A) If one root is common, then the following condition must
be satisfied.
( ab′ − a ′ b)( bc′ − b′ c) = ( ca ′ − c′ a ) 2
10
(B) If both the roots are common, then the following
condition must be satisfied.
a
b c
= =
a ′ b′ c′
Please note that the above techniques help in determining the
common roots without explicitly determining the individual
roots of the two quadratic equations.
Graphical Illustration 1 The two quadratic equations
x 2 + 2x − 3 and x 2 − 4x + 3 have two roots each where one
root is common in both of them.
20
18
16
14
12
10
8
4
–4
–3
–2
–1
0
–2
Uncommon root
–4
–6
6
4
2
–2
–1
0
1
2
3
4
5
6
–2
Common Root
Common Root
–4
–6
When both the roots are common in the two given quadratic
equations, then the graphs of two quadratic equations
intersect at exactly two points on the same coordinate plane.
It means the two graphs (parabolas), which have both the
roots common they do not overlap necessarily except at the
roots; and the two parabolas can open in opposite directions
or even in the same direction with different degree of
stretching.
Case 1 When c = c′ = 0, then the equations reduce to
2
–5
8
Additional Points about
Common Roots
6
–6
16
1
2
3
4
5
6
7
Uncommon Root
Common Root
When exactly one root is common, then the graphs of two
quadratic equations intersect at exactly one point on a
co-ordinate plane.
Graphical Illustration 2 The two quadratic equations
3x 2 − 12x + 9 and 5x 2 − 20x + 15 have two roots each where
both the roots are common in both of them.
ax 2 + bx = 0 and a ′ x 2 + b′ x = 0
⇒
x ( ax + b) = 0 and x ( a ′ x + b′ ) = 0
This implies that x = 0 is a common root.
The other root for the first equation is x = −
the second equation is x = −
b′
a′
b
and for
a
a
b
a
b c
= , but
= ≠ , then no root is
a ′ b′
a ′ b′ c′
common.
Case 2 When
776
QUANTUM
Finding the Common Root
When α is the common root, then this root must satisfy both
the equations
And
∴
aα 2 + bα + c = 0
...(i)
a ′ α 2 + b′ α + c′ = 0
...(ii)
α2
α
=
( bc′ − b′ c) ( ca ′ − c′ a )
=
1
( ab′ − a ′ b)
From Eqs. (i) and (ii), we get
( bc′ − b′ c)
α=
( ca ′ − c′ a )
CAT
the
equations
and
Exp. 3) If
x 2 + px + q = 0
2
x + qx + p = 0 have a common root, which of the
following can be true?
(a) p – q = 0 (b) p + q = − 1 (c) p q = –1
Solution Since one root is common, then
( q − p)( p 2 − q 2 ) = ( q − p) 2
⇒
( p 2 − q 2 ) = ( q − p)
⇒
[( p + q)( p − q)] = − ( p − q)
⇒
[( p + q)( p − q)] + ( p − q) = 0
⇒
( p − q)[p + q + 1] = 0
This implies that either p − q = 0
or
p + q+1=0
Therefore, (a) and (b) are valid relations.
(d) pq = 0
the
equations
and
Exp. 4) If
x 2 + px + q = 0
2
x + qx + p = 0 have a common root, which of the
following can be the common root?
And from the second and third we get
( ca ′ − c′ a )
α=
( ab′ − a ′ b)
(b) −1
−1
(c) −2
(d)
2
Solution Let us say α is the common root, then
α 2 + pα + q = α 2 + q α + p
⇒
pα + q = qα + p
⇒
α( p − q) = ( p − q)
⇒
α =1
Therefore choice (a) gives the appropriate answer.
(a) 1
Finally by equating both the values of α, we get the condition
for a common root. That is
( ab′ − a ′ b)( bc′ − b′ c) = ( ca ′ − c′ a ) 2
Exp. 1) Determine that whether there is any common
root exists between the following quadratic equations :
…(i)
x 2 − 8x + 15 = 0
…(ii)
x 2 − 9x + 18 = 0
Solution Assuming that one root is common, then
( −9 + 8)( −144 + 135) = (15 − 18) 2
⇒
( −1)( −9) = ( −3) 2
⇒
9=9
This proves that one root is common in between the given
quadratic equation.
Again assuming that both the roots of the given quadratic
equations are common, then
1 8 15
≠ ≠
1 9 18
This proves that both the roots are not common.
Thus we find that only one root is common.
Exp. 2) Determine the value of the common root of the
quadratic equations :
x 2 − 8 x + 15 = 0
…(i)
x 2 − 9 x + 18 = 0
…(ii)
Solution Since one root (or solution) is common in between the
two equations, therefore
x 2 − 8x + 15 = x 2 − 9x + 18
⇒
x=3
Therefore, the value of the common root is 3.
Exp. 5) If
the
equations
and
x 2 + bx + c = 0
2
x + dx + e = 0 have a common root, which of the
following can be the common root?
(a)
c−e
d −b
c−e
(c)
be − cd
(b)
(d)
be − cd
c−e
d −b
c−e
Solution Since one root is common, then
( d − b)( be − cd) = ( c − e) 2
be − cd c − e
=
c−e
d−b
Let us say α is the common root, then
α 2 + bα + c = α 2 + dα + e
⇒
b α + c = dα + e
⇒
α( b − d) = ( e − c)
e−c
c−e
α=
⇒α =
⇒
b−d
d−b
Now from the Eq. (i), we get
c − e be − cd
α=
=
d−b
c−e
Therefore (a) and (b) are valid relations.
…(i)
Theory of Equations
777
Exp. 6) Find the value of m if the equations
x 2 + 2x + 3m = 0 and 2x 2 + 3x + 5m = 0 have a common
root.
Exp. 9) If each pair of the three equations
x 2 + b1 x + c1 = 0, x 2 + b 2 x + c 2 = 0 and x 2 + b 3 x + c 3 = 0
have a common root, which one of the following is true?
(a) 15
(b) 8
(c) 2
(d) −1
Solution Since one root is common, then
( 3 − 4)(10m − 9m) = ( 6m − 5m) 2
⇒
− m = m 2 ⇒m = − 1
Therefore choice (d) gives the appropriate answers.
(a) ( b12 + b22 + b33 ) = 2( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 ) − 4( c1 + c2 + c3 )
(b) ( b12 + b22 + b33 )
= 2( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 ) + 4( c1 + c2 + c3 )
(c) ( b12 + b22 + b33 )
= 2( c1 ⋅ c2 + c2 ⋅ c3 + c3 ⋅ c1 ) − 4( b1 + b2 + b3 )
(d) ( b12 + b22 + b33 )
= 4 ( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 ) − 2( c1 + c2 + c3 )
Solution The three equations are
…(i)
x 2 + b1 x + c1 = 0
…(ii)
x 2 + b2 x + c2 = 0
…(iii)
x 2 + b3 x + c3 = 0
Let (α , β), (β , γ) and ( γ , α ) be the roots of Eqs. (i), (ii) and (iii)
respectively, then
…(iv)
α + β = − b1 andα ⋅ β = c1
…(v)
β + γ = − b2 andβ ⋅ γ = c2
…(vi)
γ + α = − b3 and γ ⋅ α = c3
From eqs. (iv), (v) and (vi), we get
1
α + β + γ = − ( b1 + b2 + b3 )
2
…(vii)
and(α ⋅ β + β ⋅ γ + γ ⋅ α ) = c1 + c2 + c3
Now, (α + β) 2 + (β + γ) 2 + ( γ + α ) 2 = b12 + b22 + b32
⇒ 2 (α 2 + β 2 + γ 2 ) + 2(α ⋅ β + β ⋅ γ + γ ⋅ α) = b12 + b22 + b32
⇒ 2 [(α + β + γ) 2 − 2(α ⋅ β + β ⋅ γ + γ ⋅ α )]
+ 2(α ⋅ β + β ⋅ γ + γ ⋅ α ) = b12 + b22 + b32
2
⇒
2(α + β + γ) − 2(α ⋅ β + β ⋅ γ + γ ⋅ α ) = b12 + b22 + b32
2

 1
⇒
2 − ( b12 + b22 + b32 ) − 2( c1 + c2 + c3 ) = b12 + b22 + b32

 2
1 2
⇒
[( b1 + b22 + b32 ) + 2( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 )]
2
− 2 ( c1 + c2 + c3 ) = b12 + b22 + b32
1 2
( b1 + b22 + b32 ) + ( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 )
⇒
2
− 2 ( c1 + c2 + c3 ) = ( b12 + b22 + b32 )
2
2
2
⇒
( b1 + b2 + b3 ) + 2( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 )
− 4 ( c1 + c2 + c3 ) = 2 ( b12 + b22 + b32 )
2
2
2
⇒
( b1 + b2 + b3 ) = 2( b1 ⋅ b2 + b2 ⋅ b3 + b3 ⋅ b1 ) − 4 ( c1 + c2 + c3 )
2
2
2
= ( b1 + b2 + b3 ).
Exp. 7) For any real numbers a, b and c the equations
ax 2 + bx + c = 0 and bx 2 + cx + a = 0 have a common root,
which of the following relations is NOT true?
(a) abc = 1
(b) a + b + c = 0
(c) a = b = c
(d) a 3 + b 3 + c3 = 3 abc
Solution Since one root is common, therefore
( ac − b 2 )( ab − c2 ) = ( bc − a 2 ) 2
⇒
a( a 3 + b 3 + c3 − 3 abc) = 0
But a ≠ 0 ;
…(i)
∴
a 3 + b 3 + c3 − 3 abc = 0
2
2
⇒( a + b + c)( a + b + c2 − ab − bc − ac) = 0
…(ii)
∴
a+ b+ c=0
Or
a 2 + b 2 + c2 − ab − bc − ac = 0
⇒
2a 2 + 2b 2 + 2c2 − 2ab − 2bc − 2ac = 0
⇒
( a − b) 2 + ( b − c) 2 + ( c − a) 2 = 0
⇒
a − b = 0, b − c = 0, c − a = 0
…(iii)
⇒
a=b=c
Therefore choice (a) is appropriate one.
Hint a 3 + b 3 + c3 − 3 abc = ( a + b + c)
( a 2 + b 2 + c2 − ab − bc − ac)
the
equation
and
Exp. 8) If
x 2 − ax + b = 0
2
x − a′ x + b ′ = 0, have one root in common and the second
equation has equal roots, then which one of the following
is a valid relation?
(a) 2( b′ − b) = aa′
(b) 2( a + a′ ) = bb′
(c) 2( a′ − a) = bb′
(d) 2( b + b′ ) = aa′
Solution Since the equation x 2 − a′ x + b′ = 0 has equal roots,
we can consider them α and α, then
a′
Sum of the roots = α + α = 2α = a′ ⇒α =
2
And product of the roots = α ⋅ α = α 2 = b′
Now since one root is common, then
α 2 − aα + b = α 2 − a′ α + b′
⇒
α( − a + a′ ) = ( b′ − b) ⇒α( a′ − a) = ( b′ − b)
⇒
a′ ( a′ − a) = 2( b′ − b)
⇒
( a′ ) − aa′ = 2( b′ − b)
⇒
4α 2 − aa′ = 2( b′ − b)
⇒
4 b′ − aa′ = 2( b′ − b)
⇒
aa′ = 2( b + b′ )
Therefore choice (d) is valid.
(since α =
Therefore choice (a) is the appropriate one.
Hint ( a + b + c) 2 = ( a 2 + b 2 + c2 ) + 2( a ⋅ b + b ⋅ c + c ⋅ a)
q
Exp. 10) Find the value of , if px 2 + x − 15 = 0 and
p
a′
)
2
2
(since a′ = 2α)
(since α 2 = b′ )
18x 2 + 3x + q = 0 have both the roots common.
−15
14
27
(b) − 9
(c) −
(d)
2
3
4
Solution Since both the roots are common, therefore
p
1 −15
= =
⇒ p = 6 and q = − 45
18 3
q
(a)
Therefore choice (a) is the appropriate one.
778
QUANTUM
CAT
Exp. 11) If the equation x 2 + px + qr = 0 and
x 2 + qx + rp = 0 have a common root, the other roots,
which are not common in these equations, will be the
roots of the equation :
(a) x 2 + pqx + r = 0
(c) px 2 + rx + q = 0
(b) x 2 − rx − pq = 0
(d) x 2 + rx + pq = 0
Solution The given equations are
x 2 + px + qr = 0
x 2 + qx + rp = 0
…(i)
…(ii)
(iii)
(iv)
(v)
(vi)
Let (α , β) and (α , γ) be the roots of eqs. (i) and (ii)
respectively, then
α 2 + pα + qr = α 2 + qα + rp ⇒ pα + qr = qα + rp
…(iii)
⇒
α =r
Since the given eqs. (i) and (ii) have a common root, then
( q − p)( p 2r − q 2r) = ( qr − rp) 2
⇒ ( q − p)r ( p 2 − q 2 ) = r 2 ( q − p) 2 ⇒( p 2 − q 2 ) = r( q − p)
⇒
r=
( p 2 − q2)
= − ( p + q) ⇒r = − ( p + q)
−( p − q)
…(iv)
Now from the eq. (i), the product of the roots = αβ = qr
But since one of the roots isα = r, it means the other rootβ = q.
Similarly from the eq. (ii) the product of the roots = αγ = rp
But since one of the roots is α = r, it means the other root
γ=p
Therefore the new equation whose roots are β and γ is
x 2 − (β + γ) x + βγ = 0 ⇒ x 2 − ( p + q) x + pq = 0
⇒
x 2 + rx + pq = 0
Therefore choice (d) is the appropriate one.
Exp. 12) There are six pairs of quadratic equations as
mentioned below.
1. − x 2 − x + 6 = 0 and − x 2 + 7 x − 10 = 0
2. x 2 + 3 x − 4 = 0 and − x 2 − x + 2 = 0
3. − x 2 + 8 x − 12 = 0 and x 2 − 3 x − 18 = 0
4. − x 2 + 5 x − 4 = 0 and − 3 x 2 + 15 x − 12 = 0
5. x 2 − x − 6 = 0 and − 4 x 2 + 4 x + 24 = 0
6. 4 − 2 x 2 = 0 and x 2 + 4
And there are six sets of representative quadratic graphs as
shown below, where the solid parabola depicts the first
equation and dotted parabola depicts the second equation
for each pair of quadratic equations.
(i)
(ii)
Then which one of the following is correct?
(a) 1-iv, 2-iii, 3-ii, 4-vi, 5-v, 6-i
(b) 1-v, 2-ii, 3-iii, 4-vi, 5-iv, 6-i
(c) 1-ii, 2-iii, 3-i, 4-v, 5-vi, 6-iv
(d) 1-v, 2-ii, 3-iii, 4-iv, 5-vi, 6-i
Solution Use the following tricks to figure out the right match.
Both the coefficients of x 2 in pair (1) are negative; it means
the parabolas will open downwards. But since there is only
one root common in the equations of pair (1) so the
appropriate graphs are given by graph (v).
Similarly, both the coefficients of x 2 in pair (1) are negative;
it means the parabolas will open downwards. But since both
the roots are common in the equations of (4) so the right
graphs are given by graph (vi).
Equations in pair (5) have both the roots common and one
has positive and another has negative coefficient of x 2 , so the
graphs will open in opposite directions. Thus the
appropriate graphs are represented by graph (iv).
In pair (6) one equation has non-real roots and another one
has real roots. Also, one equation has minimum 4 and
another has maximum 4. Also, the axes of symmetry for both
the equations are Y-axis only; whereas the axis of symmetry
−b
is given by x = . Thus the graph (i) is appropriate for
2a
pair (6).
Now, in pair (2) only one root is common and the first
equation gives minimum and the second equation gives
maximum, therefore graph (ii) is the appropriate match
for pair (2).
Lastly, in pair (3) only one root is common and the first
equation gives maximum and the second equation gives
minimum, therefore graph (iii) is the appropriate match
for pair (3).
Hence, choice (b) is correct.
Theory of Equations
779
14.10 Basics of Inequality
Quadratic Inequality
1. Square of any real quantity is always non-negative. In
general, even power of any real quantity is always
non-negative.
2.
3.
4.
5.
That means for every real number R , R 2n ≥ 0 ; where n is
any natural number.
If a, b, c are three real numbers such that a ≥ b, then
a ± c ≥ b ± c.
(i) If a, b, c are three real numbers such that a ≥ b and
c > 0, then ac ≥ bc.
(ii) If a, b, c are three real numbers such that a ≤ b and
c > 0, then ac ≤ bc.
(i) If a, b, c are three real numbers such that a ≥ b and
c < 0, then ac ≤ bc.
(ii) If a, b, c are three real numbers such that a ≤ b and
c < 0, then ac ≥ bc.
(i) If a and b are two real numbers and ab > 0, then
either each of a and b are positive or each of them
are negative.
(ii) If a and b are two real numbers and ab < 0, then a
and b are of opposite signs i.e., either a > 0 and
b < 0 or a < 0 and b > 0.
Essentially a quadratic expression is said to be a quadratic
inequation when ax 2 + bx + c ≠ 0. Thus ax 2 + bx + c > 0 and
ax 2 + bx + c < 0 are the quadratic inequations. A quadratic
equation talks about the zeros of y, while a quadratic
inequation talks about either positive values of y or negative
values of y.
For y = ax 2 + bx + c
y>0
y is strictly greater than 0.
y≥0
y is greater than or equal to 0. Quadratic inequation
y =0
y is strictly equal to 0.
Quadratic equation
y≤0
y is less than or equal to 0.
Quadratic inequation
y<0
y is strictly less than 0.
Quadratic inequation
Quadratic inequation
Graphically speaking a quadratic inequation states the
position of the quadratic graph either above or below the
X -axis. To know the nature (or position) of y we need to
know the following things
(i) Whether the discriminant D is positive, negative or zero
(ii) Whether the coefficient of x 2 is positive or negative
(iii) Roots of the given quadratic equation, in case of real and
distinct roots.
For ax 2 + bx + c , Such that a > 0
y > 0, for x < α and x > β
y ≥ 0, for x ≤ α and x ≥ β
D>0
ax2+bx+c>0
y < 0, for α < x < β
y ≤ 0, for α ≤ x ≤ β
ax2+bx+c<0
Where y = ax 2 + bx + c Assuming α and β are the two roots such that α < β.
y > 0, for all the values of x, except x = α = β
y ≥ 0, for all the values of x
D=0
ax2+bx+c>0
D<0
ax2+bx+c>0
Where y = ax 2 + bx + c
Assuming α and β are the two roots but they are always equal when D = 0.
y > 0, for all the values of x
Where y = ax 2 + bx + c
780
QUANTUM
CAT
For ax 2 + bx + c , such that a < 0
D>0
ax2+bx+c>0
y > 0, for α < x < β
y ≥ 0, for α ≤ x ≤ β
ax2+bx+c<0
y < 0, for x < α and x > β
y ≤ 0, for x ≤ α and x ≥ β, Where y = ax 2 + bx + c
Assuming α and β are the two roots such that α < β.
y < 0, for all the values of x, except x = α = β
y ≤ 0, for all the values of x, Where y = ax 2 + bx + c
D=0
Assuming α and β are the two roots but they are always equal when D = 0.
ax2+bx+c<0
D<0
y < 0, for all the value of x, Where y = ax 2 + bx + c
ax2+bx+c<0
Sign of Quadratic Expression ax 2 + bx + c
In the preceding discussion you must have seen that sometimes graph intersects the X -axis
and therefore some part of the graph is above the X -axis and rest of the graph is below the X -axis. So here our objective is to
determine that in general when the graph [ y = f ( x )] is positive and when it is negative.
1. If D < 0, i.e., α and β are imaginary, then ax 2 + bx + c > 0, if a > 0 and ax 2 + bx + c < 0, if a < 0
2. If D = 0, i.e., α and β are real and equal, then ax 2 + bx + c ≥ 0, if a > 0 and ax 2 + bx + c ≤ 0, if a < 0
3. If D > 0, i.e., α and β are real and unequal (α < β), then the sign of the expression ax 2 + bx + c, x ∈ R is determined as
follows :
Sign is same as that of a Sign is opposite that of a
−∞
α
Sign is same as that of a
β
+∞
Exp. 1) If a > 0 and roots are 5 + 2 i and 5 − 2 i, determine the nature of the quadratic graph.
Solution Since roots are imaginary and a > 0, therefore all the values of the quadratic function will be positive for every value of x.
Exp. 2) For a quadratic equation, a < 0 and b 2 = 4ac, determine the nature of the graph.
Solution Since D = 0 and a < 0, therefore all the values of the function will be negative except the maximum value and that will be
equal to 0.
Exp. 3) For a quadratic equation 10x 2 + x − 21 = 0, what is sign of the function for every x < −
Solution The roots of the equation are
3
every x < − .
2
3
?
2
−3
7
and and D = 841 > 0. Also a > 0, therefore all the values of function will be positive for
2
5
Theory of Equations
781
Exp. 4) Find the sign of 6x 2 − 5x + 1 for all real values of x.
Solution Q
Y
D = b 2 − 4ac ⇒ D = 25 − 4 × 6 × 1 = 1
25 − 24
1
1
i.e., and
2
3
12
∴The given expression has the same sign as the coefficient of x 2 ,
1
1
i.e., positive for all real values of x except for those which lie between and
3
2
Q The roots are real and different and the roots are
5±
X
0
(1/3, 0) (1/2, 0)
1
2
Solutions of Quadratic Inequations
Remember that in quadratic equation we are interested in knowing that for what values of x, the value of y will be zero but in
quadratic inequation we are interested in knowing that for what values of x, the value of y will be positive or negative.
Objective
Condition
Quadratic Equation
To know the values of x
So that y = 0
Quadratic Inequation
To know the values of x
So that y > 0 or y < 0
By having a look at the previous graphs, you can summarize the following facts regarding the quadratic expression
ax 2 + bx + c whose roots are α and β and R is the set of real numbers.
Discriminant
Coefficient of x
D<0
D=0
Values of y
Values of x
Alternative forms of expressing the values of x
a>0
y>0
for every x
R
a<0
y<0
for every x
R
a>0
y>0
for every x, except x = α
R − {α } or (− ∞, α ) ∪ (α , ∞ )
y≥0
for every x
R
y<0
for every x, except x = α
R − {α } or (− ∞, α ) ∪ (α , ∞ )
y≤0
for every x
R
y>0
for x < α and x > β
R − [α , β] or (−∞, α ) ∪ (β, ∞ )
y≥0
for x ≤ α and x ≥ β
R − (α , β ) or (−∞, α ] ∪ [β, ∞ )
y<0
for α < x < β
(α , β )
y≤0
for α ≤ x ≤ β
[α , β]
y>0
for α < x < β
(α , β )
y≥0
for α ≤ x ≤ β
[α , β]
y<0
for x < α and x > β
R − [α , β] or (−∞, α ) ∪ (β, ∞ )
y≤0
for x ≤ α and x ≥ β
R − (α , β ) or (− ∞, α ] ∪ [β, ∞ )
a<0
D>0
a>0
a<0
So essentially when D > 0, then only you have to use the following methods to know that for what values of x, the value of y
will be positive or negative. In other cases like D < 0 or D = 0, just by looking at the value (positive/negative) of a, you can tell
the answer. Basically there are following three methods to determine the values of x for a quadratic inequation.
(A) Graphical Method
(B) Interval Method
(C) Algebraic Method
(A) Graphical Method
Step 1 Draw the graph (parabola)
Step 2 Determine the roots (intersections at X -axis)
Step 3 Consider the appropriate values of x corresponding to the given y (or ax 2 + bx + c) as shown in the diagram by shading
the X -axis.
In the following diagrams :
782
The sign
QUANTUM
CAT
o indicates that x = α and x = β are excluded. The sign • indicates that x = α and x = β are included.
Coefficient of x 2
Inequality
a>0
a<0
ax + bx + c > 0
x < α and x > β
α < x <β
ax 2 + bx + c ≥ 0
x ≤ α and x ≥ β
α ≤ x ≤β
ax 2 + bx + c < 0
α < x <β
x < α and x > β
ax 2 + bx + c ≤ 0
α ≤ x ≤β
2
x ≤ α and x ≥ β
Exp.) Find the values of x for 3x 2 − 3x − 6 ≤ 0.
Solution Let’s consider for a while that 3 x 2 − 3 x − 6 = 0 . Then the roots of this equation are −1 and 2 . Now sketch the graph as shown
below. Since a = 3, that means a > 0, therefore the parabola will open up and it will pass through the points −1 and 2.
50
40
30
20
10
–5
–4
–3
–2
–1
0
1
2
3
4
5
–10
–20
Now you have to see where the graph of f ( x) ≤ 0. Looking at the above graph you can figure out that when − 1 ≤ x ≤ 2, then
3 x 2 − 3 x − 6 ≤ 0.
Theory of Equations
783
(B) Interval Method
Step 1 Find the roots, α and β, of the given quadratic inequations considering for a while that the given quadratic expression is
equal to zero.
Step 2 Draw a number line and mark the roots on it as you do it on the X-axis; smaller root on the left hand side and bigger root
on the right hand side.
Step 3 These two roots divide the number line in 3 contiguous regions as shown below.
Step 4 Follow the table for the final steps.
For, a > 0
++++++
–∞
For, a < 0
++++++
+∞
– – – – –
α
(α < β)
β
– – – – –
–∞
++++++
α
(α < β)
– – – – –
+∞
β
y > 0 Consider the positive region. (That means all the values less than Consider the positive region.
(That means all the values between α and β).
α and all the values greater than β.)
y < 0 Consider the negative region.
(That means all the values between α and β.)
Consider the negative region. (That means all the values
less than α and all the values greater than β.)
Exp. 1) Find the values of x for the following inequations.
(i) 3 x 2 − 3 x − 6 ≤ 0
(ii) 3 x 2 − 3 x − 6 < 0
(iii) 3 x 2 − 3 x − 6 ≥ 0
(iv) 3 x 2 − 3 x − 6 > 0
Solution The roots of given equations are −1 and 2. Now, mark the roots on the number
++++++
++++++
– – – – –
line then the line will be divided into 3 sections (or intervals) one section is before −1,
2
–∞
+∞
–1
and second section is between −1 and 2, and the third section is after 2. Since a > 0,
therefore the outer sections will be positive and the section lying between roots will be negative, so mark the sections appropriately
using + and − signs.
(i) Since you are required to find the values of x, on the number line, for which 3 x 2 − 3 x − 6 ≤ 0. That means you are looking for
non-positive region. So, it is obvious from the above that x ≥ − 1 and x ≤ 2. It implies that − 1 ≤ x ≤ 2.
(ii) In this case, it is a strict inequality, so you have to exclude the roots. For the inequation 3 x 2 − 3 x − 6 < 0, the possible values are
− 1 < x < 2.
(iii) Since, you are required to find the values of x on the number line, for which 3 x 2 − 3 x − 6 ≥ 0. That means you are looking for
non-negative region. So, it is obvious from the above that x ≤ − 1 and x ≥ 2 . It implies that x is ( −∞ , − 1] ∪ [2, ∞) or you can say
x ∈ R − ( − 1, 2).
(iv) In this case, it is a strict inequality so you have to exclude the roots. For the inequation 3 x 2 − 3 x − 6 > 0, the possible values are
x < −1 and x > 2. It implies that x is ( − ∞ , − 1) ∪ ( 2, ∞) or you can say x ∈ R − [−1, 2].
Exp. 2) Find the values of x for the following inequations.
(i) − 2 x 2 + 8 x + 10 ≥ 0
(ii) − 2 x 2 + 8 x + 10 ≤ 0
Solution The roots of this equation are −1 and 5. Now, mark the roots on the number line then
++++++
– – – – –
– – – – –
the line will be divided into 3 sections (or intervals) : one section is before −1, and second
–1
5
–∞
+∞
section is between −1 and 5, and the third section is after 5. Since a < 0, therefore the outer
sections will be negative and the section lying between roots will be positive, so mark the sections appropriately using + and – signs.
(i) Since you are required to find the values of x, on the number line, for which − 2x 2 + 8x + 10 ≥ 0. That means you are looking for
non-negative region. So it is obvious from the above that x ≥ − 1 and x ≤ 5. It implies that − 1 ≤ x ≤ 5.
(ii) Since, you are required to find the values of x on the number line, for which − 2x 2 + 8x + 10 ≤ 0. That means you are looking
for non-positive region. So it is obvious from the above that x ≤ − 1 and x ≥ 5. It implies that x is ( −∞ , − 1] ∪ [5 , ∞) or you can
say x ∈ R − ( − 1, 5).
(C) Algebraic Method
As far as CAT is concerned, I won’t recommend this method at all because it is slightly tedious and more time consuming in
comparison to the previously discussed methods. So please avoid this method for competitive exams. However, this method
may not be so oblivious to you because it has been inadvertently used on several occasions without specifying the name of it.
NOTE Once you understand the nature of a quadratic graph your most of the problem is solved. You are just expected to visualize and
draw a rough sketch of a quadratic graph just by looking at certain things of a quadratic function (or equation or inequality) like the sign of
a(+ or −) value of D (D > 0 or D = 0 or D < 0), value of roots etc. Then looking at the graph – its position, intersections at X-axis and opening
direction – you can tell a lot about the quadratic equation/inequation/function.
784
QUANTUM
CAT
Practice Exercise
Directions (for Q. Nos. 1 to 11) Find the values of x for which
of the following equations are satisfied.
1. x 2 − 4 x + 3 ≥ 0
(a) [1, 3]
(c) (− ∞ , 1) ∪ (3 , ∞ )
(b) (−∞ , − 3 ] ∪ [1, ∞ )
(d) R − (1, 3 )
2. − x 2 + 2 x + 8 ≥ 0
(a) (− 2 , 6 )
(b) (2 , − 4 )
(c) [ − 2 , 4 ]
(d) (−∞ , − 2 ] ∪ (−2 , 4 ) ∪ [ 4 , ∞ )
14. For a quadratic inequation ax 2 + bx + c ≥ 0; D < 0 and
a < 0, the values of x that satisfy the given inequation
(b) R −
(a) R +
(c) R − {0 }
(d) does not exist
3. − x 2 + 2 x + 8 < 0
(a) [ − ∞ , − 2 ) ∪ (4 , ∞ ]
(c) (−2 , 4 )
(b) (−∞ , − 2 ) ∪ (4 , ∞ )
(d) (−∞ , − 4 ) ∪ (2 , ∞ )
4. 2 x 2 + 5 x − 3 > 0

1
(a) (−∞ , − 3 ] ∪  , ∞

2
(b) (−∞ , − 3 ] ∪ [2 , ∞ ]
1

(c) R − −3 ,

2 
1

(d) R −  −3 , 

2
5. x + 2 x + 5 > 0
2
(a) (− ∞ , ∞ )
(c) (−∞ , 0 ) ∪ (0 , ∞ )
(b) [ − ∞ , ∞ ]
(d) (−∞ , − 2 ) ∪ (−5 , ∞ )
6. x 2 + x − 12 < 0
(a) − ∞ < x − 4 and 3 < x < ∞ (b) [ −4, 3]
(c) (−∞ , 4) ∪ (3, ∞ )
(d) (−4, 3)
(b) − 2 < x < 2
(d) none of these
8. x 2 + 1 < 0
(a) − 1 ≤ x ≤ 1
(c) Does not exist
3
(b) − 1 < x < 0
(d) none of these
19. Find the values of x that satisfy the inequation
x 6 − 19 x3 − 216 > 0.
(a) [ −3 , − 2 ] ∪ [2 , 3 ]
(b) (−∞ , − 2 ] ∪ [ 3 , ∞ )
(c) (− ∞ , − 2 ) ∪ (3 , ∞ )
(d) (−∞ , − 8 ] ∪ [27 , ∞ )
(b) − 3 ≤ x ≤ 3
(d) R − (− 3 , 3 )
10. 9 x 2 − 30 x + 25 ≤ 0
(a) − 3 ≤ x ≤ 5
5
(c)
3
(b) − 5 ≤ x ≤ 3
5
(d) x ≤
3
11. x 2 − 10 x − 1 ≥ 0
(a) (−∞ , ∞ )
(c) (−∞ , − 10 ) ∪ (−1, ∞ )
(d) (−∞ , 5 − 26 ] ∪ [5 +
(b) [5 − 26 , 5 +
16. For a quadratic inequation ax 2 + bx + c ≥ 0, if you
have double roots, the values of x that satisfy the given
inequation
(a) − ∞ ≤ x ≤ ∞
(b) R − R −
(c) − ∞ < x < ∞
(d) all the real numbers except one value
18. Find the values of x that satisfy the inequation
x 4 − 13 x 2 + 36 ≤ 0.
(a) [ −3 , − 2 ] ∪ [2 , 3 ]
(b) [ − 3 , 3 ]
(c) [ −2 , 2 ]
(d) (−∞ , − 3 ] ∪ [ 3 , ∞ )
9. 9 − x 2 ≥ 0
(a) − 3 ≤ x ≤
(c) 0 < x ≤ 3
15. For a quadratic inequation ax 2 + bx + c > 0, if you
have double roots, the values of x that satisfy the given
inequation
(a) − ∞ ≤ x ≤ ∞
(b) R − {α , β } where α is twice of β
(c) − ∞ < x < ∞
(d) all the real values except one value
17. Find the valid interval of x for the inequality
− 6 ≤ x 2 − 5 x ≤ 6.
(a) [ −1, 2 ] ∪ [ 3 , 6 ]
(b) (−∞ , 2 ] ∪ [ 3 , ∞ )
(c) [ −1, 6 ]
(d) none of these
7. x 2 − 4 < 0
(a) − 2 ≤ x ≤ 2
(c) (− 4 , 4 )
13. For a quadratic inequation ax 2 + bx + c > 0, having
negative discriminant, the values of x that satisfy the
given inequation
(a) R
(b) only positive real numbers
(c) does not exist
(d) Either R or none
26 ]
20. Find the values of m that satisfy the inequation
m + 3 m − 4 > 0.
(a) m ≤ − 4 and m ≥ 1
(b) m > 1
(c) (−∞ , − 4 ) ∪ (1, ∞ )
(d) (−∞ , − 16 ] ∪ [1, ∞ )
26 , ∞ )
12. For a quadratic inequation ax 2 + bx + c < 0, having
negative discriminant, the values of x that satisfy the
given inequation
(a) R −
(b) R
(c) R +
(d) can’t be determined
Answers
1. (d)
2. (c)
3. (b)
4. (c)
5. (a)
6. (d)
7. (b)
8. (c)
9. (b)
10. (c)
11. (b)
12. (d)
13. (b)
14. (d)
15. (d)
16. (c)
17. (a)
18. (a)
19. (c)
20. (b)
Theory of Equations
785
Hints
5. (−∞ , ∞ ). Since a, the coefficient of x , is positive and the
2
roots are non-real. So, for all the values of x, the inequation
is positive.
6. (−4, 3)
7. − 2 < x < 2
8. Does not exist; since x is always positive and so x + 1 is
also positive.
2
2
9. − 3 ≤ x ≤ 3
12. Since the whole graph will lie below the X-axis and we need
just the opposite.
13. Double roots means both the roots are same and since you
need to obey the ‘strict inequality’. So, you have to
EXCLUDE the zeros (roots) of this quadratic.
16. Maximum value of this expression can’t be determined as it is
keep on increasing when x < 7 or x > 7. However, minimum
value is determinable which is attained at x = 7 making
(7 − x )2 = 0, the minimum possible value of a perfect square.
Thus, the minimum value of the whole expression is −8.
17. When (7 − x )2 is minimum, i.e. zero, then 49 − (7 − x )2 will
be maximum.
14.11 Position of Roots of a Quadratic Equation
with Respect to One or Two Real Numbers
Before moving towards the core of the topic the first thing you need to be sure of is quadratic function. As in, how does a
change in x, the change in function f ( x ) occurs. The quadratic function f ( x ) is defined as f ( x ) = ax 2 + bx + c, where a ≠ 0. For
example consider an arbitrary quadratic function f ( x ) = x 2 − 8x + 12. Now put some arbitrary numerical values, preferably
integers to enjoy the convenience in the given function and then notice the values of f ( x ).
x = 0, x 2 − 8x + 12 = 12 ⇒ f (0) = 12
x =1, x 2 − 8x + 12 = 5 ⇒ f (1) = 5
x = 2, x 2 − 8x + 12 = 0 ⇒ f (2) = 0
x = 3, x 2 − 8x + 12 = − 3 ⇒ f (3) = − 3
x = 4, x 2 − 8x + 12 = − 4 ⇒ f ( 4) = − 4
x = 5, x 2 − 8x + 12 = − 3 ⇒ f (5) = − 3
x = 6, x 2 − 8x + 12 = 0 ⇒ f (6) = 0
x = 7, x 2 − 8x + 12 = 5 ⇒ f ( 7) = 5
x = 8, x 2 − 8x + 12 = 12 ⇒ f (8) = 12
It can be easily seen that f (2) = f (6) = 0. These are zeros of the function. And, f (0) > 0, f (1) > 0, f ( 7) > 0, f (8) > 0. These are
the positive values of the function. And, f (3) < 0, f (4) < 0, f (5) < 0. These are the negative values of the function.
Now, the actual point that I’m going to discuss is how to determine the position of a root (or both the roots) with respect to one
(or two) given point(s) on the X -axis. Let f ( x ) = ax 2 + bx + c, where a, b, c ∈ R , a ≠ 0. Let α, β be the two roots of this
equation such that α < β. Let k , k 1 , k 2 ∈ R such that k 1 < k 2 .
1. Position of real roots with respect to one real number k:
For, a > 0, (α < β )
If k < α < β
k
(i) D ≥ 0
If α < β < k
––––
++++++
–∞
––––
(i) D ≥ 0
(iii) k < −
b
2a
(ii) af (k ) > 0
k
+++++
(iii) K > −
b
2a
α
(i) D ≥ 0
β
+∞
β
(iii) k < −
(ii) af (k ) > 0
––––
–∞
+∞
––––
α
k
(i) D ≥ 0
+++++
β
α
+++++
––––
–∞
+∞
(ii) af (k ) > 0
++++
–∞
+++++
β
α
For, a < 0, (α < β )
–––––
+∞
k
(ii) af (k ) > 0
(iii) k > −
If α < k < β
++++
–∞
α
(i) D > 0
k
––––
+++++
–––––
β
(ii) af (k ) < 0
+∞
–∞
(i) D > 0
+++++
α
b
2a
––––
k
(ii) af (k ) < 0
β
+∞
b
2a
786
QUANTUM
CAT
2. Position of real roots with respect to two real numbers k1 , k 2:
For, a > 0, (α < β )
If k1 < α < k2 < β
Or α < k1 < β < k2
++++++
–4
++++
–4
(iii) k1 < −
(i) D > 0
k2
––––
+4
β
+++++
α
k1
(i) D ≥ a
+++++
k2
(ii) f (k1 )⋅ f (k2 ) < 0
––––
–4
+4
(ii) af (k1 ) > 0, af (k2 ) > 0
β
––––
k2 +4
(ii) af (k1 ) > 0, af (k2 ) > 0
(iii) k1 < −
b
< k2
2a
b
< k2
2a
++++
–∞
β
α
k1
(i) D > 0
+++++
–––– –
α
k1
(i) D ≥ 0
If α < k1 < k2 < β
––––
–4
+4
β
k2
(ii) f (k1 )⋅ f (k2 ) < 0
(i) D > 0
If k1 < α < β < k2
++++
––––
α
k1
For, a < 0, (α < β )
–––––
α
k1
k2 β
(ii) af (k1 ) < 0, af (k2 ) < 0
–––
++++++
–––
+++++
+∞
–∞
α
(i) D > 0
(ii) af (k1 ) < 0, af (k2 ) < 0
k1
k2
β
+∞
Now, if we consider the above example and sketch a quadratic graph, we can better understand the significance of the above
discussion.
12
10
8
6
4
2
–8
–6
–4
–2
0
2
4
6
8
10
12
–2
–4
–6
In this function f ( x ) = x 2 − 8x + 12 , a = 1, D > 0. Suppose you have to determine whether a particular point (say 8) is in
between the roots or outside the roots, then by doing some simple calculation you will be able to figure out the relative
position. Since a > 0,
b
−8
b
(i) D > 0
(ii) af ( k ) = 1 ⋅ f (8) = 1 ⋅ 12 = 12 ⇒ af ( k ) > 0
(iii) k = 8 and −
=−
= 4⇒k > −
2a
2
2a
It implies that 8 is greater than both the roots (i.e., 2 and 6), which is same as shown in the above graph.
Practice Exercise
1. What are the essential conditions to get assured that a
randomly chosen particular point k, lies between the
roots of the quadratic equation ax 2 + bx + c = 0 ?
(i) D > 0
(iii) af (k) < 0
b
(v) K > −
2a
(a) (i), (iii) and (v) only
(c) (i) and (iii) only
(ii) D ≥ 0
(iv) af (k) > 0
b
(vi) k < −
2a
(b) (i) and (iv) only
(d) (ii), (iv) and (vi) only
2. What can be the correct possible set of factors that
determines the location of roots of ax 2 + bx + c = 0 of
a quadratic equation which has a double root with
respect to a randomly chosen point k on the X-axis?
b
(a) D ≥ 0 , af (k) < 0 , k < −
2a
b
(b) D ≥ 0 , af (k) < 0 , k > −
2a
(c) D > 0 , af (k) < 0
b
(d) D ≥ 0 , af (k) > 0 , k < −
2a
3. Which of the following may not be a possible set of
determinants of relative position of roots of a
quadratic equation ax 2 + bx + c = 0 with respect to a
randomly chosen point k on the X-axis ?
b
(a) D ≥ 0 , af (k) > 0 , k < −
2a
b
(b) D ≥ 0 , af (k) < 0 , k > −
2a
(c) D > 0 , af (k) < 0
b
(d) D ≥ 0 , af (k) > 0 , k < −
2a
4. Which of the following facts are the most essential to
know if both the roots of ax 2 + bx + c = 0 are either
greater or less than the randomly chosen number k ?
(i) D
(ii) a ⋅ f (k)
(iii) Equation of axis of symmetry of the graph
(iv) Equation of the vertex of the graph
(v) Average of the real roots
(a) (i), (ii) and (iv)
(b) (i), (ii) and (iii) or (v)
(c) (i), (ii), (iii) and (iv)
(d) Only (i) and (ii)
5. If the randomly chosen point k lies between the roots of
the equation ax 2 + bx + c = 0, then which of the
following is correct ?
(a) af (k) > 0
(b) af (k) ≥ 0
(c) af (k) < 0
(d) all of these
6. If both the roots of the equation ax 2 + bx + c = 0 lie
inside the randomly chosen real points k1, k2 then
which of the following facts are the most appropriate
and valid ones?
(i) af (k1) > 0, af (k2 ) > 0
(iii) af (k1) < 0 , af (k2 ) < 0
(v) D > 0
(ii) f (k1) ⋅ f (k2 ) < 0
(iv) D ≥ 0
(vi) k1 < −
(vii) f (k1) ⋅ f (k2 ) > 0
(a) (i), (v) and (vi)
(c) (iii), (iv) and (vii)
b
< k2
2a
(b) (i), (iv) and (vi)
(d) (ii) and (v) only
7. The following facts assure that the roots of the
equation ax 2 + bx + c = 0 lie inside the randomly
chosen real points k1, k2 , then which of the following
facts are the most appropriate and valid ones ?
(a) af (k1) > 0, af (k2 ) > 0
(b) f (k1) ⋅ f (k2 ) > 0
(d) all of these
(c) af (k1) < 0 , af (k2 ) < 0
8. Which of the following facts are correct if only one of
the roots of the equation ax 2 + bx + c = 0 lies inside
the randomly chosen real points k1, k2 ?
(i) D ≥ 0
(iii) f (k1) ⋅ f (k2 ) < 0
b
(v) k1 < −
< k2
2a
(a) (i), (iii) and (iv) only
(c) (ii) and (iv) only
(ii) D > 0
(iv) f (k1) ⋅ f (k2 ) > 0
(b) (i), (iv) and (v) only
(d) (ii) and (iii) only
9. The value of axis of symmetry of the quadratic graph of
ax 2 + bx + c = 0 is required to know the position of
roots with respect to one or two real points when
(a) the randomly chosen point is either side of the
roots
(b) if both the roots lie inside the randomly chosen
points
(c) both (a) or (b)
(d) There are more possibilities other than those
mentioned in (a) and (b)
10. While determining the position of the roots with
respect to one or two randomly chosen real point(s),
the discriminant D is strictly greater than zero for the
quadratic equation ax 2 + bx + c = 0 when
(a) the randomly chosen point lies between both the
roots
(b) the randomly chosen both the points lie between
the roots
(c) only one root lies between the two randomly
chosen points
(d) all of the above
788
QUANTUM
11. Find all the parameters p for which both the roots of
the equation x 2 − 6 px + (3 − 4 p + 9 p2 ) = 0 exceed
the number 4.
3
4
19
(b) p >
(c) p < 1
(d) p >
(a) p >
4
3
9
12. Find all the values of p for which both roots of the
equation x 2 + x + p = 0 exceed the quantity p.
1
1
(a) p < −
(b) p <
(c) p > 0
(d) p < − 2
2
4
13. Determine all the values of p for which both roots of
the equation ( p2 + p − 2 )x 2 − ( p + 5 )x − 2 = 0 exceed
the greatest negative integer.
1
(a) (−∞ , − 2 ) ∪ (−1, − ) ∪ (1, ∞ )
2
(b) (−∞ , − 2 ) ∪ (1, ∞ )
1
(c) (−∞ , − 1) ∪ (− , ∞ )
2
(d) (−∞ , − 1) ∪ (1, ∞ )
14. Find the value of p for which one root of the equation
x 2 − ( p + 1)x + p2 + p − 8 = 0 is more than 2 and
another root is less than 2.
 11 
(a) [2 , 3 ]
(b)  −
, 3

 3

 11
(c)  −
, − 2

 3
(d) − 2 < p < 3
15. Find all the values of p for which roots of the quadratic
equation
are
( p2 + p + 1)x 2 + ( p − 1)x + p2 = 0
located on either side of 4.
(a) p > − 1
(b) −17 < p < 1
17
(c) −
(d) none of these
< p < 20
12
16. Find the set of values of u for which exactly one root of
the equation x 2 − ux + (u 2 + 6 u) = 0 lies in (−2 , 0 ).
(a) (−6 − 2 ) ∪ (−2 , 0 )
(b) (−8 , 0 )
(c) (−6 , − 2 ) ∪ {2 }
(d) none of these
17. If α , β are the zeros of the quadratic function
f (x) = x 2 + 2 ( p − 3 )x + 9 such that α ≠ β and
−6 < {a , β } < 1, then find the value of p.
27 

(a)  −4 ,


4
27 

(b)  −2 ,


4
 27 
(c)  6 ,


4
(d) (6 , ∞ )
CAT
18. Find all the values of m for the equation
(m − 4 ) − 2 mx + (m − 5 )x 2 = 0, so that one root is less
than 1 and another one is greater than 2.

 20
(a) 
(b) (6 , 24 )
, 24

 9

 20
(d) 
, ∞

 9
(c) (5 , 24 )
19. Find the values of w for which the equation
x 2 + 2 (w − 1)x + w + 5 = 0 has at least one positive
root.
(a) (−∞ , − 5 ) ∪ (−5 , − 1]
(b) w ≤ − 1
(c) W < 1
(d) (−∞ , − 5 ) ∪ (4 , ∞ )
20. Find the number of integral values of p if the roots of
the equation 2 x 2 + p2 − (5 x + 6 p) + 8 = 0 are of
opposite sign.
(a) 1
(b) 2
(c) 3
(d) 6
21. For what values of p exactly one root of the equation
2 p x 2 − 4 p x + 2 p − 1 = 0 lies between 1 and 2 ?


 5 + 17  
 5 − 17  

  < p < 2 − 
(a) 2 − 
4
4






 5 + 17 
 5 − 17 

 < p < log2 
(b) log2 
4
4




 5 + 17 
 5 − 17 

 < p < log5 
(c) log5 
4
4




(d)
− 5 + 17
4
< p<
5 + 17
4
Answers
1. (c)
6. (b)
11. (d)
2. (d)
7. (a)
12. (d)
3. (b)
8. (d)
13. (b)
4. (b)
9. (c)
14. (d)
5. (c)
10. (d)
15. (d)
16. (a)
17. (c)
18. (c)
19. (b)
20. (a)
21. (b)
Theory of Equations
789
Hints
11. Since both the roots exceed the arbitrary point k (or
14. Since one root is less than 2 and another is greater than
greater than k) then you must have to satisfy the
following three conditions.
b
(i) D ≥ 0
(ii) af (k) > 0
(iii) k < −
2a
2. It implies that 2 lies between the roots, then it must
satisfy the following two conditions.
So here it goes (−6p) − 4 × 1 × (3 − 4p + 9p ) ≥ 0
⇒
2
⇒
2
36p − 12 + 16p − 36p ≥ 0 ⇒ 16p ≥ 12
3
p≥
4
2
⇒
2
. ...(i)
⇒
1 × (16 − 24p + 3 − 4p + 9p2 ) > 0
⇒
9p2 − 28p + 19 > 0 ⇒ (p − 1)(9p − 19) > 0
19
, ∞ 
p ∈ (−∞ , 1) ∪ 
 9

− 6p
b
4
⇒4< −
⇒ 4 < 3p ⇒ p >
2a
3
2
19
.
The intersection of (i), (ii) and (iii) is p >
9
And k < −
15. Since 4 lies between the roots, then it must satisfy the
following two conditions.
...(iii)
(i) af (k) < 0
(ii) D > 0
Now, af (k) < 0
Where a = p2 + p + 1
The discriminant of p2 + p + 1 is negative and the
coefficient of p2 is positive, it means p2 + p + 1 > 0 ⇒ a> 0.
Similarly, f (4) = 17p2 + 20p + 12
The discriminant of 17p2 + 20p + 12 is negative and the
coefficient of p2 is positive, it means 17p2 + 20p + 12 > 0
⇒ f(4) > 0.
Since a > 0 and f(4) > 0. It implies that af(4) > 0.
af (k) > 0
⇒
1 × f (p) > 0
⇒
p ∈ (−∞ , − 2) ∈ (0, ∞ )
b
1
⇒ p< −
k< −
2a
2
⇒ p + 2p > 0 ⇒ p(p + 2) > 0
2
…(ii)
…(iii)
D > 0 ⇒ (− u)2 − 4 × 1 × (u2 + 6u) > 0
greater than k) then you must have to satisfy the following
three conditions [ − (p + 5)]2 − 4 × (p2 + p − 2) × (−2) ≥ 0
(p + 1)2 ≥ 0 ⇒ p ∈ (− ∞ , − 1) ∪ (−1, ∞ )
(p + 2)(p − 1)(p + 1)(p + 1) > 0
⇒
(p + 2)(p − 1)(p + 1)2 > 0
⇒
p ∈ (−∞ , − 2) ∪ (1, ∞ )
− (p + 5)
b
And k < −
⇒ −1 < −
2a
2(p2 + p − 2)
2p2 + 3p + 1 > 0
⇒
(p + 1)(2p + 1) > 0
p ∈ (−∞ , − 1) ∪ (−
−3u2 − 24u > 0
⇒
−3(u2 + 8u) > 0
⇒
u2 + 8u < 0 ⇒ u(u + 8) < 0
u ∈ (−8, 0)
⇒
…(ii)
⇒
⇒
(u2 + 4u + 4)(u2 + 6u) < 0
(u + 2)2 (u2 + 6u) < 0
u ∈ (−6 − 2) ∪ (−2, 0)
1
, ∞)
2
The intersection of (i), (ii) and (iii) is p ∈ (−∞ , − 2) ∪ (1, ∞ ).
…(ii)
The intersection of (i) and (ii) is (−6 − 2) ∪ (−2, 0)
Hence choice (a) is the answer.
17. Since both the roots lie between k1 and k2 therefore the
following conditions must be satisfied.
Hence choice (b) is the answer.
…(i)
And f (k1) ⋅ f (k2 ) < 0 ⇒ f (−2) ⋅ f (0) < 0
− 2(p2 + p − 2) < (p + 5)
⇒
⇒
⇒
af (k) > 0 ⇒ (p2 + p − 2) ⋅ f (−1) > 0
⇒
(ii) f (k1) ⋅ f (k2 ) < 0
Then we have,
13. Since both the roots exceed the arbitrary point k (or
(p2 + p − 2) ⋅ (p2 + 2p + 1) > 0
16. Since exactly one root lies between two randomly chosen
(i) D > 0
Hence choice (d) is the answer.
⇒
As af (k) </ 0, so 4 does not lie between the two roots.
Hence choice (d) is the answer.
points, then it must satisfy the following conditions.
The intersection of (i), (ii) and (iii) is p < − 2.
⇒
p2 − p − 6 < 0 ⇒ − 2 < p < 3
Hence choice (d) is the answer.
...(ii)
greater than k) then you must have to satisfy the following
three conditions.
1
…(i)
D ≥ 0 ⇒ 12 − 4 × 1 × p ≥ 0 ⇒ p ≤
4
⇒
…(i)
The intersection of (i) and (ii) is − 2 < p < 3
12. Since both the roots exceed the arbitrary point k (or
And
−3p2 − 2p + 33 > 0
 −11 < p < 3



 3
⇒
Hence choice (d) is the answer.
⇒
[ − (p + 1)]2 − 4 × 1 × (p2 + p − 8) > 0
And af (k) < 0 ⇒ 1 ⋅ f(2) < 0
a ⋅ f (k) > 0 ⇒ 1 ⋅ f(4) > 0
And
⇒
D>0
⇒
(i) D > 0, since it is given that α ≠ β.
(ii) af (k1) > 0, af (k2 ) > 0
b
(iii) k1 < −
< k2
2a
Then we have D > 0 ⇒ [ 2(p − 3)]2 − 4 × 1 × 9 > 0
790
QUANTUM
⇒
⇒
And
p(p − 6) > 0
…(i)
p ∈ (−∞ , 0) ∪ (6, ∞ )
af (k1) > 0 ⇒ 1. f (−6) > 0
⇒ 36 + 2(p − 3) × (−6) + 9 > 0
27
⇒
p<
4
…(ii)
Also af (k2 ) > 0
⇒
1 ⋅ f(1) > 0
⇒
1 + 2(p − 3) + 9 > 0
⇒
…(iii)
⇒
…(iv)
− 6 < − (p − 3) < 1 ⇒
2< p< 9
27 
The intersection of (i), (ii), (iii) and (iv) is  6,
.

4
Hence choice (c) is the answer.
18. Since one root is less than k1 and another root is greater
than k2 , then the following conditions must be satisfied.
(i) D > 0
(ii) af (k1) < 0, af (k2 ) < 0
Then we have
⇒
And
⇒
D > 0 ⇒ (−2m)2 − 4(m − 5)(m − 4) > 0
20
m>
9
…(i)
af (k1) < 0 ⇒ (m − 5) f (1) < 0
(m − 5) (−9) < 0 ⇒ m> 5
…(ii)
af (k2 ) > 0
Also
⇒
(m − 5)f (2) < 0
⇒
(m − 5)(m − 24) < 0
⇒
5 < m < 24
…(iii)
19. Since roots are real, so discriminant cannot be negative.
That is D ≥ 0.
…(iv)
Combining (iii) and (iv) we get w ∈ (−5, 1)
…(v)
Now the intersection of (i), (ii) and (v) is w ∈ (−∞ , − 1]
Hence choice (b) is the answer.
20. The given equation 2x 2 + p2 − (5x + 6p) + 8 = 0 can be
expressed as 2x 2 − 5x + p2 − 6p + 8 = 0
Since, one root is greater than 0 and another one is less
than 0, therefore the given equation must satisfy the
following condition.
(i) D > 0 ⇒ (−5)2 − 4 × 2 × (p2 − 6p + 8) > 0
66
⇒ − 8p2 + 48p − 39 > 0 ⇒ 3 −
< p< 3+
4
66
4
w ∈ (−∞ , − 1] ∪ [ 4, ∞ )
Case 1 When exactly one root is positive. That means
exactly one root is greater than 0. Then
Now the intersection (or common) values of (i) and (ii)
will hold true. Thus the required value of p is (2, 4). That
is 3 is the only one integral value that satisfy the required
condition. Hence choice (a) is correct.
Since exactly one root lies between 1 and 2, so f(1)(2) < 0.
Therefore, (2p − 4p + 2p − 1) = (2p 22 − 4p ⋅ 2 + 2p − 1) < 0
∴ [ 2(w − 1)]2 − 4 × 1(w + 5) ≥ 0
⇒w + 5< 0⇒w < − 5
w<1
21. Let’s assume f (x ) = 2p x 2 − 4p x + 2p − 1.
Hence choice (c) is the answer.
af(0) < 1 ⇒ 1 ⋅ f(0)
…(iii)
1 ⋅ f(0) > 0 ⇒ w + 5 > 0 ⇒ w > − 5
b
 b −2(w − 1)

And  −
> 0⇒ 1− w > 0
=
= 1 − w −
 2a
 2a
2
(ii) af (k) < 0 ⇒ 2f(0) < 0 ⇒ 2(p2 − 6p + 8) < 0 ⇒ 2 < p < 4
The common values of m from (i), (ii) and (iii) is (5, 24).
⇒
Case 2 When both the roots are positive. That means
both the roots are greater than 0. Then following three
conditions must be satisfied.
b
(i) af(0) > 0
(ii) −
>0
2a
Then we have,
⇒
p> − 2
2(p − 3)
b
And −
=−
= − (p − 3)
2a
2
b
So,
< k2
k1 < −
2a
CAT
⇒ (−4p + 2 ⋅ 2p − 1)(2p ⋅ 4 − 22 p 2 + 2p − 1) < 0
…(i)
⇒
− (4p − 2 ⋅ 2p + 1)(2p ⋅ 5 − 22 p 2 − 1) < 0
⇒
(4p − 2 ⋅ 2p + 1)(2 ⋅ 22 p − 5 ⋅ 2p + 1) < 0
⇒
(2p − 1)2 (2 ⋅ 22 p − 5 ⋅ 2p + 1) < 0
⇒ (2 ⋅ 2
2p
− 5 ⋅ 2 + 1) < 0; since (2p − 1)2 in non negative.
p
⇒ [ 2(2 ) − 5 ⋅ (2p ) + 1] < 0
 p  5 + 17  

⇒
 2 − 

4

 

p 2
 p  5 − 17  
 < 0
 2 − 

4

 

 5 − 17
5 + 17 


< 2p <


4
4


⇒
 5 − 17 
 5 + 17 
 < p < log2 

log2 



4
4




Hence choice (b) is the answer.
Theory of Equations
791
14.12 Relation Between The
Roots of Two Quadratic
Equations
Case 5 In this case, p = q, as all the four roots of both the
equations are same. It is a unique case when each equation
has double root and not only this the value of these double
roots is same.
p 1 =p 2
Let us assume that there are two quadratic equations
ap 2 + bp + c = 0 and uq 2 + vq + w = 0. Now, there can be any
of the five possibilities between p and q
(a) p < q
(b) p > q (c) p ≤ q (d) p ≥ q (e) p = q
If none of the above conditions are met, we cannot find a
conclusive relation between p and q. In order to determine
the relation between p and q, we can follow the following
method. First of all find the real roots of the equation
ap 2 + bp + c = 0 as { p1 , p2 } such that p1 ≤ p2 and the real
roots of the equation uq 2 + vq + w = 0 as {q 1 , q 2 } such that
q 1 ≤ q 2 . Now, place the roots on the number line and connect
the smallest root with the largest root of each equation. It
gives us two sections - one is p1 p2 and another one is q 1 q 2 .
Now, compare the two sections created between the smallest
and the largest roots of each quadratic equation, as shown
below.
Case 1 In this case, p < q, as all the roots of first equation are
smaller than all the roots of second equation. That is no any
point is common between these two sections. So clearly each
value of p is less than each value of q.
p1
There are following possibilities of such situations where we
cannot conclude the exact relation between the two sections.
In each such situation, either one or both the roots of a
quadratic equation lie(s) between the two roots of the other
quadratic equation. Essentially, whenever there is more than
one point common between two quadratic equations, we
cannot determine whether p < q or p > q or p ≤ q or p ≥ q.
When relation is uncertain, we can express it by the symbol
<
> to denote the relationship between p and q. If we don’t
want to use that symbol we can write it as { p = q or p <> q}.
p1
p2
q2
Case 2 In this case, p > q, as all the roots of first equation are
greater than all the roots of second equation. That is no any
point is common between these two sections. So clearly each
value of p is greater than each value of q
p1
p2
q2
Case 3 In this case, p ≤ q, as the largest root of the first
equation is equal to the smallest root of the second equation
and except this no other part of the two sections is common.
p1
p2
q2
q1
Case 4 In this case, p ≥ q, as the smallest root of the first
equation is equal to the largest root of the second equation
and except this no other part of the two sections is common.
p1
q1
Thus, we have learnt that when no value is common between
two sections, we can say either p < q or p > q. And, when
nothing is common on the number line except one value, we
can say p = q. If more than one value is common the answer
becomes indeterminable. This is where the challenge of
comparing the two sections arises.
p2
q1
q1
q1
q2
q1
q2
p1
p2
q2
q1
p1
p2
q1
q2
p1
p2
q1
q2
p1
p2
q1
q2
p1
p2
q1
p1
p2
q1
q2
p2
q2
792
QUANTUM
p2
p1
q1
q2
p1
p2
q1
q2
Exp. 1) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. p 2 − 10 p + 21 = 0
So you see, whenever there is overlapping, the relation
between p and q cannot be determined. In other words, this
kind of relation can be expressed by p <> q. Also, when the
roots of any or both of the equations are non-real, we cannot
find the exact relation between p and q.
Tricks to Identify the Signs of the Roots Let us consider a
quadratic equation ax 2 + bx + c = 0, whose roots are α and β
such that ( x − α )( x − β) = 0. The signs of the two roots will
depend upon the signs of b and c, where b denotes the
coefficient of x and c denotes the constant term of the
quadratic equation.
b
c
Signs of Roots
α
β
+
+
−
−
+
−
−
+
−
+
+
+
−
−
Both are always
negative
One is positive and
another one is
negative
Both are always
positive.
One is positive and
another one is
negative
+
−
Tricks to Determine the Relation Between the Roots of
Two Quadratic Equations
It is very useful when the given equations are such that
their roots are difficult to obtain. Let us consider two
quadratic equations:
(i) ap + bp + c = 0
2
CAT
(ii) rq + sq + t = 0
2
I. If any or both the roots of a quadratic equation lie
between the two roots of another quadratic equation, the
relation between the roots of the two quadratic
equations cannot be established.
II. If the signs of both the constant terms c and t are
negative, the relation between the roots of the two
quadratic equations cannot be established.
III. If the roots of any or both the quadratic equation(s) are
non-real, the relation between the roots of the two
quadratic equations cannot be established.
IV. When the constant terms of both the equations are
positive and the sign of coefficients of p and q are
opposite, we can definitely determine the relation
between p and q. If the sign of the coefficient of p is
positive and sign of the coefficient of q is negative, then
we have p < q.
II. q2 − 10 q + 24 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find roots of the equations.
p 2 − 10p + 21 = 0
⇒
( p − 3)( p − 7) = 0 ⇒ p = { 3 , 7}
q 2 − 10q + 24 = 0 ⇒( q − 4)( q − 6) = 0
⇒
q = {4, 6}
3
7
4
6
As we see that the two roots of the second equation lie
between the two roots of the first equation. Therefore, we
cannot determine the relation between p and q.
Hence choice (d) is correct.
Alternatively Create a table as shown below and compare
the values.
q1 = 4
q2 = 6
p1 = 3
p1 < q1
p1 < q2
p2 = 7
p2 > q1
p2 > q2
Since there is no clarity whether p < q or p > q, we cannot
determine the exact relation between p and q.
This method of using table can be applied in every problem to
test the relationship between the values of two quadratic
equations.
Exp. 2) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. p 2 − 7 p + 10 = 0
II. q2 − 10 q + 24 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations.
p 2 − 7 p + 10 = 0
⇒
( p − 2)( p − 5) = 0 ⇒ p = {2, 5}
q 2 − 10q + 24 = 0
⇒
( q − 4)( q − 6) = 0
⇒
q = {4, 6}
2
5
4
6
As we see that the one root of the one equation lies between
the two roots of the other equation. Therefore, we cannot
determine the relation between p and q.
Hence choice (d) is correct.
Theory of Equations
793
Exp. 3) There are two quadratic equations. Solve these
equations and find the relation between p and q
I. p 2 − 12 p + 32 = 0
Alternatively Create a table as shown below and
compare the values.
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations p 2 − 12p + 32 = 0
⇒
( p − 4)( p − 8) = 0 ⇒ p = {4, 8}
q 2 − 10q + 24 = 0 ⇒( q − 4)( q − 6) = 0 ⇒ q = {4, 6}
4
4
8
q2 = 9
p1 = 5
p1 = q1
p1 < q2
p2 = 9
p2 > q1
p2 = q2
6
Exp. 4) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. p 2 + 2 p − 15 = 0
II. q2 − 4 q − 12 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations p 2 + 2p − 15 = 0
( p + 5)( p − 3) = 0 ⇒ p = {−5 , 3}
q 2 − 4q − 12 = 0 ⇒( q + 2)( q − 6) = 0 ⇒ q = {−2, 6}
–5
3
6
–2
As we see that the one root of the one equation lies between
the two roots of the other equation. Therefore, we cannot
determine the relation between p and q.
Hence choice (d) is correct.
Alternatively Since the signs of the constant terms of
both the equations are negative, therefore we can definitely
say that each equation has one positive and one negative root.
It implies that at least one root of the one equation will
definitely lie between the two roots of the other equation, as
these roots lie on the both the sides of the Y-axis. That’s why
their sections will overlap.
Exp. 5) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. − p 2 + 14 p − 45 = 0
II. 2 q2 − 28 q + 90 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations.
− p 2 + 14p − 45 = 0 ⇒ ( p − 5)( p − 9) = 0 ⇒ p = {5 , 9}
2q 2 − 28q + 90 = 0 ⇒( q − 5)( q − 9) = 0 ⇒ q = {5 , 9}
5
5
Since there is no clarity whether p < q or p = q or p > q, we
cannot determine the exact relation between p and q.
Exp. 6) There are two quadratic equations. Solve these
equations and find the relation between p and q.
As we see that the one root of the second equation lies between the
two roots of the first equation. Therefore, we cannot determine the
relation between p and q. Hence choice (d) is correct.
⇒
q1 = 5
II. q2 − 10 q + 24 = 0
9
I. 3 p 2 + 5 p + 5 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations.
Since the discriminant of the 3 p 2 + 5 p + 5 = 0 is −35, which is
negative. So it does not give us real roots. Therefore, we
cannot determine the relation between p and q.
Hence choice (d) is correct.
Exp. 7) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. p 2 − 8 p + 15 = 0
II. q2 + 8 q + 12 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) Cannot be determined
Solution Let us find the roots of the equations p 2 − 8p + 15 = 0
⇒
( p − 3)( p − 5) = 0 ⇒ p = { 3 , 5}
q 2 + 8q + 12 = 0 ⇒( q + 6)( q + 2) = 0⇒ q = {−6, − 2}
3
–6
5
–2
As we see that both the roots of the first equation are greater
than both the roots of the second equation. Therefore, we can
conclude that p > q. Hence choice (b) is correct.
Alternatively Since the constant terms of both the
equations are positive and the sign of coefficients of p and q are
opposite, we can definitely determine the relation between p
and q. Further, since the sign of the coefficient of p is negative
and sign of the coefficient of q is positive, then we have p > q.
Exp. 8) There are two quadratic equations. Solve these
equations and find the relation between p and q.
I. p 2 − 5 p + 4 = 0
II. q2 + 5 q − 6 = 0
(a) p ≤ q
(b) p > q
(c) p ≥ q
(d) Cannot be determined
Solution Let us find the roots of the equations p 2 − 5 p + 4 = 0
⇒
( p − 1)( p − 4) = 0 ⇒ p = {1, 4}
q 2 + 5 q − 6 = 0 ⇒( q + 6)( q − 1) = 0 ⇒ q = {−6, 1}
1
9
As we see that there is more than one point common between
the two sections, it means the relation between p and q is
indeterminable. Therefore, we cannot determine the relation
between p and q. Hence choice (d) is correct.
II. q2 − 4 q − 12 = 0
–6
4
1
As we see that all the values of p are greater than all the values
of q, except one value. Therefore, we can conclude that p ≥ q.
Hence, choice (c) is correct.
794
QUANTUM
CAT
Practice Exercise
Directions (for Q. Nos. 1 to 15) In each of these questions, two equations (I) and (II) are given. You have to solve both the equations
and give answer regarding the relationship between p and q.
1. I. p2 − 361 = 0
(a) p < q
(e) p = q
(b) p > q
2. I. − p2 + 529 = 0
II. q2 − 40q + 399 = 0
(c) p ≤ q
(d) p ≥ q
II. 2q2 − 1058 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(e) cannot be determined
(d) p ≥ q
q2
3. I. 5 p − 120 p + 720 = 0 II. 3q − 16 =
8
(a) p < q
(b) p > q
(c) p ≤ q
(d) p ≥ q
(e) p = q
2
4. I. p2 − 17161 = 0
II. q2 + 17161 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(e) cannot be determined
5.
(d) p ≥ q
I. 25 p2 − 35 11 p − 198 = 0
II. 25 p2 + 35 11 p − 198 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) p ≥ q
(e) can’t be determined
14. I. p2 − 8 3 + 45 = 0
(a) p < q
(b) p > q
(e) p = q or p <
>q
3
2
7. I. p − 9 p + 18 p = 0
(a) p < q
(b) p > q
(e) p = q or p <
>q
2
8. I. 14 p + 27 p + 9 = 0
(a) p < q
(e) p = q
(b) p > q
9. I. p2 − 42 p − 343 = 0
(a) p < q
(b) p > q
(e) p ≤ q
>
10. I. 7 p2 + 4 7 p − 5 = 0
(c) p ≤ q
(d) p ≥ q
2
(c) p ≤ q
15. I. p2 + 2| p| + 35 = 0
II.
(d) p ≥ q
(d) p ≥ q
16
9
+
=5 q
q
q
(c) p ≤ q
(d) p ≥ q
(a) p < q
(c) p ≤ q
(e) can’t be determined
(b) p > q
(d) p ≥ q
1. (c)
2. (e)
3. (e)
4. (e)
5. (d)
6. (e)
7. (e)
8. (a)
9. (e)
10. (c)
11. (e)
12. (b)
13. (e)
14. (c)
15. (e)
Hints
1. p = {−19, 19} and q = {19, 21}
3. p = {12, 12} and q = {12, 12}
4. p = {131} and q = non-real roots
7 27 
 27 , 7 
5. p =  ,
 and q = −

5
3

3
5
6. p = {−6, 6} and q = {−6, 6}
7. p = {0, 3, 6} and q = {0, 1}
−3 3
3
3
8. p = − , −  and q =  , 
 2
7
8
7
9. p = {−7, 49} and q = {5}
II. 77q − (10 7 + 11 )q + 10 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(e) can’t be determined
2
(d) p ≥ q
(b) p > q
(d) p ≥ q

7
7
 7
11 
 2 11 9 11 
,
 and q =
5 
 5
12. p = 
 9 11
2 11 
,−
−

5
5 

 2 11
 2 11
9 11 
9 11 
,+
,−
 and q = 

5
5 
5 

 5
13. p = −
I. 25 p2 − 55 11 p + 198 = 0
II. 25 p2 + 55 11 p + 198 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(e) can’t be determined
5
1
 1 , 10 
10. p = −
,
 and q = 

11. p = {± 5, ± 7} and q = {−6.9, − 71
.}
11. I. p5 − 74 p3 + 1225 p = 0 II. q2 + 14q + 48.99 = 0
(a) p < q
(c) p ≤ q
(e) can’t be determined
II. q2 + 2q + 35 = 0
2. p = {−23, 23} and q = {−23, 23}
II. 56q2 − 3q − 9 = 0
(c) p ≤ q
(b) p > q
(d) p ≥ q
Answers
II. q − q = 0
3
II. q2 − 12 3 + 105 = 0
(a) p < q
(c) p ≤ q
(e) can’t be determined
I. 225 p2 − 1740 p + 2835 = 0
II. 225q2 + 690q − 2835 = 0
(a) p < q
(b) p > q
(c) p ≤ q
(d) p ≥ q
(e) p ≤ q
>
II. q2 = 1296
6. I. p4 − 1296 = 0
12.
13.
14. p = {3 3 , 5 3 } and q = {5 3 , 7 3 }
(d) p ≥ q
15. p = {−5, 5} and q = {−7, 5}
Theory of Equations
795
14.13 Polynomial Equations or
Functions of Higher Degree
From the discussion of the previous topics and chapters we
got to know that what a polynomial is and what the degree of
a term or polynomial is. However, I will start the topic with
some basic points.
Properties of a Polynomial Graph
˜
˜
˜
Function
Standard Form
Zero Function
f (x) = 0
Constant
Function
f ( x ) = a ; (a ≠ 0)
Degree
Terms
Undefined Undefined
Linear Function f ( x ) = ax + b; (a ≠ 0)
0
1
1
2
˜
Quadratic
Function
f ( x ) = ax + bx + c;
(a ≠ 0)
2
3
Cubic Function
f ( x ) = ax 3 + bx 2
+ cx + d ;(a ≠ 0)
3
4
Bi-quadratic
Function
f ( x )= ax 4 + bx 3
+ cx 2 + dx + e; (a ≠ 0)
4
5
2
˜
…
…
…
…
…
…
…
…
Thus the polynomial function of degree n is expressed in the
standard form as shown below f ( x ) = a n x n + a n − 1 x n − 1 +…
+ a 3 x 3 + a 2 x 2 + a 1 x + a 0 ( a n ≠ 0)
A polynomial function of degree n can be expressed as a
product of linear factors (or product of quadratic factors may
be along with linear factors) as following.
f ( x ) = ( x − α 1 )( x − α 2 )( x − α 3 )
… ( x − α n − 2 )( x − α n − 1 )( x − α n ); where α i ( i = 1, 2 … n) is
the root of the polynomial.
˜
˜
˜
˜
˜
˜
The
constants
the
a n , a n − 1 , … , a 3 , a 2 , a 1 , a 0 are
coefficients of the polynomial.
The term a n x n is the leading term and a 0 is the constant term.
The highest power of variable x is called the ‘degree’ and
n is a non-negative integer.
A polynomial function of degree n has at most ( n −1) local
extrema (or lumps).
A polynomial function of degree n has at most n zeros (or
roots or solutions).
The function is continuous. Thus the graph has no discrete
jumps or breaks, or no sharp corners.
It has at most n roots. Thus the graph cuts the X-axis in at
most n points. In general, the graph cuts any X-axis in at
most n points.
The function has at most n −1 critical points. It follows
that it has at most n −1 local extreme points, called
extrema.
The domain is the whole real line. In particular, the graph
has no vertical asymptotes.
The asymptotes are defined and explained in the
upcoming topics.
Since you have already learnt about the low degree
polynomials such as linear and quadratic functions, so in
the upcoming discussion I will primarily zero in on the
polynomials of higher degree, such as degree 3, 4, 5, etc.
Deriving the Common Polynomials From the
Standard Polynomial of Degree n
By substituting n =1 in the Standard Polynomial Expression
you can obtain the linear function as shown below.
f ( x ) = a 1 x + a 0 ( a 1 ≠ 0)
Similarly by substituting n = 2 in the Standard Polynomial
Expression you can obtain the quadratic function as shown
below.
f ( x ) = a 2 x 2 + a 1 x + a 0 ( a 2 ≠ 0)
Again, by substituting n = 3 in the Standard Polynomial
Expression you can obtain the cubic function as shown
below.
f ( x ) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 ( a 3 ≠ 0)
Once again, by substituting n = 4 in the Standard Polynomial
Expression you can obtain the biquadratic function as shown
below.
f ( x ) = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 ( a 4 ≠ 0)
In this way, using the standard polynomial function you can
express the polynomial of any degree.
796
QUANTUM
CAT
14.14 End (or Long Term) Behaviour of a Polynomial Function
If you look at any of the following polynomial graphs you will notice two vital things :
(1) The typical graph of a polynomial function looks like a wave because of the roots of the polynomials. These roots are
responsible for this wave-like graph. However, this wave shape ceases to exist as soon as the graph moves beyond the
region of roots. Here region of the roots imply the stretch (or section) of the X -axis that falls between the smallest and the
largest root. The numerically smallest root is always the left-most root and the largest root is always the right-most root
along the X -axis.
(2) As soon as the graph moves beyond the region of the roots (that means towards the extreme values of X-axis) the graph
keeps on moving away from the X-axis and never returns back towards it. Thus it ends the phenomenon of wave-like
graph after a certain point. And this is your area of interest at this time.
So you got it right that you are going to study that what happens or how does a graph behave when you keep on moving
endlessly on the either side of the X-axis. The importance of this end behaviour of the graph is critical to sketch the
complete graph (or to connect all the roots). If you know the end behaviour of a polynomial graph, it becomes pretty easy
to sketch the complete graph; especially it becomes obvious to differentiate between peaks and valleys. The positive
extrema are called the peaks and the negative extrema are called the valleys.
In this section you will get to know the tricks to find out the direction in which the two ends of the graph move - either both
ends go up or both ends go down, or one end goes up and the other end goes down. Remember that we are here not
interested in knowing that what happens between the two end roots (the smallest and the biggest root) of the polynomial.
The following illustrative graphs suggest that the long term behaviour of a polynomial function basically depends on the
leading term a n x n of the polynomial.
a n x n + a n − 1 x n − 1 + .... + a 2 x 2 + a1 x + a 0 .
an > 0
n
Odd
an < 0
50
50
40
40
30
30
20
20
10
–5
–4
–3 –2
–1
0
10
1
–10
2
3
4
5
–5
–4
–3 –2
–1
0
1
2
–10
–20
–20
–30
–30
–40
–40
–50
–50
f ( x ) = 2x − x − 16 x + 5
3
2
5

⇒f ( x ) = 2( x + 3)( x − 1) x − 

2
f ( x ) = −2x 3 + x 2 + 16 x − 15
5

⇒f ( x ) = −2( x + 3)( x − 1) x − 

2
3
4
5
Theory of Equation
797
an > 0
n
Odd
–5
–4 –3 –2
an < 0
1000
1000
800
800
600
600
400
400
200
200
–1
0
1
2
3
4
5
–5
–4 –3
–200
⇒
an < 0
2500
2500
2500
2000
2500
1500
2500
1000
2500
500
1
2
3
4
5
–500
–5
–4
–3 –2
–1 0
–500
–1500
–1500
–2000
–2000
2
3
4
5
–2500
–2500
f ( x ) = 3x7 + 12x 6 − 42x 5 − 168 x 4
1
–1000
–1000
⇒
5
f ( x ) = −8 x 5 − 28 x 4 + 64 x 3 +188 x 2 − 56 x − 160
5

f ( x ) = −8( x + 4)( x + 2)( x + 1)( x − 1) x − 

2
an > 0
0
4
–1000
f ( x ) = 8 x 5 + 28 x 4 − 64 x 3 −188 x 2 + 56 x + 160
5

⇒ f ( x ) = 8( x + 4)( x + 2)( x + 1)( x − 1)  x − 

2
–1
3
–800
–1000
–2
2
–600
–800
–4 –3
1
–400
–600
–5
0
–200
–400
n
Odd
–2 –1
+147 x 3 + 588 x 2
− 10 8 x 2 − 422
f ( x ) = 3( x + 4)( x + 3)( x + 2)
( x + 1)( x − 1)( x − 2)( x − 3)
f ( x ) = −3x7 − 12x 6 + 42x 5 + 168 x 4 −147 x 3 − 588 x 2
+108 x + 422
⇒f ( x ) = −3( x + 4)( x + 3)( x + 2)( x + 1)( x − 1)
( x − 2)( x − 3)
798
QUANTUM
CAT
Even
−5
−4
−3
−2
50
50
40
40
30
30
20
20
10
10
−1 0
1
2
3
4
5
−5
−4
−3
−2
−1 0
1
2
3
4
5
−10
−10
−20
−20
−30
−30
−40
−40
−50
−50
⇒ f ( x ) = − 2x 4 − 3x 3 + 18 x 2 + 17 x − 30
f ( x ) = 2x 4 + 3x 3 − 18 x 2 − 17 x + 30
5

⇒f ( x ) = 2( x + 3)( x + 2)( x − 1) x − 

2
⇒ f ( x ) = − 2( x + 3)( x + 2( x − 1)( x − 5 / 2)
Even
−5
−4
−3
5000
5000
4000
4000
3000
3000
2000
2000
1000
1000
−2 −1
0
1
2
3
4
5
−5
−4
−3 −2
−1 0
−1000
−1000
−2000
−2000
−3000
−3000
−4000
−4000
−5000
−5000
f ( x ) = 12x 6 − 243x 4 + 54 x 3 + 1071 x 2 − 54 x − 840
5 
7

⇒ f ( x ) = 12( x + 4)( x + 2)( x + 1)( x − 1)  x −   x − 

2 
2
1
2
3
4
5
f ( x ) = −12x 6 + 243x 4 − 54 x 3 − 1071 x 2 + 54 x + 840
5 
7

⇒ f ( x ) = −12( x + 4)( x + 2)( x + 1)( x − 1) x −   x − 

2 
2
Theory of Equation
799
an > 0
n
even
an < 0
5000
5000
4000
4000
3000
3000
2000
2000
1000
1000
–5
–5
–4
–3
–2
–1
0
1
2
3
4
–4
–3
–2
5
–1
0
1
2
3
4
5
–1000
–1000
–2000
–2000
–3000
–3000
–4000
–4000
–5000
–5000
f ( x ) = −2x 8 + x7 − 56 x 6 − 14 x 5 + 490 x 4 + 49 x 3
f ( x ) = 2x 8 + x7 − 56 x 6 − 14 x 5
3
+ 490 x + 49 x −1444 x − 36 x + 1008
4
2
⇒f ( x ) = 2( x + 4)( x + 3)( x + 2)
7

( x + 1)( x − 1)( x − 2)( x − 3)( x − 3)  x − 

2
−1444 x 2 − 36 x+1008
⇒ f ( x ) = 2( x + 4)( x + 3)( x + 2)( x + 1)( x − 1)( x − 2)
7

( x − 3) x − 

2
The graphical description of the end behaviour of the graph can be summarised as below, where a n x n is the leading term of the
polynomial y = a n x n + a n − 1 x n − 1 + ..... + a 2 x 2 + a1 x + a 0 .
an > 0
xn
an < 0
n is ODD
As x → +∞, y increases
n is EVEN
As x → − ∞, y decreases
As x → +∞, y increases
As x → −∞, y increases
As x → +∞, y decreases
As x → − ∞, y increases
As x → +∞, y decreases
As x → − ∞, y decreases
Sketchy Illustration
The circles in the following diagram hide all the real roots of the polynomial in order to give a clear picture of how the two ends
of the graph move, when the values of x are extremely high on either side of the X-axis.
n
an > 0
an < 0
For extremely high
values of x, when n
is ODD, the two
ends of the graph
move
in
the
OPPOSITE
direction
Y
Y
–X
Root
–Y
X
–X
Root
X
–Y
Thus beyond the range of the roots when you increase Thus beyond the range of the roots when you increase
x, then y will also increase and when you decrease x, x, then y will decrease and when you decrease x, then y
then y will also decrease.
will increase.
800
QUANTUM
n
an > 0
an < 0
For extremely high
values of x, when n
is EVEN, the two
ends of the graph
move in the SAME
direction.
Y
Y
–X
Root
X
–Y
–X
Root
CAT
X
–Y
Thus beyond the range of the roots when you either Thus beyond the range of the roots when you either
increase x, or decrease x, the value of y is always increase x, or decreases x, the value of y will always
decrease.
increase.
Exp. 1) Determine the end behaviour of
f ( x) = 10 x + 2 x 3 − 7 x 2 + 24
Solution The end behaviour of any polynomial graph depends on
the leading term which is supposed to be the term with the highest
power, so for the purpose of convenience you may rearrange the
terms in decreasing order as f ( x) = 2x3 − 7 x2 + 10x + 24.
Now since the power of leading term 2x 3 is 3, i.e. odd, therefore
the two ends of the graph will go in opposite directions. Further
the coefficient of x 3 is 2 (i.e., positive) therefore when x increases,
y also increases and likewise when x decreases, y also decreases,
provided x is considerably far from the roots.
Solution The end behaviour of any polynomial graph depends on
the leading term, which is supposed to be the term with the
highest power.
Now since the power of leading term −7 x5 is 5, i.e. odd, therefore
the two ends of the graph will go in opposite directions. Further
the coefficient of x5 is −7 (i.e., negative) therefore when x increases,
y decreases and likewise when x decreases, y increases, provided
x is considerably far from the roots. Hence, choice (a) is true.
14.15 Solutions (or Roots) of a
Polynomial Equation
The values of the variable satisfying the given polynomial
Exp. 2) For the extremely high values of x, what will be
equation are called its roots or solutions or zeros. That is α is a
the tendency of the graph of −3x 4 − 2x − 11?
root of the polynomial equation f ( x ) = 0 if f (α ) = 0.
Solution The end behaviour of any polynomial graph depends on
the leading term, which is supposed to be the term with the
highest power. Now since the power of leading term −3 x 4 is 4 (i.e.,
even) therefore the two ends of the graph will move in the same
direction. Further the coefficient of x 4 is −3 (i.e., negative)
therefore either when x increases or it decreases, y always
decreases, provided x is considerably far from the roots.
There are various methods to find out the roots of the
polynomial equations, however in your competitive exams
you are not expected to know these sophisticated methods
at all.
What all you are expected to do is use your reasoning, logic
and common sense to figure out the correct answer regarding
Exp. 3) When x approaches −∞, then what will be the polynomial expressions.
tendency of the graph of 5x 8 + 3x 7 − 18x 4 − 7 x − 30?
Examiners are not interested in knowing the higher
Solution The end behaviour of any polynomial graph depends on mathematics much into depth; rather they try to test your
the leading term, which is supposed to be the term with the logical prowess in this kind of problems. That’s why I’m not
highest power. So for the purpose of convenience you may going to unnecessarily discuss the methods of finding the
rearrange the terms in decreasing order as
roots.
8
7
4
f ( x) = 5 x + 3 x − 18x − 7 x − 30
Now since the power of leading term 5 x 8 is 8 (i.e., even) therefore
the two ends of the graph will go in the same direction. Further
the coefficient of x 8 is 5 (i.e., positive) therefore when x decreases,
y also decreases, provided x tends to −∞.
Exp. 1) In the following diagram you can see the graph of
a cubic polynomial f ( x) = x 3 − 6x 2 + 3x + 10 . Since the
degree of this polynomial is 3, therefore it has three roots
namely −1, 2, 5.
This shows that if you substitute x = − 1 or 2 or 5 in the given
Exp. 4) Which one of the following is true about the function f ( x) = x 3 − 6 x 2 + 3 x + 10, then you will get f ( x) = 0.
polynomial function y = −7 x5 + 3x 3 − 18x 2 − 7 x + 23?
(a) y → + ∞, when x → −∞
(c) y → + ∞, when x → +∞
(b) y → − ∞, when x → −∞
(d) none of the above
And there are two local extrema too, one above the X-axis and
one below the X-axis.
Theory of Equation
801
20
6. For an even degree equation (i.e., if n is even) whose
constant term is negative and a n > 0 (i.e., the coefficient
of highest degree term is positive) there must be at least
two real roots, one positive and one negative.
7. A real root must intersect or touch the X-axis.
8. The number of intersections or touching points will
always be less than or equal to the number of real roots.
If it is less than the real roots, it means there is a
multiplicity of the roots.
15
10
5
–2
0
–1
1
2
3
4
5
6
–5
–10
–15
–20
Exp. 1) If two of the roots of the polynomial equation
Exp. 2) In the following diagram you can see the graph
of a biquadratic polynomial f ( x) = x 4 − 6x 3 − 7 x 2
+ 36x + 36. Since the degree of this polynomial is 4,
therefore it has four roots namely −2, − 1, 3, 6.
This show that if you substitute x = − 2 or −1 or 3 or 6 in the
given functionf ( x) = x 4 − 6 x 3 − 7 x 2 + 36 x + 36, then you will
get f ( x) = 0. And there are three local extrema too, one above
the X-axis and two below the X-axis. That means there are
one local maximum and two local minima.
60
50
40
30
20
10
–3
–2
–1
0
1
2
3
4
5
6
–10
–20
–30
–40
–50
–60
–70
–80
–90
Basic Properties of Roots
1. An equation of degree n has exactly n roots; may be all
real or all imaginary or the combination of both.
2. Imaginary roots (also known as complex roots) occur in
conjugate pairs. That is if a + ib is a root of the
polynomial, then a − ib is also a root of that polynomial.
3. Irrational roots always occur in pairs. That means if
a + b is a root of a polynomial equation then a − b is
also a root of that polynomial.
4. Total number of roots n = Real roots (R) + Complex
roots (2K); where K = 0, 1, 2, 3, K ⇒ Number of real
roots = n − 2K , since complex roots occur in pairs.
5. For an odd degree equation (i.e., if n is odd) there must
be at least one real root. Since n is odd and 2K is even.
ax 4 + bx 3 + cx 2 + dx + e = 0, a ≠ 0, are 7 + 5 2 and 5 − 7 2 ,
then what will be the sum of all the roots of this polynomial?
Solution Since this a polynomial of degree four, so it will have
maximum four roots. Further since there are two distinct
irrational roots, it implies that there must be the conjugates of
these two roots. The conjugate of these two roots will be 7 − 5 2
and5 + 7 2 respectively. Therefore the sum of all the roots is 24.
Exp. 2) A polynomial equation of degree 5 has complex
roots and irrational roots. How many real roots does this
equation have?
Solution Since complex roots and irrational roots occur in
conjugate pairs, so if there is minimum one pair of each, then
there will be two complex roots and two irrational roots. Since
there is only one root unknown, so it must be real and rational
root. However since irrational roots are also the real roots. So
there are total 3 real roots and two non-real roots.
Exp. 3) What cannot be the exact number of real roots
in a polynomial of degree 9?
(a) 5
(b) 6
(c) 7
(d) 9
Solution Since the total number of roots are odd and the non-real
roots (i.e., complex roots) occur in pairs (i.e., 0, 2, 4, 6...) therefore
the number of real roots will be either 9 or 7 or 5 or 3 or 1,
depending on the number of pairs of complex roots.
Refer the following table.
Total Number of
Roots
9
9
9
9
9
Number of
Complex Roots
0
2
4
6
8
Number of Real
Roots
9
7
5
3
1
Thus, we can say that there cannot be exctly 6 real roots.
Multiplicity of a Root of a Polynomial
Function
A normal polynomial equation in which all the roots are real
and distinct can be expressed in the form of linear factors as
( x − α 1 )( x −α 2 )( x −α 3 ) .... ( x − α n ).
802
QUANTUM
However when certain roots are same, i.e., roots are repeated
then we call it the multiplicity of the root. For example
( x −α ) ( x −α ) ( x −α ) ( x −α ) ( x − β) ( x − β) ( x − β)
( x − γ 1 )....... ( x − γ k )
= ( x − α ) 4 ( x −β) 3 ( x − γ 1 ) ..... ( x − γ k )
In the above example α occurs 4 times, so α is a root of
multiplicity 4 and β occurs 3 times so β is a root of multiplicity 3.
Definition If f is a polynomial function and ( x − α ) m is a
factor of f but ( x − α ) m + 1 is not, then α is a root of
multiplicity m of f .
Exp.) The following graph shows the x-intercepts of
( x + 2) 3 ( x − 1) 2 . Even though there are five real roots, but
due to multiplicity of roots you see that there are only two
points on the X-axis where the graph intersects or touches
the X-axis.
12
10
8
6
4
2
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
–2
–4
–6
CAT
Sign Rules of a Polynomial Equation
For a polynomial of any degree written in the decreasing
order – highest degree term first and lowest degree term in
the last – you can determine the maximum possible positive
and negative real roots by the following rules.
1. The maximum number of POSITIVE real roots of a
polynomial equation f ( x ) = 0 is the same (or less than by
an even number) as the number of changes of signs –
from positive to negative and/or negative to positive – in
the coefficients of f ( x ). If the maximum nuber of
changes in the signs is p, the maximum number of
positive real roots of this polynomial equation would be
p − 2k ≥ 0, such that k = 012
, , ,3, .....
2. The maximum number of NEGATIVE real roots of a
polynomial equation f ( x ) = 0 is the same (or less than
by an even number) as the nuber of changes of
signs-from positive to negative and/or negative to
positive-in the coefficients of f ( −x ). If the maximum
number of changes in the signs is n, the maximum
number of positive real roots of this polynomial
equation would be n − 2k ≥ 0, such that k = 012
, , ,3, ....
Exp. 1) Find the possible number of positive and
negative real roots of x 4 + 7 x 3 − 4x 2 − x − 7 = 0.
Solution For positive real roots, you have to determine the
number of times the change happens in the sign of coefficients of
the given polynomial when written in the decreasing order of
degree of its terms. So write down the signs of coefficients as
shown below and mark the change in sign.
–8
+
+
–
–
–
–10
Change
–12
Odd Multiplicity If a polynomial function f has a real root
α of odd multiplicity, then the graph of f crosses the X-axis
at (α, 0) and the value of f changes sign at x = α.
In the polynomial function f ( x ) = ( x + 2) 3 ( x −1) 2 , x = − 2
is a root of multiplicity 3, which is a root of odd multiplicity.
Therefore at x = − 2 the function (or graph) crosses the X-axis
and changes the sign.
Even Multiplicity If a polynomial function f has a real root
α of even multiplicity, then the graph of f does not cross the
X-axis at (α, 0) and the value of f does not change the sign at
x = α. In the polynomial function f ( x ) = ( x + 2) 3 ( x − 1) 2 ,
x =1 is a root of multiplicity 2, which is a root of even
multiplicity. Therefore at x =1 the function (or graph) does
not cross the X-axis; however the graph kisses the X-axis and
bounces back at x =1.
Since there is only one change in sign, therefore there will be
maximum one positive real root. Now if you try to replace the
real roots with complex roots then you have to replace them in
pairs. It means one positive real root cannot be replaced by 2
complex roots. So finally there will have to be exactly one real
root. For negative real roots, you have to determine the number
of times the change occurs in sign of coefficients of the
polynomial, where x is replaced by –x and the terms are written
in the decreasing order of degree.
So first you have to replace every x with −x as shown blow.
( − x) 4 + 7( − x) 3 − 4( − x) 2 − ( − x) − 7 = 0
⇒
x 4 − 7 x 3 − 4x 2 + x − 7 = 0
Now write down the signs of coefficients as shown below and
mark the change in sign.
+
–
Change
–
+
–
Change Change
Theory of Equation
803
Since there are three changes in the sign, therefore there will
be maximum three negative real roots. So finally there will
be either 3 or 1 negative real root.
Total Number
of Roots
Number of
Complex
Roots
Number of
Positive
Roots
Number of
Negative
Roots
4
4
0
2
1
1
3
1
NOTE Please take cognizance of the fact that if you consider 4
complex roots then you have to have 1 negative root, which is
impossible.
+ + + +++
Since there is no change in sign, so there won’t be any positive
real root. For negative real roots, you have to determine the
number of times the change occurs in sign of coefficients of the
polynomial, where x is replaced by −x and the terms are written
in the decreasing order of degree.
So first you have to replace every x with −x as shown bleow.
( − x)5 + ( − x) 4 + 4( − x) 3 + 3( − x) 2 + ( − x) + 1 = 0
⇒
− x5 + x 4 − 4x 3 + 3 x 2 − x + 1 = 0.
Now write down the signs of coefficients as shown below and
mark the change in sign.
–
Exp. 2) Find the possible number of positive and
negative real roots of 2x 4 − x 3 + 4x 2 − 5x + 3 = 0.
Solution For positive real roots, you have to determine the
number of times the change happens in the sign of coefficients of
the given polynomial when written in the decreasing order of
degree of its terms. So write down the signs of coefficients as
shown below and mark the change in sign.
+
–
Change
+
Change
–
Change
+
Change
Since there are only 4 changes in sign, therefore there will be
maximum four positive roots. Thus there could be 4, or 2 or 0
positive real roots.
For negative real roots, you have to determine the number of
times the change occurs in sign of coefficients of the polynomial,
where x is replaced by −x and the terms are written in the
decreasing order of degree.
So first you have to replace every x with −x as shown below.
2( − x) 4 − ( − x) 3 + 4( − x) 2 − 5( − x) + 3 = 0
⇒
2x 4 + x 3 + 4x 2 + 5 x + 3 + 0
+
Change
Total Number
of Roots
Number of
Complex
Roots
Number of
Positive
Roots
Number of
Negative
Roots
4
4
4
0
2
4
4
2
0
0
0
0
Exp. 3) Find the possible number of positive and
negative real roots of x5 + x 4 + 4x 3 + 32x 2 + x + 1 = 0.
Solution For positive real roots, you have to determine the
number of times the change happens in the sign of coefficients of
the given polynomial when written in the decreasing order of
degree of its terms. So write down the signs of coefficients as
shown below and mark the change in sign.
Change
+
–
Change
Change
+
Change
Since there are five changes in the sign, therefore there will be
maximum five negative real roots. So finally there will be either
5 or 3 or 1 negative real root.
Total
Number of
Roots
Number of
Complex
Roots
5
5
5
0
2
4
Number of
Number of
Positive Real Negative Real
Roots
Roots
0
0
0
5
3
1
Exp. 4) Find the possible number of positive and
negative real roots of x5 − x 4 + 3x 3 + 9x 2 − x + 5 = 0.
Solution For positive real roots, you have to determine the
number of times the change happens in the sign of coefficients of
the given polynomial when written in the decreasing order of
degree of its terms. So write down the signs of coefficients as
shown below and mark the change in sign.
+
Now write down the signs of coefficients as shown below and
mark the change in sign.
+ + + + +
Since there is no change in sign, so there won’t be any negative
real root.
–
–
Change
+
+
Change
–
Change
+
Change
Since there are four changes in sign, so there will be at most four
positive real roots. Thus finally there will be either 4 or 2 or 0
positive real root. For negative real roots, you have to determine
the number of times the change occurs in sign of coefficients of
the polynomial, where x is replaced by −x and the terms are
written in the decreasing order of degree.
So first you have to replace every x with −x as shown below.
( − x)5 − ( − x) 4 + 3( − x) 3 + 9( − x) 2 − ( − x) + 5 = 0
⇒ − x5 − x 4 − 3 x 3 + 9x 2 + x + 5 = 0.
Now write down the signs of coefficients as shown below and
mark the change in sign.
–
–
–
+
+
+
Change
Since there is only one change in the sign, therefore there will be
maximum one negative real root.
804
QUANTUM
So finally there will be exactly 1 negative real root.
Total Number of
Roots
Number of
Complex
Roots
Number of
Positive
Real Roots
Number of
Negative
Real Roots
5
5
5
0
2
4
4
2
0
1
1
1
Exp. 5) Find the possible number of positive and
negative real roots of
4 x 7 + 3 x 6 + x5 + 2 x 4 − x 3 + 9 x 2 + x + 1 = 0
Solution For positive real roots, you have to determine the
number of times the change happens in the sign of coefficients of
the given polynomial when written in the decreasing order of
degree of its terms. So write down the signs of coefficients as
shown below and mark the change in sign.
+
+
+
+
–
+
+
+
CAT
In a sense you have to find out the conditions when the graph
(or function) will be positive, negative, non-negative, or
non-positive. To get the valid values of x, first of all
determine the real roots and mark them on the number line as
per convention; means from left to right in increasing order.
Then you will get n +1 intervals, if there are n real roots. In
the next step what you have to do is to pick any convenient
number in each interval and substitute it for x in the given
function and see whether this test gives a positive number or
negative number.
Then you can decide which intervals are to be considered for
your answer and it depends on your given inequation.
f (x) > 0
All positive values of x except the roots
f (x) ≥ 0
All positive values of x including the roots
f (x) < 0
All negative values of x except the roots
f (x) ≤ 0
All negative values of x including the roots
Change Change
Since there are two changes in sign, so there will be at most two
positive real roots. Thus finally there will be either 2 or 0 positive
real root. For negative real roots, you have to determine the
number of times the change occurs in sign of coefficients of the
polynomial, where x is replaced by −x and the terms are written
in the decreasing order of degree.
So first you have to replace every x with −x as shown blow.
4( − x) 7 + 3( − x) 6 + ( − x)5 + 2( − x) 4 − ( − x) 3 + 9( − x) 2 + ( − x) + 1 + 0.
⇒ −4x 7 + 3 x 6 − x5 + 2x 4 + x 3 + 9x 2 − x + 1 = 0
Now write down the signs of coefficients as shown below and
mark the change in sign.
–
+
–
+
+
Change Change Change
+
–
+
Change Change
Since there are five changes in the sign, therefore there will be
maximum five negative real roots. So finally there will be either
5 or 3 or 1 negative real root.
Total
Number of
Roots
7
7
7
7
7
7
Number of
Complex
Roots
0
2
2
4
4
6
Number of
Positive
Real Roots
2
0
2
0
2
0
Number of
Negative Real
Roots
5
5
3
3
1
1
14.16 Polynomial Inequality
In a polynomial equation you have to find the values of x,
which make f ( x ) equal to zero. However, in a polynomial
inequation you have to find the values of x which make f ( x )
less than or greater than zero.
How to Express Intervals
When roots are to be included in the desired interval
[α 1 , α 2 ]or α 1 ≤ x ≤ α 2
When roots are not to be included in the desired interval:
(α 1 , α 2 ) or α 1 < x < α 2
When only root α 1 is to be included in the desired interval
[α 1 , α 2 ) or α 1 ≥ x > α 2
When only root α 2 is to be included in the desired interval:
(α 1 , α 2 ]or α 1 > x ≥ α 2
Infinity is not an exact point or position, so it’s always
expressed as ( −∞, ∞ ) or − ∞ < x < ∞.
How to Express Various Intervals and
Individual Values Jointly
Let us assume that α 1 < α 2 < α 3 < α 4 <.... < α n .
[α 1 , α 2 ] ∪ [α 2 , α 3 ] = [α 1 , α 3 ] =α 1 ≤ x ≤ α 3
(α 1 , α 2 ) ∪ (α 2 , α 3 ) = (α 1 , α 3 ) − {α 2 }
=α 1 < x < α 3 except α 2
(α 1 , α 2 ) ∪ {α 2 } ∪ (α 2 , α 3 ) = (α 1 , α 3 ) =α 1 < x < α 3
(α 1 , α 2 ] ∪ [α 2 , α 3 ) = (α 1 , α 3 ) =α 1 < x < α 3
(α 1 , α 2 ) ∪ (α 3 , α 4 ) = (α 1 , α 4 ) − [α 2 , α 3 ]
=α 1 < x ≤ α 4 , except α 2 ≤ x ≤ α 3
[α 1 , α 2 ] ∪ [α 3 , α 4 ] = [α 1 , α 4 ] − (α 2 , α 3 )
=α 1 ≤ x ≤ α 4 , except α 2 < x < α 3
( −∞, α 1 ) ∪ (α 2 , ∞ ) = ( −∞, ∞ ) − [α 1 , α 2 ]
( −∞, α 1 ] ∪ [α 2 , ∞ ) = ( −∞, ∞ ) − (α 1 , α 2 )
( −∞, α 1 ] ∪ [α 2 , ∞ ) = [ −∞, ∞ ] − (α 1 , α 2 )
{α 1 } ∪ {α 2 } ∪ {α 3 } ∪....{α n } = {α 1 , α 2 , α 3 ..... α n }
Theory of Equation
805
Exp. 1) Find the values that satisfy the polynomial
inequation x( x + 3)( x − 5) > 0.
Exp. 4) Find the values that satisfy the polynomial
inequation ( x 2 + 7)( 2x 2 + 1) < 0.
Solution The roots of this inequation are −3, 0 and 5. So, now you
have to mark these roots on the number line. Since there are three
real roots, so there will be four intervals on the number line as
shown below.
Solution Since ( x 2 + 7) > 0 for all real numbers, and, similarly
( 2x 2 + 1) > 0 for all real numbers, therefore the product of two
positive real numbers can never be negative. Therefore there is
no solution for this inequation or you can say the solution set is
empty. Graphically speaking the graph of this function never
exists below the X-axis; even it does not touch or cross the
X-axis at all.
–
–
+
–∞
–3
+
0
+∞
5
The four intervals are ( −∞ , −3),( −3 , 0) , ( 0, 5) and (5 , ∞). Now check
the first interval. Consider any value, say−5 which lies in the first
interval ( −∞ , − 3), you will get x( x + 3) ( x − 4) = − 5( −2)( −9)
= − 90 < 0. So the first intervval gives negative value.
Similarly you can consider x = − 1 to test the second interval, since −1
lies in the second interval ( −3 , 0). This will give you
x( x + 3)( x − 4) = ( −1)( 2)( −5) = 10 > 0. So the second interval is
positive. Again the third interval test yields negative values and the
fourth interval test yields the positive values.
Therefore you see that the second interval and fourth interval satisfy
the given inequality. Thus the required answer is ( −3 , 0) ∪ (5 , ∞) or
−3 < x < 0 and 5 < x < ∞.
Exp. 2) Find the values that satisfy the polynomial
inequation ( x + 1)( x − 3) 2 > 0.
Solution The roots of this inequation are −1 and 3. So now you have
to mark these roots on the number line. Since there are only two
distinct real roots, so there will be exactly three intervals on the
number line as shown below.
–
–∞
+
+
–1
+∞
3
The three intervals are ( −∞ , − 1), ( −1, 3) and ( 3 , ∞). Now to check the
first interval you may consider any value between that interval, say
−2, then you will get ( x + 1)( x − 3) 2 = ( −1)( 25) = − 25 < 0. So the first
Exp. 5) Find the values that satisfy the polynomial
inequation ( x 2 − 3x + 3)( 2x + 5) 2 ≥ 0.
Solution The given inequation is ( x 2 − 3 x + 3)( 2x + 5) 2 ≥ 0.
If you can assume for a while that it’s an equation then there
will be total 4 roots, out of them only one
−5 

root  x =
 is a real root of multiplicity 2. The even

2
−5
multiplicity is an indicator of the fact that the graph at x =
2
will not intersect the X-axis, instead it will bounce back.
Further, if you simplify the expression then the leading term
will be 4x 4 . The even power of x suggests that both the ends
of the graph will move in the same direction, provided x is
very-very large.
Further the coefficient of x 4 is 4, which is positive. That
means when x approaches ∞, y also approaches ∞. Also,
when x approach −∞, y again approaches +∞.
That means whole graph never goes below the X-axis.
Therefore, for every x, such that −∞ ≤ x ≤ ∞ you will have
( x 2 − 3 x + 3)( 2x + 5) 2 ≥ 0. Or other way round, the solution
of ( x 2 − 3 x + 3)( 2x + 5) 2 ≥ 0 is −∞ ≤ x ≤ ∞.
interval( −∞ , − 1) is negative.
Similarly you can consider any value from the second interval, say 0,
then you will get ( x + 1)( x − 3) 2 = (1)( 9) = 9 > 0. It means the second
200
175
150
interval ( −1, 3) is positive.
Finally you have to test the third and last interval by considering
any value from that interval, say 4, then you will get
( x + 1)( x − 3) 2 = (5)(1) = 5 > 0. Thus the third interval is also positive.
Since you are looking for positive values of the function, therefore
only second and third intervals satisfy the inequation. Thus the
required answer is ( −1, 3) ∪ ( 3 , ∞).
125
100
75
50
–4
–3
–2
–1
25
0
1
2
3
–25
NOTE At x = 3, the inequation is invalid. That’s why you can’t include
–50
3 in this solution set.
Exp. 3) Find the values that satisfy the polynomial
inequation ( x 2 + 7)( 2x 2 + 1) > 0.
Solution The given inequation is ( x + 7)( 2x + 1) > 0. Since
( x 2 + 7) > 0 for all real numbers. Similarly ( 2x 2 + 1) > 0 for all real
numbers. Therefore ( x 2 + 7)( 2x 2 + 1) > 0 for all real numbers. Thus
the desired values of x are ( −∞ , ∞).
2
2
−5
and
2
( 2x + 5) 2 ≥ 0 for all real numbers. And ( x 2 − 3 x + 3)
Alternatively
You see that ( 2x + 5) 2 = 0 for x =
produces two complex roots. But the complex roots never
change the sign, since they never intersect the X-axis.
806
QUANTUM
Thus you can conclude that the graph of the product
( x 2 − 3 x + 3)( 2x + 5) 2 never goes below the X-axis. Thus,
( x 2 − 3 x + 3)( 2x + 5) 2 ≥ 0, for all real numbers. That means
−∞ ≤ x ≤ ∞.
Exp. 6) Find the value that satisfy the polynomial
inequation ( x 2 − 3x + 3)( 2x + 5) 2 > 0.
Solution Since you know that ( x 2 − 3 x + 3)( 2 x + 5) 2 ≥ 0 for
every x ∈ ( −∞ , ∞) and ( x 2 − 3 x + 3)( 2 x + 5) 2 = 0 at x =
Therefore ( x 2 − 3 x + 3)( 2 x + 5) 2 > 0
−5
.
2
−5
x ∈ ( −∞ , ∞) −  
 2
when
Exp. 7) Find the value that satisfy the polynomial
inequation 2( x − 2) 3 ( x + 3) 2 < 0.
Solution The roots of this inequation are −3 and 2. So now you
have to mark these roots on the number line. Since there are two
roots, so there will be three intervals on the number line as
shown below.
–
–∞
–
–3
+
2
+∞
The three intervals are ( −∞ , − 3), ( −3 , 2), ( 2, ∞). Now to check the
first interval you may consider any value between that interval,
say
then
you
will
get
−4,
2( x − 2) 3 ( x + 3) 2
= 2( −216)(1) = − 432 < 0. So the first interval( −∞ , − 3) is negative.
Similarly you can consider any value from the second interval,
say 0, then you will get 2( x − 2) 3 ( x + 3) 2 = 2( −8)( 9) = − 144 < 0. It
means the second interval (−3, 2) is also negative.
Finally you have to test the third and last interval by considering
any value from that interval, say 4, then you will get
2( x − 2) 3 ( x + 3) 2 = 2( 8)( 49) = 784. Thus the third interval is
positive.
Since you are looking for strictly negative values of the function,
therefore only first and second intervals satisfy the inequation.
Thus the required answer is ( −∞ , −3) ∪ ( −3 , 2), which you can
express as (−∞, 2)−{−3}
Alternatively Here 2 is the root of multiplicity 3 (i.e., odd
multiplicity) so the graph will intercept at x = 2 that will lead to
change in sign of the graph (or function). Again, −3 is the root of
multiplicity 2 (i.e. even multiplicity) so the graph will bounce
back at x = − 3 that will not lead to any change in sign of the
graph.
The leading term of this function is 2x5 , which suggests that the
one end of the graph will be below X-axis and one end will be
above the X-axis. Further the coefficient of the leading term is
positive, so the graph will be positive when x is positive and
greater than 2. Similarly the graph will be negative when x is less
than 2, except when x is −3. Thus it can be concluded that the
graph will be negative for x = ( −∞ , 2) − {−3}
CAT
Exp. 8) Find the values that satisfy the polynomial
inequation ( x + 3)( x 2 + 1)( x − 4) 2 < 0.
Solution The irreducible quadratic factor ( x 2 + 1) gives complex
conjugate roots. The only two distinct real roots are -3 and 4. So
now you have to mark these two real roots on the number line.
Since there are two roots, so there will be three intervals on the
number line as shown below.
–
–∞
+
+
–3
+∞
4
The three intervals are ( −∞ , −3), ( −3 , 4), ( 4, ∞). Now to check the
first interval you may consider any value between that interval,
say −4 , then you will get ( x + 3)( x 2 + 1)( x − 4) 2 = ( −1)(17)( 64)
= − 1088 < 0. So the first interval ( −∞ , − 3) is negative. Similarly
you can consider any value from the second interval, say 0, then
you will get ( x + 3)( x 2 + 1)( x − 4) 2 = ( 3)(1)(16) = 48 > 0. It means
the second interval ( −3 , 4) is positive.
Finally you have to test the third and last interval by considering
any value from that interval, say 5, then you will get
( x + 3)( x 2 + 1)( x − 4) 2 = ( 8)( 26)(1) = 208 > 0. Thus the third
interval is also positive.
Since you are looking for strictly negative values of the function,
therefore only first interval satisfies the inequation. Thus the
required answer is ( −∞ , − 3).
Alternatively At x = 4, due to even multiplicity of the root
the graph the graph touches the x- axis and bounces back
retaining the sign. At x = − 3, due to odd multiplicity the graph
intersects the X-axis and changes the sign. Since the leading
term, after simplification into standard polynomial form, is x5 ,
which implies that if x increases y also increases; and when x
decreases y also decreases. Thus you can conclude that the graph
will be negative when x < −3 and graph will be non-negative
when x > −3. So the answer is x < −3.
Exp. 9) Find the values that satisfy the following
polynomial inequation ( x + 5) 2 ( x + 1)( x − 1) ≤ 0.
Solution This function gives three real distinct roots −5 , − 1 and
1. So now you have to mark these roots on the number line. Since
there are three roots, so there will be four intervals on the
number line as shown below.
+
–∞
–
+
–5
–1
+
1
+∞
The four intervals are ( −∞ , − 5),( −5 , − 1), ( −1,1)(1, ∞).
Now check the first interval. Consider any value, say −6 which
lies in the first interval ( −∞ , − 5), you will get
( x + 5) 2 ( x + 1)( x − 1) = (1)( −5)( −7) = 35 > 0
So the first interval gives positive value. Similarly you can
consider x = − 2 to test the second interval, since −2 lies in the
second interval ( −5 , −1).
This will give you ( x + 5) 2 ( x + 1)( x − 1) = ( 9)( −1)( −3) = 27 > 0
So the second interval is also positive.
Theory of Equation
807
Again you can consider x = 0 to test the third interval, since 0 lies
in the third interval ( −1, 1). This will give you
( x + 5) 2 ( x + 1)( x − 1) = − 25 < 0.
So the third interval gives negative value. Finally you can
consider x = 2 to test the fourth interval. since 2 lies in the fourth
interval (1, ∞).
It will give you ( x + 5) 2 ( x + 1)( x − 1) = ( 49)( 3)(1) = 147.
So the forth interval is again positive. Therefore you see that the
third interval and all the roots satisfy the given inequality. Thus
the required answer is {−5} ∪ [−1, 1] or x = − 5 and −1 ≤ x ≤ 1.
Alternatively At x = − 5, due to even multiplicity of the root
the graph touches the xaxis and bounces back retaining the sign.
At x = − 1, due to odd multiplicity the graph intersects the X-axis
and changes the sign. Similarly at x = 1, due to odd multiplicity
the graph intersects the X-axis and changes the sign once again.
Since the leading term, after simplification into standard
polynomial form, is x 4 , which implies that both the ends will
move in the same direction when you move on the X-axis too far
in either direction. Further if x increases y also increases; but
when x decreases y increases. Thus you can conclude that the
graph will be non-positive when x = − 5 and −1 ≤ x ≤ 1 and
graph will be positive when x < −5 and −5 < x < −1 and x > 1. So
the answer is {−5} ∪ [−1, 1].
Exp. 10) Find the values that satisfy the polynomial
inequation 3( x − 2) 2 2( x + 4) 3 ( − x 2 − 2) < 0.
Solution You can modify the given inequation as,
−3( x − 2) 2 ( x + 4) 3 ( x 2 + 2). The given inequation if equated with
zero, it gives only two real distinct roots −4 and 2. The irreducible
quadratic factor ( x 2 + 2) yields only non real zeros, which don’t
affect the sign of the graph. So now you have to mark these two
real roots on the number line. Since there are two roots, so there
will be three intervals on the number line as shown below.
+
–∞
–
–
–4
2
+∞
The three intervals are ( −∞ , −4),( −4, 2), ( 2, ∞). Now to check the
first interval you may consider any value between that interval,
say −5, then you will get
−3( x − 2) 2 ( x + 4) 3 ( x 2 + 2) = − 3( 49)( −1)( 27) = 3969 > 0.
So the first interval ( −∞ , − 4) is positive. Similarly you can
consider any value from the second interval, say 0, then you will
get −3( x − 2) 2 ( x + 4) 3 ( x 2 + 2) = − 3( 4)( 64)( 2) = −1536 < 0. It means
the second interval (−4, 2) is negative.
Finally you have to test the third and last interval by considering
any value from that interval, say 3, then you will get
−3( x − 2) 2 ( x + 4) 3 ( x 2 + 2) = − 3(1)( 343)(11) = − 11319 < 0. Thus the
third interval is also negative.
Since you are looking for negative values of the function,
therefore only second and third intervals satisfy the inequation.
Thus the required answer is ( −4, 2) ∪ ( 2, ∞).
At x = − 4, due to odd multiplicity the graph
intersects the X-axis and changes the sign. At x = 2, due to even
multiplicity of the root the graph touches the X-axis and bounces
back retaining the sign. Since the leading term, after
simplification into standard polynomial form, is −3 x 7 which
implies that the two ends of the graph will be in opposite
directions and if x increases y decreases; and when x decreases y
increases. Thus you can conclude that the graph will be positive
when x < −4 and it will be negative when −4 < x < 2 and
2 < x < ∞. So the answer is ( −4, 2) ∪ ( 2, ∞) or ( −4, ∞) − {2}.
Alternatively
Exp. 11) Find the values that satisfy the following
polynomial inequation −( x + 2) 4 ( x + 1)( 2x 2 + x + 1) ≥ 0.
Solution The given inequation if equated with zero, it gives only
two real distinct roots −2 and−1. The irreducible quadratic factor
( 2x 2 + x + 1) yields only non-real zeros, which don’t affect the
sign of the graph. So now you have to mark these two real roots
on the number line. Since there are two roots, so there will be
three intervals on the number line as shown below.
+
+
–∞
–2
–
–1
+∞
The three intervals are ( −∞ , − 2),( −2, − 1), ( −1, ∞). Now after
testing each of the three intervals for their respective signs you
will get that the first and second intervals are positive and the
last interval is negative; also the graph touches at x = − 2. Since
you are looking for non-negative values of the function,
therefore only first and second intervals will satisfy the
inequation. Thus the required answer is
( −∞ , − 2) ∪ {−2} ∪ ( −2, − 1) ∪ {−1} or ( −∞ , −1]
Alternatively At x = − 2, due to even multiplicity of the root
the graph touches the X-axis and bounces back retaining the
sign. At x = −1, due to odd multiplicity the graph intersects the
X-axis and changes the sign. Since the leading term, after
simplification into standard polynomial form, is − 2x 7 which
implies that the two ends of the graph will be in opposite
directions and if x increases y decreases; and when x decreases y
increases. Thus you can conclude that the graph will be positive
when −∞ < x < −2 and −2 < x < −1 and it will be negative when
x > −1. So the answer is ( −∞ , − 1].
Practice Exercise
1. Find the values that satisfy the polynomial inequation
x 4 + 4 x3 − 12 x2 ≤ 0.
2. Find the values that satisfy the polynomial inequation
x 4 + 4 x3 − 12 x2 < 0.
3. Find the values that satisfy the polynomial inequation
x 4 + 4 x3 − 12 x2 > 0.
4. Find the values that satisfy the polynimial inequation
x 4 + 4 x3 − 12 x2 ≥ 0.
Answers
1. [ −6, 2]
2. (−6, 0) ∪ (0, 2)
3. (−∞ , − 6) ∪ (2, ∞ )
4. (−∞ , 6] ∪ {0} ∪ [ 2, ∞ )
808
QUANTUM
14.17 Relation Between Roots and
Coefficients in a Polynomial
Formation of a Polynomial Equation from the Given
Roots If α 1 , α 2 , α 3 , ..., α n be the roots of the nth degree
polynomial and S k ( k =12
, ,3, ... n) denotes the sum of the
products of roots taken k at a time, then the equation will be
If f ( x ) = a n x n + a n −1 x n −1 +...+ a 3 x 3 + a 2 x 2 + a1 x + a 0
x n − S 1 x n −1 + S 2 x n − 2 − S 3 x n − 3 + ... + ( −1) n S n = 0.
(where a n ≠ 0)
And if α 1 , α 2 , α 3 , ..., α n be the roots of the polynomial, then
a
S 1 = α 1 + α 2 + α 3 +...+ α n = ∑ α i = − n −1
an
S 2 = α 1α 2 + α 1α 3 +... s = ∑ α i α j = ( −1) 2
i≠ j
Formation of a Quadratic Equation from the Given Roots
x 2 − S 1 x + S 2 = 0 ⇒ x 2 − (α + β) x + (α β) = 0
Formation of a Cubic Equation from the Given Roots
a n −2
an
x 3 − S1 x 2 + S 2 x − S 3 = 0
⇒ x 3 − (α + β + γ )x 2 +(αβ + αγ + βγ ) x − (αβγ ) = 0
S 3 = α 1α 2α 3 + α 2α 3α 4 +...
=
CAT
∑ α iα jα k
i ≠ j ≠k
= ( −1) 3
a n −3
an
Formation of a Biquadratic Equation from the Given Roots
x 4 − S1 x 3 + S 2 x 2 − S 3 x + S 4 = 0
⇒ x 4 − (α + β + γ + δ ) x 3 +(αβ + αγ + αδ + βγ + βδ + γδ ) x 2
... ... ... ... ... ... ... ...
−(αβγ + αβδ + αγδ + βγδ ) x + (αβγδ ) = 0
... ... ... ... ... ... ... ...
S n −1 = α 1α 2 ... α n −1 + α 2α 3 ... α n
a
∑ α i α j ..α n −1 = (−1) n −1 a 1
=
i ≠ j ≠... ≠ ( n − 1)
n
n
S n = α 1α 2α 3 ... α n = ∏ a i = ( −1) n
i =1
a0
an
Thus a quadratic equation ax 2 + bx + c = 0, having the
roots α and β, then
b
c
S 1 = α + β = − and S 2 = αβ =
a
a
Similarly, a cubic equation ax 3 + bx 2 + cx + d = 0, having
the roots α, β and γ, then
b
c
S 1 = α + β + γ = − and S 2 = αβ + βγ + γα =
a
a
d
S 3 = αβγ = −
a
Similarly, a biquadratic equation
ax 4 + bx 3 + cx 2 + dx + e = 0
having α, β, γ and δ, then
S1 = α + β + γ + δ = −
b
a
S 2 = αβ + αγ + αδ + βγ + βδ + γδ =
S 3 = αβγ + αβδ + αγδ + βγδ = −
S 4 = αβγδ =
e
a
d
a
c
a
Exp. 1) Let P( x) = x 3 + ax 2 + bx + c be a polynomial with
real coefficients, c ≠ 0 and x1 , x 2 , x 3 be the roots of P ( x).
Determine the polynomial Q( x) whose roots are
1 1 1
,
,
.
x1 x 2 x 3
Solution
x1 + x 2 + x 3 = − a , x1 x 2 + x 2 x 3 + x 3 x1 = b
x1 x 2 x 3 = − c
1
1
1
x x + x 2 x 3 + x 3 x1
b
∴
+
+
= 1 2
=−
x1 x 2 x 3
x1 x 2 x 3
c
1
1
1
x1 + x 2 + x 3 a
+
+
=
=
x1 x 2 x 2 x 3 x 3 x1
x1 x 2 x 3
c
1
1
and
=−
x1 x 2 x 3
c
 −b
 a
 1
3
Therefore
Q ( x) = x −   x 2 +   x −  − 
 c
 c
 c
and
⇒
Q ( x) = cx 3 + bx 2 + ax + 1
Exp. 2) The real number x1 , x 2 , x 3 satisfying the
equation x 3 − x 2 + bx + c = 0 are in AP. Find the intervals
in which b and c lie.
Solution Let the roots be a − d , d , a + d , then
Sum of roots = ( a − d) + ( a) + ( a + d) = 1
1
⇒
a=
3
and ( a − d) ( a) + a( a + d) + ( a − d)( a + d) = b
 1
3 a2 − d 2 = b ⇒ 3   − d 2 = b
⇒
 9
1
⇒
d 2 = − b [since the minimum value of d 2 = 0]
3
1
⇒
−b ≥ 0 ⇒b ≤1/ 3
3
Now product of roots = ( a − d)( a)( a + d) = − c
Theory of Equation
⇒
⇒
⇒
⇒
809
a( a 2 − d 2 ) = − c
1 1
2
 − d  = −c

39
Exp. 1) Find the critical numbers of the rational
x 2 − 6x + 8
.
polynomial
x+2
d2 1
−
3 27
1
c≥−
27
c=
[since the minimum value of
Solution
d2
= 0]
3
1
Therefore −∞ < b ≤ 1 / 3 and −
≤c<∞
27
14.18 Rational Polynomials
Conceptually, a rational polynomial function is a quotient of
polynomials. Therefore a rational polynomial function f ( x )
is expressed as following
p( x )
f (x ) =
q (x )
Where p( x ) and q ( x ) are the polynomial functions, such that
q ( x ) ≠ 0.
˜
˜
The rational function f(x) is not defined for any x which is
the root of q(x).
The numerator and denominator can be polynomials of
any order.
Critical Numbers
The critical numbers for a rational function are all the values
of x where the function f ( x ) is zero or undefined, That is, all
the x where the numerator p (x) and denominator q ( x ) are
zero.
˜
˜
˜
˜
x 2 − 6x + 8 ( x − 2)( x − 4)
=
x+2
( x + 2)
Since the roots of x 2 − 6x + 8 are {2, 4} and that of x + 2 is {−2}.
Therefore the critical points are −2, 2 and 4.
Exp. 2) Find the critical numbers of the rational
x 2 + 4x + 3
.
polynomial 2
x − x − 20
Solution
x 2 + 4x + 3 ( x + 3)( x + 1)
=
x 2 − x − 20 ( x + 4)( x − 5)
Since the roots of x 2 + 4x + 3 are {−3, −1} and that of x 2 − x − 20
are {−4, 5}. Therefore the critical points are −1, −4, −3 and 5.
Exp. 3) Find the critical numbers of the following
x 2 + 2x − 15
.
rational polynomial
x2 − 9
Solution
x 2 + 2x − 15 ( x + 5)( x − 3)
=
( x + 3)( x − 3)
x2 − 9
Since the roots of x 2 + 2x − 15 are ( −5 , 3) and that of x 2 − 9 are
( −3 , 3). Therefore that critical points are −5 , − 3 and 3.
Exp. 4) Find the critical numbers of the rational
x2 +1
polynomial 2
.
x −x−6
Solution
x2 + 1
( x 2 + 1)
.
=
x 2 − x − 6 ( x + 2)( x − 3)
Since the roots of x 2 + 1 are non-real, so these are not
considerable. But the roots of x 2 − x − 6 are ( −2, 3), which are
real. Therefore the critical points are −2 and 3.
A rational expression will be zero only when its numerator
is zero and the denominator is not zero for the same value of
x. That is f ( x ) = 0, if p ( x ) = 0 but q ( x ) ≠ 0 for the same x.
Root or Solution of a Rational
Polynomial Equation
For critical points you have to consider only real numbers.
So, if upon solving the polynomial in the numerator you
will get non-real numbers, these non-real values won’t be
considered as critical points.
The root of a rational polynomial equation f ( x ) = 0 is the
value of x, for which p ( x ) = 0, but q ( x ) ≠ 0.
It’s not necessary that all the functions will have critical
points; it may occur when a polynomial function has no real
root and the denominator is defined for all the values of the
given interval.
However, in this course most of the functions that you will
be looking at do have critical points. That is only because
those problems make for more interesting examples. Do
not let this fact lead you to expect that a function will
always have critical points.
NOTE The terms root, zero and solution of a rational
polynomial can be interchangeably used for one another. And the
root/zero/solution can be called x-intercept too, if the root/zero/
solution is a real number.
Exp. 1) Find the roots of the rational polynomial
( 2x + 1)
equation
= 0.
( x + 3)( x − 1)
Solution
( 2x + 1)
= 0 ⇒ 2x + 1 = 0 ⇒ x = −1 / 2.
( x + 3)( x − 1)
Therefore there is only one solution of the above rational
equation, which is x = −1 / 2 .
810
QUANTUM
Exp. 2) Find the roots of the rational polynomial
x 2 − 6x + 8
equation
=0
x+2
Solution
x 2 − 6x + 8
( x − 4)( x − 2)
=0⇒
=0
x+2
( x + 2)
⇒
( x − 2)( x − 4) = 0 ⇒ x = 2 or 4.
Therefore there are two solutions of the above rational
equation, which are x = 2 and 4 .
NOTE
1.
2.
3.
Exp. 3) Find the solutions of the rational polynomial
x2 − 4
equation 3
=0
x − x 2 − 6x
Solution
⇒
( x + 2)( x − 2)
x2 − 4
=0⇒
⇒0
3
2
x( x + 2)( x − 3)
x − x − 6x
( x − 2)
=0 x
x( x − 3)
⇒
( x − 2) = 0
⇒
x =2
Therefore there is only one solution of the above rational
equation, which is x = 2.
Exp. 4) Find the solutions of the rational equation
x 2 ( x 4 − 81)( x + 2) 3
=0
x5 − x 4 − 6 x 3
x ( x − 81)( x + 2)
=0
x5 − x 4 − 6x 3
x 2 ( x 2 + 9)( x + 3)( x − 3)( x + 2) 3
=0
x 3 ( x − 3)( x + 2)
2
Solution
⇒
4
3
⇒
( x 2 + 9)( x + 3)( x + 2) 2
=0
x
( x 2 + 9)( x + 3)( x + 2) 2 = 0
⇒
( x 2 + 9) = 0 or ( x + 3) = 0 or ( x + 2) 2 = 0
⇒
If x 2 + 9 = 0 ⇒ x = ± 3i , which are complex (i.e, non-real)
roots. And if x + 3 = 0 ⇒ x = −3 which is a real root.
Again , if ( x + 2) 2 = 0 ⇒ x = −2, which is a real root.
Therefore there are two non-real roots and two real roots of
the above rational equation and thus there are total four
solutions.
NOTE Whenever there is a non-real root, the graph or the
function does not intersect or touch the X-axis.
Domain of a Rational Polynomial
Equation
The domain of a rational function includes only points where
both the numerator and the denominator are defined for real
values of x.
Essentially, the domain of a rational function does not
contain any value that would make the denominator equal to
zero.
CAT
Any value of x that makes both the numerator and
denominator 0, simultaneosy, will either create a ‘hole’ a
‘vertical asymptote’. And it happens when some factors
are common between numerator and denominator.
A hole in the graph shows the absence of the graph or a
miniscule break in the graph, without changing its
direction.
The vertical asymptote is nothing but a virtual line that
creates a boundary for the graph and thus the graph
cannot cross the boundary and it runs parallel to this
boundary. So a vertical asymptote changes the normal
path of the graph apart from braking it.
Exp. 1) Find the domain of the rational polynomial
x 2 + x − 20
.
f ( x) = 2
x − 3x − 18
Solution
x 2 + x − 20 ( x + 5)( x − 4)
=
x 2 − 3 x − 18 ( x + 3)( x − 6)
Since at x = −3 and 6, the denominator will be zero, i.e. the
rational polynomial will be undefined at these points. So the
domain of this polynomial will be all the real numbers except
x = −3 and 6.
That is
R − {−3 , 6 } or ( −∞ , −3) U ( −3 , 6) U ( 6, ∞).
Exp. 2) Find the domain of the rational polynomial
2x 2 − 5x − 3
.
f ( x) = f ( x) =
( x 2 + 4)
2x 2 − 5 x − 3
Solution
=
( x 2 + 4)

( x − 3)  x +

1

2
( x 2 + 4)
Since ( x + 4) cannot be equal to zero for any real value of x. So
2
this factor ( x 2 + 4), in the denominator, gives non-real zeros
and a non-real zero never affects the domain. So the domain of
this rational function will include all the real numbers. That is
R or ( −∞ , ∞).
Exp. 3) Find the domain of the rational polynomial
x2 −1
.
f ( x) = 2
2x − 5x
Solution
( x + 1)( x − 1)
x2 − 1
.
=
x( 2x − 5)
2x 2 − 5 x
Since at x = 0 and 5/2, the denominator will be zero, i.e. the
rational polynomial will be undefined at these points.
So the domain of this polynomial will be all the real numbers
except
5
x = 0 and
2
 5
R - 0, 
That is
 2
 5  5 
or
( −∞ ,0) U  0,  U  , ∞ .
 2  2 
Theory of Equation
Exp. 4) Find the domain of the rational polynomial
x2 − 4
.
f ( x) = 3
x − x 2 − 6x
Solution.
( x + 2)( x − 2)
x2 − 4
=
3
2
x − x − 6x x ( x + 2)( x − 3)
Since at x = 0, −2 and 3, the denominator will be zero, i.e. the
rational polynomial will be undefined at these points.
So the domain of this polynomial will be all the real numbers
except x = 0, −2 and 3.
That is
R- {−2, 0, 3}
or
( −∞ , − 2) ∪ ( −2, 0) ∪ ( 0, 3) ∪ ( 3 , ∞).
Exp. 5) Find the domain of the rational polynomial
x 3 + 2x 2 − 4x − 8
f ( x) =
x 4 − x 3 − 6x 2
Solution f ( x) =
x 3 + 2x 2 − 4x − 8 ( x + 2) 2 ( x − 2)
= 2
x 4 − x 3 − 6x 2
x ( x + 2)( x − 3)
Since at x = 0, − 2 or 3, the denominator will be zero, i.e. the
rational polynomial will be undefined at these points.
So the domain of this polynomial will be all the real number
except x = 0, − 2 and 3.
That is R - {0, − 2, 3} or ( −∞ , 0) ∪ ( 0, − 2) ∪ ( −2, 3) ∪ ( 3 , ∞).
Exp. 6) Find the domain of the rational polynomial
( x + 3)( x + 2)( x − 2) 3
.
f ( x) =
( x + 2) 2 ( x − 2)( x − 3)
Solution f ( x) =
( x + 3)( x + 2)( x − 2) 3
( x + 2) 2 ( x − 2)( x − 3)
Since at x = −2, 2 and 3, the denominator will be zero, i.e. the
rational polynomial will be undefined at these points.
So the domain of this polynomial will be all the real numbers
except x = −2, 2 and 3.
That is R - {−2, 2, 3} or ( −∞ , −2) ∪ ( −2, 2) ∪ ( 2, 3) ∪ ( 3 , ∞).
Graphical Features of a Rational
Polynomial
A rational function is a function that can be written as a
fraction of two polynomials where the denominator is not
zero.
Let the rational polynomial function be
f (x ) =
p ( x ) a n x n + a n −1 x n −1 + ...+ a1 x + a 0
=
q ( x ) bm x m + bm−1 x m−1 +...+ b1 x + b0
Or in the factor form a rational function can be expressed as
( x − α 1 )( x − α 2 )... ( x − α n −1 )( x − α n )
f (x ) =
( x − β1 )( x − β 2 )... ( x − β m−1 )( x − βm)
Where n is the degree of numerator p ( x ) and m is the degree
of denominator q ( x ) remove comma and α i and β i are the
roots of the polynomials p ( x ) and q ( x ), respectively.
811
Asymptote
An asymptote is an imaginary line that a curve approaches but
does not touch. There are basically two types of asymptotes:
vertical and non-vertical asymptotes. The non-vertical
asymptotes are also called as end-behavior asymptotes.
Asymptote
Condition
Value of Asymptote
Vertical
Vertical asymptote, x = c
Roots of the
denominator of the
rational polynomial,
in the lowest term.
Horizontal
If n < m
Horizontal asymptote, y = 0
If n = m
If n > m
Horizontal asymptote,
y = an / bm
No horizontal asymptote
Oblique
If n > m,
but n = m +1
Oblique asymptote, y = k ( x ) ;
wherep ( x ) = q ( x ) k( x ) + r ( x )
Curvilinear
If n > m, but
n > m +1
Curvilinear asymptote,
y = k ( x ); Where
p( x ) = q ( x ) k ( x ) + r ( x )
NOTE A rational polynomial in the 'lowest term’ implies that all
the common factors have been cancelled out and thus there is no
common factor between the numerator and denominator in the
rational polynomial when it is in the reduced or lowest term.
Mind the fact the even after cancelling out the common
factors the relation between n and m remains intact, since
equal number of factors get cancelled out from the numerator
and the denominator.
Vertical Asymptotes
1. A rational polynomial function may have none, one, or
several vertical asymptotes.
2. The graph of the rational function never intersects the
vertical asymptote.
3. The vertical asymptotes occur at all the zeros of the
denominator when the rational function is expressed in
the lowest term.
4. To determine the vertical asymptote factorize both the
numerator and denominator. If there are any common
factors between the numerator and denominator, then
cancel out the common factors and then set the
denominator equal to zero and solve. Any real solution
will represent a vertical asymptote. The equation of a
vertical asymptote is x = c.
Holes
Sometimes, as mentioned above, a factor may appear in both the
numerator and the denominator. Then there is a possibility that
there may not be a vertical asymptote and so it may create a hole
in the graph. A hole is simply a single point on the graph that
doesn’t actually exist due to a restriction.
812
QUANTUM
Let us assume that the factor ( x − c) u is in the numerator and
( x − c) is the denominator. If
v
(i) u < v, there will be a vertical asymptote x = c.
(ii) u > v, there will be a hole in the graph on the X-axis at
x = c. there is no vertical asymptote there.
(iii) u = v, there will be a hole in the graph at x = c, but not on
the X -axis. The y-value of the hole can be found by
cancelling the factors and substituting x = c in the
reduced function.
NOTE A rational function may have both - holes and vertical
asymptotes-for different values of x. However, for a certain value of x
the function cannot have both the hole and the vertical asymptote.
Exp. 1) Find all the vertical asymptotes and holes
( x + 3)( x − 7)
of
.
( 2x − 5)
Solution Since 2x − 5 = 0 ⇒ x = 5 / 2
Therefore there is only one vertical asymptotes, x = 5 / 2 .
Since there is no common factor in numerator and
denominator, so there won’t be any hole in the graph.
Exp. 2) Find all the vertical asymptotes and holes of
( 2 x + 9)(5 x − 4)
( x − 3)( x − 5)( 2 x − 11)
Solution Since ( x − 3)( x − 5)( 2x − 11) = 0 ⇒ x = 3 or 5 or 11/2.
Therefore there are three vertical asymptotes, x = 3 , x = 5
and x = 11 / 2.
Exp. 3) Find all the vertical asymptotes and holes of
( x + 2)( x + 1)( x − 3)
.
x 2 ( x − 3)( 3 x + 8)
Solution Since ( x − 3) is common in both the numerator and
denominator, so you have to cancel it out.
Then the simplified (or reduced) rational function is
( x + 2)( x + 1)
.
x 2 ( 3 x + 8)
Now since x 2 ( 3 x + 8) = 0 ⇒ x = 0 or x = −8/ 3.
Therefore there are two vertical asymptotes, x = 0 and
x = − 8 / 3. If you look at the original rational function in
which the power of x in the numerator is lower than that of
denominator, so there will be no hole in the graph.
The other way to look at this issue is that the reduced function
has no common factor in numerator and denominator, so
there won’t be any hole in the graph.
The possibility of hole arises only when numerator and
denominator of the reduced function has some common
factor. In this example, the reduced function has no common
factor between numerator and denominator.
CAT
Exp. 4) Find the vertical asymptotes and holes of
3 ( 2x + 5) 3 ( 8x − 3) 2 ( x − 3)
.
2 ( 8x − 3) 2 ( 2x + 5)
Solution. First of all cancel out all the common factors between
the numerator and denominator.
Then the simplified rational function is
3( 2x + 5) 2 ( x − 3) 3
= ( 2x + 5) 2 ( x − 3)
2
2
Since, there is no factor or root in the denominator of
simplified polynomial. Therefore there is no vertical
asymptote.
Since the power of ( 2x + 5) in the numerator is higher than that
of denominator, so there will be a hole in the graph on the
X-axis at x = −5 / 2. Again since the power of ( 8x − 3) in the
numerator is same as that of denominator, so there will be a
hole in the graph at x = 3 / 8, but not on the X-axis. The
y-value of the hole can be found by canceling the factors and
substituting x = 3 / 8 in the reduced function as following.
2
3
3
3
 3

y = ( 2x + 5) 2 ( x − 3) =  2 × + 5  − 3
 8

2
2
8
−33327
=
= −130.18
⇒
256
Exp. 5) Find all the vertical asymptotes and holes of
( x − 3) 2 ( x − 4)
.
( x − 5)( x − 3)
Solution. Since ( x − 3) is common in both the numerator and
denominator, so you have to cancel it out.
( x − 3)( x − 4)
Then the simplified rational function is
.
( x − 5)
Now, since ( x − 5) = 0 ⇒ x = 5. Therefore there is only one
vertical asymptote, x = 5. Since the power of ( x − 3) in the
numerator is higher than that of denominator, so there will be
a hole in the graph on the X-axis at x = 3.
Exp. 6) Find all the vertical asymptotes and holes of
( x + 2)( x + 4) 3
.
( x + 4)( x + 2) 3 ( 2x + 3)( x − 4)
Solution. First of all cancel out all the common factors between
the numerator and denominator.
( x + 4) 2
Then the simplified rational function is
( x + 2) 2 ( 2x + 3)( x − 4)
Now, Since ( x + 2) 2 ( 2x + 3) ( x − 4) = 0
⇒
x=−2
or
x = −3 / 2 or x = 4.
Therefore there are three vertical asymptotes,
x = −2, x = −3 / 2 and x = 4.
However, since the power of ( x + 4) in the numerator is higher
than that of denominator, so there will be a hole in the graph
on the X-axis at x = −4.
Theory of Equation
Exp. 7) Find the vertical asymptotes and holes of
x( x + 5) 4 ( x − 2) 6
.
( x + 5) 6 ( x − 2) 3 ( x − 5)
Solution. First of all cancel out all the common factors between
the numerator and denominator.
x( x − 2) 3
Then the simplified rational function is
( x + 5) 2 ( x − 5)
Now, since ( x + 5) 2 ( x − 5) = 0 ⇒ x = −5 or x = 5.
Therefore there are two vertical asymptotes, x = −5 and x = 5.
However, since the power of ( x − 2) in the numerator is higher
than that of denominator, so there will be a hole in the graph
on the X-axis at x = 2.
Exp. 8) Find the vertical asymptotes and holes of
x 2 ( x + 7) 2 ( x − 2) 3
.
x( x 2 + 7)
Solution. First of all cancel out all the common factors between
the numerator and denominator.
x( x + 7) 2 ( x − 2) 3
Then the simplified rational function is
x( x 2 + 7)
Now, since x 2 + 7 = 0 ⇒ x = ± −7 , which is not a real
solution.
Therefore there is no vertical asymptote.
Looking at the original function we can say that the power of x
in the numerator is higher than that of denominator, so there
will be a hole in the graph on the X-axis at x = 0.
Exp. 9) Find the vertical asymptotes and holes of
( x − 10)( x − 20) 2 ( x − 30) 3 ( x − 40)
.
( x − 10) 3 ( x − 20) 2 ( x − 30)( x − 50)
Solution. First of all cancel out all the common factors between
the numerator and denominator.
( x − 30) 2 ( x − 40)
Then the simplified rational function is
( x − 10) 2 ( x − 50)
Now since ( x − 10) 2 ( x − 50) = 0
⇒
x = 10 or x = 50.
Therefore, there are two vertical asymptotes, x = 10and x = 50.
Looking at the original function, we can say that the power of
(x − 30) in the numerator is higher than that of denominator,
so there will be a hole in the graph on the X −axis at x = 30.
Again since the power of (x − 20) in the numerator is same as
that of denominator, so there will be a hole in the graph at
x = 20, but not on the X-axis.
The y-value of the hole can be found by canceling out the
common factors and substituting x = 20 in the reduced
function as following.
( x − 30) 2 ( x − 40) ( 20 − 30) 2 ( 20 − 40) 2
=
y=
( x − 10) 2 ( x − 50)
( 20 − 10) 2 ( 20 − 50)
2
= = 0.67
3
813
Non-Vertical Asymptotes
(or End Behaviour Asymptotes)
Since the non-vertical asymptotes determines the end
behaviour or long term behaviour of a rational function,
that’s why they are called as end behaviour asymptotes.
These asymptotes give the glimpse of what a rational graph
look like for higher values of x or in short these asymptotes
give you a big picture of a rational function.
1. There are three types of End-behaviour asymptotes:
Horizontal, Slant and Curvilinear.
2. A rational polynomial function may have at most one
non-vertical asymptote.
3. The graph of the rational polynomial may intersect the
non-vertical asymptote.
(a) Horizontal Asymptotes
The graph of a rational function may also have horizontal
asymptote. To determine the horizontal asymptotes look for
the highest power in the numerator ( i, e., n) and denominator
( i, e., m) of the rational function.
1. If the degree of the numerator is higher than that of
denominator ( i, e., n > m), then there is no
horizontal asymptote.
2. If the degree of the numerator is less than the degree
of denominator ( i. e., n < m), then the X -axis is the
horizontal asymptote. That means the horizontal
asymptote is y = 0
3. If the degree of the numerator and the denominator
is same ( i. e., n = m), then the horizontal asymptote
is given by the following formula.
 the coefficient of the highest power 


in the numerator

y=
 the coefficient of the highest power in 


the denominator


That means if the rational Function is
P ( x ) a n x n + ... + a1 x + a 0
f (x ) =
=
q ( x ) bm x m + ... + b1 x + b0
Then the horizontal asymptote is y =
an
bm
Exp. 1) Find the horizontal asymptote of
Solution.
( x + 3)( x − 7) x 2 − 4x − 21
.
=
( 2x − 5)
2x − 5
( x + 3)( x − 7)
( 2x − 5)
Since n = 2 and m = 1. It implies that n > m. So there won’t be
any horizontal asymptote.
814
QUANTUM
Exp. 2) Find the horizontal asymptote of
Exp. 2) Find the slant asymptote of
( 2 x + 9)(5 x − 4)
( x − 3)( x − 5)( 2 x − 11)
Solution.
( 2x + 9)(5 x − 4)
10x 2 + 37 x − 36
= 3
( x − 3)( x − 5)( 2x − 11) 2x − 27 x 2 + 118x − 165
Since n = 2 and m = 3. It implies that n < m. Then X-axis
(i. e., y = 0) is the horizontal asymptote.
Exp. 3) Find all the vertical asymptotes of
( x − 2) 2 ( x + 2)( x + 1)( x − 3)
x 3 ( x − 3)( 3 x + 8)
Solution
( x − 2) 2 ( x + 2) ( x + 1) ( x − 3) 7 x5 + ...
=
3 x5 + ...
x 3 ( x − 3) ( 3 x + 8)
Since n = 5 and m = 5. It implies that n = m.
7
Then the horizontal asymptote is y = .
3
Exp. 4) Find The horizontal asymptote of
( 2 x − 3) 2 (5 x − 4)
x( 3 x − 5)(7 x − 3)
Solution
( 2x − 3) 2 (5 x − 4) 20x 3 + ...
=
x( 3 x − 5)(7 x − 3) 21x 3 + ...
Since n = 3 and m = 3. it implies that n = m
20
Then the horizontal asymptote is y =
21
(b) Slant (or Oblique) Asymptotes
The graph of a rational function may also have a slant
asymptote. When the degree of the numerator is exactly one
more than the degree of the denominator (i.e. n = m +1),
the graph of the rational function will have an oblique
asymptote.
To find the equation of the oblique asymptote, perform long
division (or synthetic division if it works).
As x gets very large (this is the far left or far right position),
the remainder portion becomes very small, almost zero. So,
to find the equation of the oblique asymptote, perform the
long division and discard the remainder. The slant asymptote
is y = k ( x ); where p ( x ) = q ( x ) k ( x ) + r ( x ). Here k ( x ) is a
linear function.
Exp. 1) Find the slant asymptote of ( x + 3)( x + 7)
( x − 5)
Solution
( x + 3)( x + 7) x 2 + 10x + 21
=
( x − 5)
x −5
Since n = 2 and m = 1. It implies that n = m + 1. So a slant
asymptote is possible. After dividing the numerator by
denominator you will get quotient k ( x) = x + 15. Therefore
the slant asymptote is y = x + 15.
Hint x 2 + 10x + 21 = ( x − 5)( x + 15) + 96
CAT
( x + 3)( x − 7 )( 2 x + 3) 2
( x − 5) 2 ( 2 x + 3)
Solution
( x + 3)( x − 7)( 2x + 3) 2 ( x + 3)( x − 7)( 2x + 3)
=
( x − 5) 2
( x − 5) 2 ( 2x + 3)
2x 3 − 5 x 2 − 54x − 63
x 2 − 10x + 25
Since n = 3 and m = 2. It implies that n = m + 1. So a slant
asymptote is possible. After dividing the numerator by
denominator you will get quotient k( x) = 2x + 15 , Therefore
the slant asymptote is y = 2x + 15
Hint 2x3 − 5x2 − 54x − 63 = ( x2 − 10x + 25)( 2x + 15) + ( 46x − 438)
=
(c) Curvilinear Asymptotes
The graph of a rational function may also have a curvilinear
asymptote. When the degree of the numerator is higher than
the degree of the denominator and the difference in the
degree is more than one (i.e. n > m +1), the graph of the
rational function will have a curvilinear asymptote.
To find the equation of the curvilinear asymptote, perform
long division (or synthetic division if it works). As x gets
very large (this is the far left or far right position), the
remainder portion becomes very small, almost zero. So, to
find the equation of the curvilinear asymptote, perform the
long division and discard the remainder.
The
slant
asymptote
is
;
where
y = k (x )
p ( x ) = q ( x ) k ( x ) + r ( x ). Here k ( x ) is a polynomial function
of at least degree 2.
Exp. 1) Find the curvilinear asymptote of
( x − 3)( x + 7 )( x − 8)
( x − 5)
Solution
( x + 3)( x + 7)( x − 8) x 3 + 2x 2 − 59x − 168
=
( x − 5)
( x − 5)
−288
= ( x 2 + 7 x − 24) +
( x − 5)
Since n = 3 and m = 1, It implies that n > m + 1. So a curvilinear
asymptote is possible. After dividing the numerator by
denominator you will get quotient k ( x) = x 2 + 7 x − 24.
Therefore the curvilinear asymptote is y = x2 + 7x − 24.
Hint ( x + 3)( x + 7)( x − 8) = x 3 + 2x 2 − 59x − 168
= ( x − 5)( x 2 + 7 x − 24) − 288.
Exp. 2) Find the curvilinear asymptote of
3 x5 − x 4 + 2 x 2 + x + 1
( x 2 + 1)
Solution
4x − 2
3 x5 − x 4 + 2x 2 + x + 1
.
= ( 3 x 3 − x 2 − 3 x + 3) + 4
2
( x + 1)
( x + 1)
Theory of Equation
815
Since n = 5 and m = 2. It implies that n − m > 1. So a curvilinear
asymptote is possible. After dividing the numerator by
denominator you will get quotient k( x) = 3 x 3 − x 2 − 3 x + 3.
Therefore the curvilinear asymptote is y = 3 x 3 − x 2 − 3 x + 3
Now, we see that for x = −2, 1, 4 the numerator of the function
is zero, but for all these values of x = −2, 1, 4 the denominator
is also zero.
Therefore, we don’t have any x-intercepts for the given
Hint
function.
3x5 − x4 + 2x2 + x + 1 = ( x2 + 1)( 3x3 − x2 − 3x + 3) + ( 4x − 2)
X-intercepts and Y-intercepts
These are the points where the rational polynomial graph
intercepts the X−axis and Y−axis.
x-intercepts
y-intercepts
What to determine
Values of x
Value of y
Favorable condition
at p ( x ) = 0
at x = 0
but q( x ) ≠ 0, for the same x
if q (0) ≠ 0
Restriction
X-intercepts
These occur at the real zeros of the numerator, which are not
the zeros of the denominator too. That is x-intercepts are the
values of x when p( x ) = 0, but q ( x ) ≠ 0 for the same value of
x. Graphically speaking, these are the points on the X−axis
where the rational function intercepts or touches the X −axis.
Look at the given rational function, that is, original function
in which you have not yet cancelled the common factors, if any,
between numerator and denominator. So whenever there is a
common factor between numerator and denominator of the
given rational function there will be a discontinuity ( either a
Hole or a Vertical Asymptote) in the graph of the rational
function. And, wherever, there is a Hole or a Vertical
Asymptote there cannot be any x-intercept. Therefore, while
finding the x-intercepts we need to be careful about the
discontinuity of the graph. If there are holes and asymptotes
coinciding with x-intercepts, please discard these holes and
asymptotes, as they cannot be considered as the x-intercepts.
( x + 3)( x − 7)
Exp. 1) Find all the x −intercepts of
.
( 2x − 5)
Solution Fox x −intercepts p ( x) = 0
That means ( x + 3)( x − 7) = 0
⇒
( x + 3) = 0 or ( x − 7) = 0 ⇒ x = −3 or x = 7
Therefore there are two x −intercepts namely −3 and 7.
Exp. 2) Find all the x −intercepts of
f ( x) =
( x + 2)( x − 1)( x − 4) 2
( x + 2) 2 ( x − 1)( x − 4)
( x + 2)( x − 1)( x − 4) 2
( x + 2) 2 ( x − 1)( x − 4)
p( x)
For x-intercepts, f ( x) =
= 0, such that p( x) = 0. but for the
q( x)
same values of x q( x) ≠ 0.
Solution Given that f ( x) =
NOTE Since the power of (x + 2) in the numerator is less than
that of denominator, therefore at x = −2, there is a vertical
asymptote.
Since the power of (x − 4) in the numerator is higher than that
of denominator, therefore at x = 4, there is a hole.
Since the power of ( x − 1) in the numerator is same as that of
denominator, so there will be a hole in the graph at x = 1, but
not on the X-axis.
The Y-value of the hole can be found by canceling the factors
and substituting x = 1 in the reduced function as following.
( x − 4)
y=
( x + 2)
(1 − 4)
=
(1 + 2)
−3
=
= −1
3
Exp. 3) Find all the x- intercepts of
Solution For x-intercepts p ( x) = 0
( x 2 + 2)( x 4 + 4)
( 2x − 25)
That means ( x 2 + 2)( x 4 + 4) = 0
⇒
( x 2 + 2) = 0 or ( x 4 + 4) = 0
Now you can see that there are no real roots in the numerator.
Therefore there are no x-intercepts given by this function.
Exp. 4) Find all the x-intercepts of
f ( x) =
Solution Given that f ( x) =
⇒
( x − 3) 4 ( x − 6) 4 ( x − 4)(5 x − 8)
.
( 2 x − 5)( x − 3)( x − 6) 2
( x − 3) 4 ( x − 6) 4 ( x − 4)(5 x − 8)
( 2x − 5)( x − 3)( x − 6) 2
f ( x) =
( x − 3) 3 ( x − 6) 2 ( x − 4)(5 x − 8)
( 2x − 5)
For x-intercepts p ( x) = 0
That means ( x − 3) 3 ( x − 6) 2 ( x − 4)(5 x − 8) = 0
⇒
( x − 3) 3 = 0 or ( x − 6) 2 = 0
or
( x − 4) or (5 x − 8) = 0
⇒
x = 3 or x = 6
or
x = 4 or x = 8 / 5
But x = 3 and x = 6 cannot be the x-intercepts, as when x = 3
and x = 6 the denominator of the given rational function
becomes 0 (zero). Essentially, at x = 3 and x = 6, there are
holes on the X-axis.
8
Thus there are only two x-intercepts namely 4 and .
5
816
QUANTUM
Y-intercept
This is the value of f (0), if defined. Graphically speaking,
this is the point on the Y-axis where the rational function
intercepts the Y-axis.
Exp. 1) Find the y-intercept of
So the quotient polynomial is k ( x) = x. That is y = x is a slant
asymptote. The only zero (root) of the numerator is 0, as
f ( 0) = 0 and thus we see that the point (0,0) is the only
x-intercept and the y-intercept of the graph of f ( x).
( x + 3)( x − 7)
( 2x − 5)
( 0 + 3) × ( 0 − 7 ) 21
Solution For y −intercept, f ( 0) =
=
( 2 × 0 − 5)
5
–20
Therefore the y −intercept is 21/5.
Exp. 2) Find the y −intercept of
Solution For y-intercept f ( 0) =
( x + 2)( x + 1)( x − 3)
x 2 ( x − 3)( 3x + 8)
That means the given function is not defined at x = 0.
Therefore there is no y-intercept for this rational function.
Exp. 3) Find the y -intercept of
x 2 ( x + 7) 2 ( x − 2) 3
x( x 2 + 7)
.
0( 0 + 7) 2 × ( 0 − 2) 3
0( 0 + 7)
0 × ( 49)( −8) 0
=
=
0 × (7)
0
Solution For Y-intercept f ( 0) =
That means the given function is not defined at x = 0
Therefore there is no Y-intercept for this, rational function
Graphing the Rational Functions
To sketch the graph of a rational function you may need to
follow the ensuing pointers in order to keep it infallible and
glitch free.
1. Find the domain
2. Find the Vertical Asymptotes and holes
3. Find the Horizontal/Slant/Curvilinear asymptote
4. Find the X-intercepts and Y-intercepts
5. Find the points where graph intercepts the non-vertical
asymptote
6. Finally determine where the graph will be positive or
negative in different intervals to draw the graph then
you can easily sketch it.
Exp. 1) Find the asymptotes, intercepts and sketch the
graph of the function f ( x) =
–15
–10
–5
.
( 0 + 2)( 0 + 1)( 0 − 3)
−6
=
0
( 0) 2 ( 0 − 3)( 3 × 0 + 8)
x3
x2 − 9
Solution. Factoring the denominator, you will get the roots of the
denominator −3 and 3. Consequently, x = −3 and x = 3 are the
vertical asymptotes of the given function f ( x).
The degree of the numerator is greater than the degree of the
denominator by 1, so there will be a slant asymptote. Using
polynomial long division, we obtain
9x
x3
=x+ 2
f ( x) = 2
x −9
x −9
CAT
30
25
20
15
10
5
0
–5
–10
–15
–20
–25
–30
5
10
15
20
.
The graph of f ( x) in the above figure passes through (0,0)
where x = −3 and x = 3 are the vertical asymptotes and the
slant asymptote is y = x.
Exp. 2) Find the asymptotes, intercepts and sketch the
graph of the function f ( x) =
2x + 6
.
x 2 − 2x − 15
Solution The given rational function can be expressed as
following.
2x + 6
2 ( x + 3)
=
f ( x) = 2
x − 2x − 15 ( x + 3)( x − 5)
(i) The Domain: ( −∞ , −3) ∪ ( −3 ,5) ∪ (5 , ∞). Since, at x = −3 and
x = 5, the given function is not defined.
(ii) Vertical Asymptotes/Holes: One asymptote is there at
x = 5. Also, there is one hole in the graph. Since the power
of (x + 3) in the numerator is same as that of denominator,
so there will be a hole in the graph at x = −3, but not on
the X-axis. The y-value of the hole can be found by
canceling out the common factors and substituting x = −3
in the reduced function as following.
2
2
1
y=
=
=
= 0.25
( x − 5) ( −3 − 5) −4
(iii) Non-vertical Asymptote: Since the degree of numerator is
less than that of denominator, therefore there will be a
horizontal asymptote.
Then the horizontal asymptote is the X-axis itself, That is
y = 0.
(iv) X-intercepts: It is the value of x when P( x) = 0 , but
q ( x) ≠ 0, for the same value of x. Therefore, no x-intercept
is there because at x = −3 the function f ( x) is not defined.
(v) Y-intercept: It is the value of y when the value of x = 0.
2( 0 + 2)
6
2
Therefore y-intercept is y =
=−
=− .
( 0 + 3)( 0 − 5)
15
5
 0, − 2 
Thus the coordinates of y −intercept are 
.
 5 
(vi) Points where the graph intercepts the non-vertical
asymptote: Since the horizontal asymptote is X-axis itself.
Theory of Equation
817
So the horizontal asymptote overlaps the X-axis. Now,
since there are no x-intercepts, so there is no such point
where graph intercepts the horizontal asymptote.
6
5
4
3
2
1
–6
–5
–4
–2 –1 0
1
2
3
4
5
6
7
8
9
10
11
(iv) X-intercepts: It is the value of x when p ( x) = 0, but
q( x) ≠ 0, for the same value of x. Therefore x-intercept is
obtained at x = −2 . Thus the coordinates of x-intercept are
{−2, 0}.
(v) Y-intercept: It is the value of y when the value of x = 0.
Therefore y-intercept is y = ( 0 + 2) = 2. Thus the
coordinates of y-intercept are {0, 2}.
(vi) Points where the graph intercepts the non-vertical
asymptote: Since the reduced function is same as the
equation for slant asymptote, it implies that the graph of
the given function will overlap the slant asymptote.
–1
–2
–3
4
–4
–5
3
–6
2
The above graph manifests that when x = −3 there is a
hole in the graph at y = − 0.25 and being a vertical
asymptote at x = 5, the graph approaches the Y-axis but
does not intercept it.
For, all the values of x < 5, all the values of y < 0 and for all
the values of x > 5, all the values of y > 0.
1
–6 –5
–4
–3
–2
–1
f ( x) =
x 2 − x − 6 ( x + 2)( x − 3)
=
= ( x + 2)
x−3
( x − 3)
Since the power of ( x − 3) in the numerator is same as that
of denominator, so there will be a hole in the graph at
x = 3, but not on the X-axis. The value of the hole can be
found by canceling out the common factors and
substituting x = 3 in the reduced function as following.
y = ( x + 2) = ( 3 + 2) = 5
(iii) Non-vertical Asymptote: Since the degree of numerator is
one more than the denominator, therefore the graph will
experience a slant asymptote, not the horizontal one.
The value of the slant asymptote is equal to the quotient of
the function when numerator is divided by the
denominator. So, the quotient = ( x + 2). Therefore, the
slant asymptote is also y = ( x + 2).
2
3
4
5
–3
x2 − x − 6
.
graph of the function f ( x) =
x− 3
(i) The Domain: ( −∞ , 3) ∪ ( 3 , ∞). Since, at x = 3, the function
is not defined.
(ii) Vertical Asymptotes/Holes: There is no vertical
asymptote since after canceling out the common factors,
there is no factor at all in the denominator.
1
–2
Exp. 3) Find the asymptotes, intercepts and sketch the
Solution The given rational function can be expressed as
following.
x 2 − x − 6 ( x + 2)( x − 3)
f ( x) =
=
x−3
( x − 3)
0
–1
–4
–5
The only difference between the slant asymptote of the
rational function and the rational function itself is that the
rational function isn't defined at x = 3. To account for this,
you have to leave an open circle in the graph
corresponding to the point x = 3, which indicates that this
point is not actually included on the graph, because of the
zero in the denominator of the rational.
Exp. 4) Find the asymptotes, intercepts and sketch the
graph of the function f ( x) =
x 3 − 4x
.
x − 4x + 4
2
Solution The given rational function can be expressed as
following.
x 3 − 4x
x( x + 2)( x − 2)
=
f ( x) = 2
( x − 2) 2
x − 4x + 4
(i) The domain: ( −∞ , 2) ∪ ( 2, ∞). Since, at x = 2, the given
function is not defined.
(ii) Vertical Asymptotes/Holes: There is one vertical
asymptote x = 2
(iii) Non-vertical Asymptote: Since the degree of numerator is
one more than the denominator, therefore the graph will
experience a slant asymptote, not the horizontal one.
818
QUANTUM
The value of the slant asymptote is equal to the quotient of
the function when numerator is divided by the
denominator. Since
x 3 − 4x = ( x 2 − 4x + 4)( x + 4) + ( 8x − 16),
so, the quotient is ( x + 4). Therefore, the slant asymptote is
also y = (x + 4).
(iv) X-intercepts: It is the value of x when p ( x) = 0, but
q ( x) ≠ 0, for the same value of x. Therefore x-intercept is
obtained at x = 0 and x = −2. Thus the coordinates of
x-intercept are {0,0} and {−2, 0}.
(v) Y-intercept: It is the value of y when the value of x = 0.
Therefore y-intercept is
y = [0( 0 + 2)( 0 − 2)] / [( 0 − 2)( 0 − 2) = 0.
Thus the coordinates of y-intercept are {0,0}.
(vi) Points where the graph intercepts the non-vertical
asymptote: To determine such points you need to equate
the given function with the equation of slant asymptote.
x 3 − 4x
Then 2
=x+4
x − 4x + 4
Since you don't get any solution here, it shows that the
graph never intersects the slant asymptote.
60
50
40
(v) Y-intercept: It is the value of y when the value of x = 0.
Therefore y-intercept is y = 9 / 9 = 1. Thus the coordinates
of y-intercept are {0, 1}.
(vi) Points where the graph intercepts the non-vertical
asymptote: To determine such points you need to equate
the given function with the equation of horizontal
x2 + 9
asymptote. Then 2
=1⇒ x = 0
x − 6x + 9
Therefore the graph of the given function will intercept
the horizontal asymptote at {0,1}.
7
6
5
4
3
2
1
-30
–8
–4
0
–10
–20
0
10
20
Please mind the fact that this rational function has two
non-real roots and no real roots. That's why in the absence
of any real root the graph does not have any x-intercepts.
graph of the function f ( x) =
4
8
12
16
20
–30
–40
Exp. 5) Find the asymptotes, intercepts and sketch the
graph of the function f ( x) =
-10
-1
10
–12
-20
Exp. 6) Find the asymptotes, intercepts and sketch the
30
20
–16
CAT
x2 + 9
x 2 − 6x + 9
Solution The given rational function can be expressed as
x2 + 9
x2 + 9
following f ( x) = 2
=
x − 6x + 9 ( x − 3) 2
(i) The domain: ( −∞ , 3) ∪ ( 3 , ∞). Since, at x = 3, the given
function is not defined.
(ii) Vertical Asymptotes/Holes: There is one vertical
asymptote x = 3.
(iii) Non-vertical Asymptote: Since the degree of numerator
and denominator is same, therefore there will be a
horizontal asymptote y = 1.
(iv) X-intercepts: It is the value of x when p ( x) = 0, but q ( x) ≠ 0,
for the same value of x. Now, when x 2 + 9 = 0, you won't get
the real roots, so there won't be any x-intercept by this graph.
x2 − 4
.
x 3 − 4x 2
Solution The given rational function can be expressed as
( x + 2)( x − 2)
x2 − 4
following f ( x) = 3
=
2
x − 4x
x 2 ( x − 4)
(i) The domain: ( −∞ , 0) ∪ ( 0, 4) ∪ ( 4, ∞) . Since, at x = 0, 4 the
given function is not defined.
(ii) Vertical Asymptotes/Holes: There are two vertical
asymptotes x = 0 and x = 4.
(iii) Non-vertical Asymptote: Since the degree of numerator is
less than that of denominator, therefore there will be a
horizontal asymptote y = 0.
(iv) X-intercepts: It is the value of x when p ( x) = 0, but
q ( x) ≠ 0, for the same value of x . Therefore x-intercept is
obtained at x = −2 and x = 2 . Thus the coordinates of
X-intercept are {−2, 0} and {2, 0}.
(v) Y-intercept: It is the value of y when the value of x = 0.
Since at x = 0, the function is not defined, therefore there
won't be any y-intercept.
(vi) Points where the graph intercepts the non-vertical
asymptote: To determine such points you need to equate
the given function with the equation of horizontal
asymptote.
Then
x2 − 4
=0⇒x =± 2
x 3 − 4x 2
Theory of Equation
819
1.4
1.2
1
0.8
0.6
0.4
0.2
–14 –12 –10 –8 –6 –4 –2 0
–0.2
2
6
8
10 12 14 16 18 20
–0.4
Therefore x-intercept is obtained at x = −3 , x = −7 and
x = 8. Thus the coordinates of x-intercept are {−3 , 0},{−7 , 0}
and {8, 0}.
(v) Y-intercept: It is the value of y when the value of x = 0.
Therefore the y-intercept is 168/5. And so the respective
coordinates are {0, 168/5}.
(vi) Points where the graph intercepts the horizontal/slant/
curvilinear asymptote: To determine such points you need
to equate the given function with the equation of
horizontal asymptote.
x 3 + 2x 2 − 59x − 168
= x 2 + 7 x − 24
x −5
There is no solution for the above equation. Therefore the
graph of the given function does not intercept the
curvilinear asymptote.
Then
–0.6
–0.8
–1
–1.2
–1.4
400
Therefore the graph of the given function will intercept
the horizontal asymptote at {0, −2} and {0, 2}.
Exp. 7) Find the asymptotes, intercepts and sketch the
graph of the function f ( x) =
300
x + 2x − 59x − 168
x −5
3
2
200
Solution The given rational function can be expressed as
following
x 3 + 2x 2 − 59x − 168 ( x + 3)( x + 7)( x − 8)
f ( x) =
=
x −5
( x − 5)
(i) The domain: ( −∞ ,5) ∪ (5 , ∞). Since, at x = 5 the given
function is not defined.
(ii) Vertical Asymptotes/Holes: There is one vertical
asymptote x = 5.
(iii) Non-vertical Asymptote: Since the degree of numerator is
2 more than that of denominator, therefore there will be a
curvilinear asymptote. Now since
f ( x) =
x 3 + 2x 2 − 59x − 168
−288
= ( x 2 + 7 x − 24) +
x −5
( x − 5)
100
–30
–20
–10
0
–100
–200
Therefore the curvilinear asymptote is y = x 2 + 7 x − 24.
(iv) X-intercepts: It is the value of x when p ( x) = 0, but
q ( x) ≠ 0, for the same value of x .
–300
10
20
30
820
QUANTUM
CAT
Practice Exercise
1. Which of the following is not true about the features of
a rational function?
(i) Can be written as a polynomial over polynomial
(like a fraction)
(ii) Variables do not have fractional or imaginary
exponents
(iii) A variable must at least be in the denominator of
a ratio
(iv) Variables cannot be exponents themselves
(v) Variables cannot exist within absolute value delimiters
(vi) Variables cannot be applied to a trigonometric
function
(a) (i) and (iii)
(b) (iv) and (vi)
(c) (ii) and (v)
(d) none of these
2. Which of the following is not true regarding a rational
function?
(a) The equation of vertical asymptote is x = k, for
some constant k
(b) The equation of horizontal asymptote is y = k, for
some constant k
(c) The equation of slant asymptote is y = mx + c, for
some constants m and c
(d) There can't be both the vertical and horizontal
asymptotes together
3. Which of the following is not true regarding a rational
function?
(a) There can be only one of the asymptotes, either
horizontal or slant or curvilinear at a time
(b) It's not necessary that there is always a
non-vertical asymptote
(c) There cannot be more than one non-vertical
asymptote
(d) The rational graph cannot intersect a non-vertical
asymptote
4. Which of the following is true for the following rational
function?
8 x7 − 32 x 6 + 3 x5 − 12 x 4 − 10 x + 40
f (x) =
2 x2 − 32
(a) There is one vertical asymptote at x = −4
(b) The horizontal asymptote is y = 4.
(c) The slant asymptote is y = 7 /2.
(d) There are two vertical asymptotes.
5. Which of the following is correct for the rational
function
3 x2 + 27 x + 25
f (x) =
3x + 4
(a) The horizontal asymptote is y = 0
(b) The slant asymptote is y − 1 = 0
(c) The vertical asymptote is y = − 4/3
(d) The slant asymptote is x + 7.
6. The number of times the graph intercepts the
horizontal asymptote of the following function.
(x2 − 4 )(x2 − 25 )
f (x) =
x5 (x2 − 4 x)
(a) 0
(b) 1
(c) 4
(d) none of these
7. Which of the following is/are always true regarding a
polynomial rational function?
(i) There is no need to know about the numerator
since y-intercepts can be determined just by
finding all the zeros of denominator.
(ii) All the roots of the numerator yield the
x-intercepts of the rational function.
(iii) In case of the X-axis acting as the horizontal
asymptote and you know the x-intercepts, then
there is no need to further work out the points of
interception between the graph and the
horizontal asymptote as they are already known.
(iv) There is at most one point of interception
between the graph of the rational function and
horizontal/slant/curvilinear asymptote.
(v) Whenever the X-axis acts as a horizontal
asymptote there would not be any x-intercept.
(a) Only (iv)
(b) (ii) and (v)
(c) (i), (ii) and (iv)
(d) Only (iii)
8. Which of the following is/are never true about the
rational function?
(i) Wherever there is no domain there is either a
vertical asymptote or a hole in the graph.
(ii) All the critical points either have x-intercept or
vertical asymptote or a hole in the graph.
(iii) Wherever there is no domain there are vertical
asymptotes and holes together for the same
value of x.
(iv) Union of all the roots of numerator and
denominator forms the set of critical numbers.
(v) A rational graph may intercept or touch the
X-axis at x = k, even when there is a hole on the
X-axis at x = k .
(a) (iv) and (v)
(b) (iii), (iv) and (v)
(c) (i), (ii) and (iii)
(d) (ii) and (iii)
9. For the following rational function which one of the
given statements is/are incorrect?
(x − 1)(x − 3 )2 (x − 7 )3
f (x) =
(x − 1)3 (x − 3 )2 (x − 7 )
(i) Domain does not include x = 1, 3 and 7.
(ii) Critical points are x = 1, 3 , 7
(iii) The only vertical asymptote is x = 1
Theory of Equation
821
(iv) There is one hole in the graph at x = 7 and
another hole is at {x, y} = {3, 4}
11. There are some facts presented in the table regarding
three different rational functions.
(v) There is only one root or one x-intercepts at x = 7.
(vi) There is one horizontal asymptote y = 1.
Critical points
(vii) The y-intercept is y = 49.
(a) (iii) and (iv)
f (x )
g (x )
h (x )
(i) Domain
R − { 0}
R − { 0}
R − { 0}
(ii) Critical
points
0
0
0
No
No
x=0
(iv) Holes
(v) Roots
(vi) x-intercepts
2/3, 3/5
−9, 0, 1, 6
R − {−5 / 2,−5 / 3}
R − { 2 / 3}
R − {−9, 0, 1}
x = −5 / 2
No
x = {−9, 1}
Holes
x = −5 / 3
x=0
{x , y }
j (x ) k (x )
R
R R − {−2,2}
−2, 2 No
No
l (x )
−2, 2
No x = − 2, 2
±4 i
Roots
3/5
6
X-intercepts
No
3/5
6
Horizontal
asymptote
y=0
y = 125
No
Slant asymptote
No
No
No
Curvilinear
asymptote
No
No
y = x 2 − 5x − 14
Y-intercept
4
3/2
No
Which of the following functions represent the given
description correctly?
(2 x + 5 )(3 x + 5 )2 (x2 + 16 )
(a) f (x) =
,
(3 x + 5 )(2 x + 5 )3 (x 4 + 64 )
(3 x − 2 ) (5 x − 3 )5
x7 − 7 x 6 + 6 x5
,h (x) = 4
g (x) =
4
(3 x + 34 ) (3 x − 2 )
(x + 9 x3 ) (x − 1)2
(2 x + 5 )2 (3 x + 5 )2 (x2 + 16 )2
,
(3 x + 5 )(2 x + 5 )3 (x 4 + 4 )
(3 x − 2 )3 (5 x − 3 )4
x7 − 7 x 6 + 6 x5
, h (x) = 3
g (x) =
4
(4 x + 4 ) (3 x − 2 )
(x + 9 x2 ) (x − 1)
(b) f (x) =
x = 0 {x , y } = {0,1} No
No
No
(vii) Horizontal
asymptote
No
(viii) Slant
asymptote
y=x
No
No
y=1
No
No
No
y=0
No
No
No
No
−2, 2, No
± 2i
No
−2, 2, No
± 2i
No
No
No
y=1
No
(c) f (x) =
No
No
No
No
x − 4 No
(x) y-intercept
{x , y }
No
No
No
{0,−4} {0,1} {0,−1 / 4}
2
No
(2 x + 5 )(3 x + 5 )2 (x2 + 16 )
,
(3 x + 5 ) (2 x + 5 )2 (x 4 + 4 )
g (x) =
(3 x − 2 )(5 x − 3 )4
,
(5 x 4 + 54 ) (3 x − 2 )
h (x) =
x7 − 7 x 6 + 6 x5
(x3 + 9 x2 ) (x − 1)2
y=0
(ix) Curvilinear
asymptote
(a) 6
(b) 7
(c) 8
(d) 10
−5 / 2, − 5 / 3
= {2 /3,1/242}
10. There are six different rational functions as shown below.
x2
x
x
f (x) =
, g (x) = , h (x) = 2
x
x
x
x 4 − 16
x2 + 4
, k(x) = 2
j (x) = 2
x +4
x +4
x2 + 4
l (x) = 4
x − 16
The following table gives the facts from (i) to (x)
regarding the above functions. How many facts are
correct for all the functions?
(iii) Vertical
asymptotes
Function g (x ) Function h (x )
Vertical
asymptotes
Domain
(b) (v) and (vii)
(c) (iv) and (v)
(d) (v)
Facts
Function f (x )
(d) f (x) =
g (x) =
(2 x − 5 )(3 x − 5 )2 (x2 − 16 )
,
(3 x + 5 ) (2 x + 5 )2 (x 4 + 4 )
(3 x + 2 ) (5 x − 3 )4
x7 − 7 x 6 − 6 x5
,
(
)
=
h
x
(5 x 4 + 54 ) (3 x + 2 )
(x3 + 9 x2 ) (x − 1)2
Answers
1. (d)
6. (c)
11. (c)
2 (d)
7. (d)
3. (d)
8. (b)
4. (a)
9. (d)
5. (d)
10. (d)
822
QUANTUM
14.19 Rational Polynomial
Inequalities
There are four different ways of representing the rational
polynomial inequations:
p (x )
p (x )
(i)
(ii)
>0
≥0
q (x )
q (x )
(iii)
p (x )
<0
q (x )
(iv)
p (x )
≤0
q (x )
Here p ( x ) and q ( x ) are polynomials of same or varying
degrees:
Solving Rational Inequalities
1. First of all bring all the values towards the left side of the
inequality sign and leave just 0 in the right side.
2. Now, simplify all the terms in the left side of the
inequality sign making a single rational function.
3. Now what you need to do is to factorize numerator and
denominator, assuming that the numerator and
denominator each are equal to 0; and determine the
critical points. Critical points are all those points which
are either the real roots of the rational function or the
points on the X-axis where the rational function is not
defined.
4. Then mark all the critical points on the number line in
the conventional order, which in turn these critical
points will create intervals.
5. Now test these intervals for signs; which intervals give
positive and which ones give negative values.
6. If your inequality is positive, choose all the positive
intervals and if it is negative inequality, choose the
negative intervals. And then combine all such desired
intervals.
7. Do not forget to consider the limit of the interval; if it's a
strict inequality, do not include limits of the interval and
if it's not a strict inequality, include the limits of the
interval.
NOTE Coincidently, you may notice that a critical point has
alternating signs on the different sides of it. However, it is very
uncommon in the rational functions. So make sure that you have
tested the sign in each interval before concluding the results.
Exp. 1) Find the values of x which satisfy this rational
x +5
inequation
≤0
x−6
Solution. The critical points of this rational function are − 5 and 6.
+
–∞
–
–5
+
–6
+∞
CAT
After arranging the critical points on the number line, we are
now supposed to test the sign in each interval.
x + 5 −10 + 5 5
Let us assume x = −10 , then
=
=
x − 6 −10 − 6 4
Therefore the given function is positive for the first interval.
Again let us assume x = 0, then
x +5 0+5
5
=
=−
x −6 0−6
6
Therefore the given function is negative for the second
x + 5 10 + 5 15
interval. Now let us assume x = 10, then
=
=
x − 6 10 − 6
4
Therefore the given function is positive for the third and last
interval. Since, our inequality is non-positive so we will not
consider the first and the third intervals. Further since it's not
a strict inequality, so we have to include the boundary values
in the interval.
Therefore the required values of x are [−5 , 6) or − 5 ≤ x < 6.
NOTE Since at x = 6, the rational function is undefined so we
cannot say that whether the function will be positive or negative.
Therefore it's better to ignore this boundary value from the interval
consideration. However at x = 5 the given function is zero, which is
acceptable by the inequality so we have to include it.
Exp. 2) Find the values of x which satisfy this rational
inequation
5
−3
.
<
x + 3 x −1
Solution. The given inequality can be simplified as shown
below.
5
5
3
4 ( 2x + 1)
−3
<
+
< 0⇒
⇒
<0
x + 3 x −1 x + 3 x −1
( x + 3) ( x − 1)
1
The critical points of this rational function are − , −3 and 1.
2
–
–∞
–
+
–3
–1/2
+
1
-∞
After arranging the critical points on the number line, we are
now supposed to test the sign in each interval.
Let us assume x = −5, then
4( −10 + 1)
= −3
( −5 + 3) ( −5 − 1)
Therefore the given function is negative for the first interval.
4( −4 + 1)
Again let us assume x = −2, then
=4
( −2 + 3)( −2 − 1)
Therefore the given function is positive for the second
interval.
4 ( 0 + 1)
−4
Now let us assume x = 0, then
=
( 0 + 3) ( 0 − 1)
3
Therefore the given function is negative for the third interval.
Finally let us assume x = 2 ,
4 ( 4 + 1)
then
=4
( 2 + 3) ( 2 − 1)
Therefore the given function is positive for the fourth and last
interval.
Theory of Equation
823
Since, our inequality is negative so we will consider the first
and the third intervals only. Further since it's a strict
inequality, so we have to exclude the boundary values from
the interval.
Therefore the required values of x are
 1 
( −∞ , −3) ∪  − ,1
 2 
Exp. 5) Find the values of x which satisfy this rational
inequation
Solution The given inequality can be simplified as shown below.
x 3 + 4x 2 + 4x
x( x + 2) 2
>
⇒
0
>0
( x − 1)( x − 5)
x 2 − 6x + 5
The critical points of this rational function are −2, 0,1, and 5.
Exp. 3) Find the values of x which satisfy this rational
inequation
–∞
solution The given inequality can be simplified as shown below.
x 2 − 81
≤0
x 2 − 25
( x + 9) ( x − 9)
⇒
≤0
( x + 5) ( x − 5)
The critical points of this rational function are −9, 9, − 5, and 5.
−∞
–
−9
–
+
–5
+
+∞
9
5
Now testing each of the 5 intervals we get the results as shown
in the above number line.
Here we need non-positive intervals so we exclude all the
positive intervals and combine the others keeping in mind the
restriction on limits of the intervals.
Therefore the valid values of x are [−9, −5) ∪ (5 , 9].
NOTE The critical points −5, 5 are not included in the domain
since the function is not defined there and −9 , 9 are included since
they satisfy the given relation.
Exp. 4) Find the values of x which satisfy this rational
inequation
–
x 2 − 81
≤ 0.
x 2 − 25
+
x 2 + 5x + 4
≥ 0.
x 2 − 16
Solution The given inequality can be simplified as shown below.
x2 + 5x + 4
≥0
x 2 − 16
( x + 4) ( x + 1)
≥0
⇒
( x + 4) ( x − 4)
–∞
–
+
–4
–1
+
4
Now testing each of the 4 intervals we get the results as shown
on the above number line.
Here we need non-negative intervals so we exclude all the
negative intervals and combine the others keeping in mind
the restriction on limits of the intervals. Therefore the valid
values of x are
( −∞ , −4) ∪ ( −4, −1] ∪ ( 4, ∞)
NOTE The critical points −4 , 4 are not included in the domain
since the function is not defined there and −1 is included since it
satisfies the given relation.
–
+
0
1
+
+∞
5
Now testing each of the 4 intervals we get the results as shown
on the above number line.
Here we need positive intervals so we exclude all the negative
intervals and combine the others keeping in mind the
restriction on limits of the intervals.
Therefore the valid values of x are ( 0, 1) ∪ (5 , ∞).
NOTE The critical points −1 and 5 are not included in the
domain since the function is not defined there and the critical point 0
is also not included because there is a strict in equality.
Exp. 6) Solve the rational inequality
x − 2 2x − 3
>
x + 2 4x − 1
Solution
below
⇒
or
The given inequality can be simplified as shows
x − 2 2x − 3
>
x + 2 4x − 1
x − 2 2x − 3
−
>0
x + 2 4x − 1
( x − 2)( 4x − 1) − ( x + 2)( 2x − 3)
>0
( x + 2)( 4x − 1)
2( x 2 − 5 x + 4)
>0
( x + 2)( 4x − 1)
2( x − 1)( x − 4)
>0
( x + 2)( 4x − 1)
or
or
The critical points of this rational function are −2,
−
+
−∞
+∞
–
–2
The critical points of this rational function are −4, −1, and 4.
+
x 3 + 4x 2 + 4x
> 0.
x 2 − 6x + 5
−2
+
1/4
+
−
1
1
, 1 and 4.
4
4
+∞
Now testing each of the 5 intervals we get the results as shown
on the above number line.
Here we need positive intervals so we exclude all the negative
intervals and combine the others keeping in mind the
restriction on limits of the intervals.
1 
Therefore the valid values of x are ( −∞ , − 2) ∪  , 1 ∪ ( 4, ∞).
4 
NOTE The critical points −2 and 1/4 are not included in the
domain since the function is not defined there. And, the critical point
1 and 4 are also not included because there is a strict inequality.
824
QUANTUM
Exp. 7) Determine the range of values of x for which
x 2 − 2x + 5 1
> .
3x 2 − 2x − 5 2
Solution The given inequality can be simplified as shown
below
1
1
x 2 − 2x + 5
x 2 − 2x + 5
or
>
− >0
2
3 x − 2x − 5 2
3 x 2 − 2x − 5 2
or
or
+
−
−5
+
−1
5/3
5
and 3.
3
+
−
+∞
3
Now testing each of the 5 intervals we get the results as shown
on the above number line.
Here we need the negative intervals so we exclude all the
positive intervals and combine the others keeping in mind the
restriction on limits of the intervals. Therefore the valid values
5 
of x are ( −5 , − 1) ∪  , 3
3 
NOTE The critical points–1 and 5/3 are not included in the
domain since the function is not defined there. And, the critical point
−5 and 3 are also not included because there is a strict inequality.
Exp. 8) Find all the real values of x for which
x−1
x− 3
.
<
( 4 x + 5) 4 x − 3
Solution The given inequality can be simplified as shown
x −1
x−3
x −1
x−3
below
<
−
⇒
<0
4x + 5 4x − 3
4x + 5 4x − 3
( x − 1)( 4x − 3) − ( x − 3)( 4x + 5)
⇒
<0
( 4x − 3)( 4x + 5)
or
or
−∞
−5/4
2.
x2 − 16
<0
(x − 1)2
2. ( −4 , 1) ∪ (1 , 4)
14.20
Maximum and Minimum Value
of a Rational Expression
Exp. 1) For any real value of x, find the minimum value
x2 + x +1
of the rational expression 2
.
x − x +1
(a) −1 / 6
(c) 2/9
(b) 1/3
(d) 1
x2 + x + 1
=k
x2 − x + 1
⇒ x 2 + x + 1 = k ( x 2 − x + 1) ⇒(1 − k) x 2 + (1 + k) x + (1 − k) = 0
Solution Let
Since x is a real number, so the discriminant D ≥ 0. That is
(1 + k) 2 − 4(1 − k)(1 − k) ≥ 0
⇒
⇒
−3 k 2 + 10k − 3 ≥ 0 ⇒ 3 k 2 − 10k + 3 ≤ 0
( k − 3) ( 3 k − 1) ≤ 0
1
⇒
≤k≤3
3
Hence the minimum value of the given expression is 1/3.
Exp. 2) If x be real, find the maximum value of
( x + 2)
.
2
( 2x + 3x + 6)
x+2
2x + 3 x + 6
2
2x 2 y + ( 3 y − 1) x + 6y − 2 = 0
For x to be real, ( 3 y − 1) 2 − 8y( 6y − 2) ≥ 0
+
3/4
x2 + 4 x + 3
>0
x −1
x−8
3.
+x≤3
x
1.
then
The critical points of this rational function are −5 / 4 and 3 / 4.
−
the following equations are satisfied.
Solution Let y =
( 4x 2 − 7 x + 3) − ( 4x 2 − 7 x − 15)
<0
( 4x + 5)( 4x − 3)
18
<0
( 4x + 5)( 4x − 3)
+
Directions (for Q. Nos. 1 to 3) Find the values of x for which of
1. ( −3 , −1) ∪ (1 , ∞)
3. ( −∞ , 2] ∪ ( 0 , 4]
The critical points of this rational function are −5 , − 1,
−∞
Practice Exercise
Answers
x 2 + 2x − 15
3 x 2 − 2x − 5
− x 2 − 2x + 15
>0 ⇒
3 x 2 − 2x − 5
( x + 5) ( x − 3)
<0
( x + 1)( 3 x − 5)
CAT
or (1 + 13 y)(1 − 3 y) ≥ 0 or (13 y + 1)( 3 y − 1) ≤ 0
Putting each factor equal to zero, we get y = −
+∞
Now testing each of the 3 intervals we get the results as shown
on the above number line.
Here we need the negative intervals so we exclude all the
positive intervals and combine the others keeping in mind the
restriction on limits of the intervals.
 5 3
Therefore the valid values of x are  − ,  .
 4 4
NOTE The critical points −5/ 4 and 3/ 4 are not included in the
domain since the function is not defined there.
1 1
,
13 3
1
, (1 + 13 y)(1 − 3 y) < 0
13
1
1
If
−
< y < , (1 + 13 y)(1 − 3 y) > 0
13
3
1
If
y> ,
(1 + 13 y)(1 − 3 y) < 0
3
1
1
Thus, y will lie between − and .
3
13
1
1
Hence the maximum value of y is and minimum value is − .
3
13
If
y<−
Theory of Equation
825
Exp. 3) If x is real, prove that the value of the expression
( x − 1)( x + 3)
4
cannot be between and 1.
( x − 2)( x + 4)
9
Solution Let
( x − 1)( x + 3)
= y, then x2 + 2x − 3 = y( x2 + 2x − 8)
( x − 2)( x + 4)
∴ we get y =
x 2 (1 − y) + 2x(1 − y) + ( 8y − 3) = 0
or
4
,
9
4

(1 − y)  − y > 0
9

4
< y < 1,
9
4

(1 − y)  − y < 0
9

If y <
If
4
,1
9
4

(1 − y)  − y > 0
9

If y > 1
For x to be real, its discriminant D ≥ 0
or
[2(1 − y)]2 − 4(1 − y)( 8y − 3) ≥ 0

4
or (1 − y)( 4 − 9y) ≥ 0 or (1 − y)  − y ≥ 0

9
It shows that(1 − y) and( 4/ 9 − y) should be of the same sign.
Hence for x to be real, y,(i.e., the value of the given expression)
4
cannot lie between and 1.
9
Introductory Exercise 14.1
Directions (for Q. Nos. 1 to 12)
Find the roots of the following
9.
equations.
1. 15 x2 − 7 x − 36 = 0 :
5
4
(a) , −
9
3
9
3
(c) , −
5
4
2. 7 y2 − 6 y − 13 7 = 0 :
(a) 7 , 2 7
(a) 2
(b)
10.
9
4
,−
5
3
13
(d) none of these
,− 7
7
3. 6 x2 + 40 = 31x :
3 2
3 3
8 5
8
(b) ,
(c) 0 ,
(d) ,
(a) ,
8 5
8 2
3 2
3
 y − 3
y−3
1
4. 6 
,y ≠ :
 +1=5
2y + 1
2
 2 y + 1
11
13
(a) 4,
(b) 4 ,
2
2
13 9
(c)
(d) none of these
,
2 2
5. (x + 2 )(x − 5 )(x − 6 )(x + 1) = 144 :
x+5 +
(a) 1
(b) 7 , − 3 , 2
(d) none of these
x + 21 =
6 x + 40 :
(b) 2
(c) 4
1
1


7. 2  x2 + 2  − 3  x +  − 1 = 0 :


x
x 
3
1
(a) 2,
(b) −2 , 4
(c) 2 ,
2
2
8.
x
+
1−x
(a)
13
,5
4
1−x
x
(b)
=
(c)
4
,6
13
(d) 5
x − 12 x + 32 = 2 x − 25 x + 68 :
(b) 3 , 9
2
(c) 2, 4
(d) 2, 6
2
1
1


11.  x −  + 8  x +  = 29, x ≠ 0 :



x
x
(b) 3, 2
3 + 5 −11 ± 3 13
(c) 1, 3
(d)
,
2
2
 1
1   1
1 
12. 
+
+
 =
 :
x + 4
 x + 1 x + 5  x + 2
(a) 7
(b) −3
(c) −5
(d) 4
13. Determine k such that the quadratic equation
x2 + 7 (3 + 2 k) − 2 x(1 + 3 k) = 0 has equal roots :
(a) 2, 7
(b) 7, 5
−10
(d) none of these
(c) 2 ,
9
14. Discriminant of the equation −3 x2 + 2 x − 8 = 0 is :
(a) −92
(b) −29
(c) 39
(d) 49
15. The nature of the roots of the equation x2 − 5x + 7 = 0 is
(a) no real roots
(c) can’t be determined
(b) 1 real root
(d) none of these
16. The roots of a2x2 + abx = b2 , a ≠ 0 are :
(d) 6
(a) equal
(c) unequal
(b) non-real
(d) none of these
17. The equation x2 − px + q = 0 ,p, q ∈ R has no real roots
(d) none
13
:
6
4 9
,
13 13
x − 9 x + 20 −
(c) 4
2
(a) −5 , 4
2
7
(c)
6.
(b) 3
2
(a) 4 , 9
(d) none of these
(b) 3 ,
(a) −1, − 2 , − 3
(c) 2 , − 3 , 5
2 x2 − 2 x + 1 − 2 x + 3 = 0 :
(d) none
if:
(a) p2 > 4q
(c) p2 = 4q
(b) p2 < 4q
(d) none of these
18. Determine the value of k for which the quadratic
equation 4 x2 − 3 kx + 1 = 0 has equal roots :
 2
 4
(a) ±  
(b) ±  
(c) ± 4
(d) ± 6
 3
 3
826
QUANTUM
19. Find the value of k such that the equation
x2 − (k + 6 )x + 2 (2 k − 1) = 0 has sum of the roots
equal to half of their product
(a) 3
(b) 4
(c) 7
(d) 10
20. Find the value of k so that the sum of the roots of the
quadratic equation is equal to the product of the roots :
(a) –2
(k + 1)x2 + 2 kx + 4 = 0
(b) − 4
(c) 6
(d) 8
21. If –4 is a root of the quadratic equation x − px − 4 = 0
2
and the quadratic equation x2 − px + k = 0 has equal
roots, find the value of k :
(a) 9/4
(b) 1
(c) 2.5
(d) 3
22. Find the value of k such that the sum of the squares of
the roots of the quadratic equation x2 − 8 x + k = 0 is
40 :
(a) 12
(b) 2
(c) 5
(d) 8
23. Find the value of p for which the quadratic equation
x2 + p(4 x + p − 1) + 2 = 0 has equal roots :
2
4
3
(b) 3, 5
(c) 1, −
(d) , 2
(a) −1,
3
3
4
α and β are the roots of the
3 x2 + 7 x + 3 = 0 Find the value of αβ :
(a) 1
(b) 0
(c) 2
(d) 3
24. If
equation
25. If α and β are the roots of the equation x2 − 3x + 2 = 0.
Find the quadratic equation whose roots are − α and − β
(a) x2 − 3 x + 2 = 0
(b) x2 + 3 x + 2 = 0
2
(c) x + 3 x − 2 = 0
(d) none of these
26. Find the quadratic equation whose roots are 3 and
2 3:
(a) x2 + 3 3 x − 6 = 0
(b) x2 − 3 3 x + 6 = 0
2
(c) x + 3 3 x + 5 = 0
(d) none of these
27. If α and β are the roots of equation 6 x2 + x − 2 = 0, find
α
β
:
the value of +
β
α
25
12
4
(b) −
(c) −
(d) none
(a) −
12
75
9
28. If a and c are such that the quadratic equation
ax2 − 5 x + c = 0 has 10 as the sum of the roots and
also as the product of the roots, find a and c
respectively :
1
1
(b) , 5
(a) , 6
2
2
1
(c) , 8
(d) none of these
2
29. If α and β are the roots of the equation x2 − x − 4 = 0 ,
1 1
find the value of + − αβ :
α
β
16
8
15
(b)
(c)
(d) none
(a)
7
5
4
CAT
30. If α and β are the roots of the equation x2 − 2 x − 1 = 0,
find the value of α 2β + αβ2
(a) −8
(b) −5
(c) 6
(d) −2
31. If α and β are the roots of x2 − x − 2 = 0, find the
quadratic equation in x whose roots are (2α + 1) and
(2β + 1) :
(a) x2 + 5 x − 4 = 0
(b) x2 − 4 x − 5 = 0
2
(c) x + 4 x − 5 = 0
(d) none of these
32. If α, β be the roots of the quadratic equation
3 x2 − 6 x + 4 = 0 , find the value of
β
α
 1 1
 +  + 2  +  + 3αβ :
β
α
α
β
(a) 6
(b) 8
(c) 7
(d) 5
33. If α , β be the roots of the quadratic equation
x2 − 5 x + k = 0 , find the value of k such that α − β = 1 :
(a) 2
(b) 4
(c) 6
(d) 8
34. If one root of the quadratic equation ax2 + bx + c = 0
is double the other then which one of the following is
correct?
(a) b2 = 3 ac
(b) 2 b2 = 5 ac
2
(c) 2 b = 9 ac
(d) 2 b2 > 9 ac
35. The length of a hypotenuse of a right triangle exceeds
the length of its base by 2 cm and exceeds twice the
length of the altitude by 1 cm. Find the length of each
side of the triangle (in cm) :
(a) 6, 8, 10
(b) 7, 24, 25
(c) 8, 15, 17
(d) 7, 40, 41
36. A two digit number is such that the product of its digits
is 12. When 9 is addded to the number, the digits
interchange their places, find the number :
(a) 62
(b) 34
(c) 26
(d) 43
37. A plane left 40 minutes late due to bad weather and in
order to reach its destination, 1600 km away in time, it
had to increase its speed by 400 km/h from its usual
speed. Find the usual speed of the plane :
(a) 600 km/h
(b) 750 km/h
(c) 800 km/h
(d) none of these
38. The sum of the squares of two consecutive positive
odd numbers is 290. Find the sum of the numbers :
(a) 14
(b) 24
(c) 28
(d) none
39. A shopkeeper buys a number of books for ` 80. If he
had bought 4 more for the same amount, each book
would have cost ` 1 less. How many books did he buy?
(a) 8
(b) 16
(c) 24
(d) 28
40. Two square have sides x cm and (x + 4 ) cm. The sum of
their areas is 656 cm2. Find the sides of the square :
(a) 8 cm, 12 cm
(b) 12 cm, 15 cm
(c) 6 cm, 10 cm
(d) 16 cm, 20 cm
CAT-Test
Questions Helping you bell the CAT
LEVEL 01 > BASIC LEVEL EXERCISE
1 If p and q (≠ 0) are the roots of the equation
8 The number of quadratic equations, which remain
unchanged by squaring their roots, is :
(a) 0
(b) 2
(c) 4
(d) infinitely many
x + px + q = 0, then the least value of x + px + q ( x ∈ R )
2
2
is :
(a) −
1
4
(b)
1
4
(c)
−9
4
(d)
9
4
2 The value of p for which the sum of the squares of the roots
of the equation x 2 − ( p − 2)x − p − 1 = 0 assumes the least
value is :
(a) –1
(b) 1
(c) 0
(d) 2
3 If the equation x 2 + 2( p + 1)x + 9 p − 5 = 0 has only
negative roots, then :
(a) p ≤ 0
(b) p ≤ − 6
(c) p ≥ 6
(d) p ≥ 36
4 The ratio of the roots of the equation ax 2 + bx + c = 0 is
same as the ratio of the roots of the equation
px 2 + qx + r = 0. If D1 and D 2 are the discriminants of
ax 2 + bx + c = 0 and px 2 + qx + r = 0, respectively, then
D1 : D 2 is equal to
a2
(a) 2
q
The value of 6 + 6 + 6 + 6 + K ∞ is :
(a) 2
(c) 4
(b) 3
(d) 5
are
the
roots
of
the
equation
α , β, γ
x 3 + a0 x 2 + a1 x + a2 = 0, then (1 − α 2 )(1 − β 2 )(1 − γ 2 ) is
10 If
equal to :
(a) (1 − a1 )2 + (a0 − a2 )2
(b) (1 + a1 )2 − (a0 + a2 )2
(c) (1 + a1 ) + (a0 + a2 )
(d) none of these
2
11 If x ∈ R , and K =
(a) x ≤ 0
(c) k ≥ 5
2
( x − x + 1)
, then :
( x 2 + x + 1)
1
(b) ≤ k ≤ 3
3
(d) none of these
2
2
(b)
a
p2
12 If x = 2 + 2 + 2 + 2 + K ∞ then x is :
2
b
(c) 2
q
9
(d) none of these
5 If every pair from among the equations x + px + qr = 0
2
x 2 + qx + rp = 0 and x 2 + rx + pq = 0 has a common root,
then the sum of the three common roots is :
(a) 2( p + q + r)
(b) p + q + r
(c) − ( p + q + r)
(d) pqr
6 If the roots of the equation, ax + bx + c = 0, are of the
(a) 1
(c) 3
(b) 2
(d) none of these
13 If α,β, γ are the roots of the equation x 3 − 3x + 11 = 0, then
the equation whose roots are (α + β ), (β + γ )and (γ + α )is :
(a) x 3 + 3x + 11 + 0
(b) x 3 − 3x + 11 = 0
(c) x 3 + 3x − 11 = 0
(d) x 3 − 3x − 11 = 0
2
α
α+1
then the value of (a + b + c)2 is :
and
form
α −1
α
(a) b − 2ac
(b) b − 4ac
(c) 2b2 − ac
(d) 4b2 − 2ac
2
2
7 Equation ax 2 + 2x + 1 has one double root if :
(a) a = 0
(c) a = 1
(b) a = − 1
(d) a = 2
14 If x 2 − ax − 21 = 0 and x 2 − 3ax + 35 = 0; a > 0 have a
common root, then a is equal to :
(a) 1
(b) 2
(c) 4
(d) 5
15 If α , β, γ are such that α + β + γ = 2, α 2 + β 2 + γ 2 = 6,
α 3 + β 3 + γ 3 = 8, then α 4 + β 4 + γ 4 is equal to :
(a) 10
(c) 18
(b) 12
(d) none of these
828
QUANTUM
16 The real values of a for which the quadratic equation
2x − (a + 8a − 1)x + a − 4a = 0
2
3
2
possesses
roots
of
opposite signs are given by :
(a) a > 6
(b) a > 9
(c) 0 < a < 4
(d) a < 0
the roots of px 2 + qx + r = 0, then k is equal to
 a p
1  a p
(a) −  − 
(b)  − 
 b q
2  b q
k ∈ R lies between α and β, if :
(a) a2k 2 + bk + c < 0
(b) a2k 2 + abk + ac < 0
(d) None of these
19 If the equation (3x ) + (27 × 3
1/ k
− 15)x + 4 = 0 has equal
roots, then k is equal to
(c)
1
2
(d) 0
(a) 1
(c) 4
(b) 3
(d) 6
21 The number of real solutions of the equation 23x 2 − 7 x + 4 = 1
is :
(a) 0
(c) 2
(b) 4
(d) infinitely many
solutions
(a) 0
(c) 2
30 If (2 + 3)x 2 − 2x + 1 + (2 − 3)x 2 − 2x − 1 =
2
, then x is
2− 3
(b) 1
(d) both (a) and (c)
number of real roots
( x − 1)2 + ( x − 2)2 + ( x − 3)2 = 0 :
(a) 0
(c) 3
of
the
of
the
one root in common are :
1 2
1
(b) 0,
(a) ,
2 9
2
1 2
2
(d) 0, ,
(c)
2 9
9
32 The
integer
k
for
which
the
inequality
x 2 − 2(4k − 1)x + 15k 2 − 2k − 7 > 0 is valid for any x is :
(a) 2
(b) 3
(c) 4
(d) 6
33 The values of a for which 2x − 2(2a + 1)x + a(a + 1) = 0
2
equation
(b) 4
(d) none of these
23 The
equation
(b) 2
(d) 6
24 The equation x + 1 − x − 1 = 4 x − 1 has :
(a) no solution
(b) one solution
(c) two solutions
(d) more than two solutions
may have one root less than a and other root greater than a
are given by :
(a) −1 < a < 0
(b) 0 < a < 1
(c) a ≥ 0
(d) a < − 1 or a > 0
34 The condition that x 3 − ax 2 + bx − c = 0 may have two of
its roots equal to each other but of opposite signs is :
2
(a) ab = c
(b) a = bc
3
(c) a2b = c
(d) none of these
35 If the roots of ax 2 + bx + c = 0, a > 0, be each greater than
25 If α, β are the roots of the equation 8 x 2 − 3x + 27 = 0, then
the value of [(α 2 / β )1/ 3 + (β 2 /α )1/ 3] is :
(a) 1/3
(c) 1/5
(b) 1
(d) none of these
(1 − 2a)x 2 − 6ax − 1 = 0 and ax 2 − x + 1 = 0 have at least
x 2 − 2(a − 1)x + (2a + 1) = 0 has both the roots positive is :
number of real
2| x|2 − 5| x| + 2 = 0 is :
(a) 0
(c) 2
31 The values of a for which the quadratic equations
20 The least possible integral value of a for which the equation
22 The
29 The number of solutions of the equation 3x −1 + 5x −1 = 34 :
equal to :
(a) 0
(c) 2
1
(b) −
2
(a) –2
(b) a > b > c
(d) none of these
(a, b, c) > 0, then α and β are :
(a) rational numbers
(b) real and negative
(c) negative real parts
(d) none of these
18 If α , β ∈ R are the roots of the equation ax 2 + bx + c = 0,
2
for every x ∈ R , if :
(a) a < b < c
(c) c lies between a and b
28 If α and β are the roots of the equation ax 2 + bx + c, where
(d) (ab − pq)
(c) ak 2 + abk + ac > 0
possesses roots of opposite sign, then a lies in :
(a) (− ∞, 0) (b) (− ∞, 1) (c) (1, 2)
(d) (4, 9)
( x − c)
17 If α , β are the roots of ax + bx + c = 0 and α + k, β + k are
1  b q
 − 
2  a p
26 The quadratic equation 3x 2 + 2(a2 + 1)x + a2 − 3a + 2 = 0
27 The expression ( x − a)( x − b) will assume all real values
2
(c)
CAT
(b) 1/4
(d) 1/6
unity, then :
(a) a + b + c = 0
(c) a + b + c < 0
(b) a + b + c > 0
(d) none of these
1
1
36 The number of real solutions of x − 2
is :
= 2− 2
x −4
x −4
(a) 0
(c) 2
(b) 1
(d) infinite
Theory of Equation
37 The
x/2
2
829
number of real solutions
+ ( 2 + 1)x = (5 + 2 2)x / 2 is :
(a) one
(c) six
of
the
equation
48 If a2 + b2 + c2 = 1, then ab + bc + ac lies in the interval :
49 The integral values of x satisfying the equation
2|x + 1| − 2x = |2x − 1| + 1 is :
38 If xy = 2( x + y ), x ≤ y and x, y ∈ N , the number of
solutions of the equation :
(a) 2
(c) 0
(a) (− ∞, 0)
(c) [ 0, ∞ )
(b) 3
(d) infinitely many
39 If a, b, c ∈ R and equality ax 2 + bx + c = 0 has complex
roots which are reciprocal to each other, then one has :
(a) a = c
(b)| b| ≤ | c|
(c)| b| ≤ | a|
(d) all of these
solutions for x if a belongs to :
(a) (−∞, − 1] ∪ [ 3, ∞ )
(b) [1 − 5, 1 + 5]
(c) [1 − 5, − 1] ∪ [ 3, 1 + 5]
(d) none of the above
of values of x satisfying the equation
(15 + 4 14 )t + (15 − 4 14 )t = 30, where t = x 2 − 2| x| :
(a) 0
interval :
(a) (−4, 3)
(c) (1, 2)
(c) x 2 + x − 1 = 0
(a) ± 2
(c) ± 6
54 If x > 0, xy = 1, minimum value of x + y is :
x
equation
(b) x ≥ − 1
(d) ( x ≥ − 1) ∪ ( x = − 3)
+ (5 − 2 6 )
(b) ± 2, ± 4
(d) 2, 2, 3
= 10
46 If a, b ∈{1, 2, 3, 4}, then the number of quadratic equations
of the form ax 2 + bx + 1 = 0, having real roots is :
(a) 6
(c) 8
(b) 7
(d) none of these
47 The minimum value of the expression a + 1 ; a > 0 is :
a
(a) 0
(c) 2
(b) 1
(d) 4
(a) 64
(c) 184
56
If x ∈ R , and α =
y
(b) 128
(d) 194
x2
, then :
(1 + x 4 )
(a) 0 ≤ α ≤ 2
1
(c) 0 ≤ α ≤
4
(b) 0 ≤ α ≤ 1
1
(d) 0 ≤ α ≤
2
57 The set of values for which x 3 + 1 ≥ x 2 + x is :
45 The values of x which satisfy the expression :
(5 + 2 6 )
(b) 0
(d) 2
55 If x = 7 + 4 3 and xy = 1, then the value of 1 + 1 is :
2
2
the
x2 − 3
(b) ± 4
(d) ± 8
(a) 1
(c) 2
(b) (1 − 2)a, ( 6 − 1)a
(d) none of these
for
(d) x 2 + x + 1 = 0
the equation x 2 + px + 8 = 0 is 2 are :
43 The roots of the expression x 2 − 2a| x − a| − 3a2 = 0 ; a < 0 :
(a) ± 2, ± 3
(c) ± 2, ± 2
(b) (−1, 2)
(d) (3, 4)
53 The value of p for which the difference between the roots of
two distinct negative roots if :
(a) p < 1
(b) p = 0
3
 1 1
(d) p ∈  − , 
(c) p >
 2 2
4
x2 − 3
(d) 6
equation whose roots are α 19, β7 is :
(a) x 2 − x − 1 = 0
(b) x 2 − x + 1 = 0
42 The expression x 4 + 2px 3 + x 2 + 2px + 1 = 0 has at least
(a) x = − 3
(c) x = 0
(c) 4
52 Let α and β be the roots of the equation x 2 + x + 1 = 0. The
p + qk1/ 3 + rk 2/ 3 = 0, then :
q
(a) p =
(b) p = q1/ 3 ≠ r2/ 3
r
(c) p = q = r = 0
(d) none of these
possible values of x
2|x + 2| − |2x + 1 − 1| − 2x + 1 = 1 is :
(b) 2
51 The inequality | 2x − 3| < 1 is valid when x lies in the
41 p, q, r ∈ R and k is a prime number such that
44 The
(b) [1, ∞ )
(d) 2n : n ∈ I
50 Number
40 The equation | x + 1|| x − 1| = a2 − 2a − 3 can have real
(a) (1 + 2)a, (−1 − 6 )a
(c) (1 ± 2)a
(b) [ −1 / 2, 1]
(d) [ 2, − 4]
(a) [1 , 2/ 3]
(c) [ −1 , 1 / 2]
(b) four
(d) infinite
(a) x ≥ 0
(c) x ≥ − 1
(b) x ≤ 0
(d) −1 ≤ x ≤ 1
58 If a2 = 5a − 3 and b2 = 5b − 3, (a ≠ b) find the quadratic
equation whose roots are
(a) 3x 2 − 19 x + 1 = 0
(c) 3x 2 − 19 x + 3 = 0
a
b
and .
b
a
(b) 2x 2 − 39 x + 27 = 0
(d) 3x 2 + 19 x + 3 = 0
59 If f ( x ) = ax 2 + bx + c and g( x ) = − ax 2 + bx + c, where
ac ≠ 0, the new equation f ( x ) ⋅ g( x ) = 0 has n real roots.
Find the minimum value of n.
(a) 0
(b) 1
(c) 2
(d) 4
830
QUANTUM
60 Given that x 2 + ux + v = 0 and x 2 + jx + k; where
uj = 2(v + k ) and u, v, j, k are real numbers. Also,
f ( x ) = x 2 + ux + v and g( x ) = x 2 + jx + k. Then which
one of the following is always correct?
(a) f ( x ) has real roots
(b) g( x ) has real roots
(c) f ( x ) and g( x ) both have real roots
(d) either f ( x ) or g( x ) have real roots
61 If one root of a quadratic equation ax 2 + bx + c = 0 is
only one common root, the which one of the following is
necessarily true?
(a) p − q = 1
(b) p + q = −1
(c) p − q = −1
(d) p + q = 1
x, then find the value of a.
(a) − 1
(c) 1
(b) 0
(d) 3
72 Consider the following equation.
( x + b) ( x + c) ( x + c)( x + a) ( x + a)( x + b)
+
+
=1
(b − a)(c − a)
(c − b)(a − b)
(a − c)(b − c)
(d) ac(1 + r)2 = br2
63 If c, d are the roots of the quadratic equation
( x − a)( x − b) − k = 0, then, a , b are the roots of the
equation
(a) ( x − c)( x − d ) − k = 0
(b) ( x + c)( x + d ) − k = 0
(c) ( x − c)( x − d ) + k = 0
(d) ( x + a)( x + b) + k = 0
64 In a quadratic polynomial a2 x 2 + a1 x + a0, the roots are r1
and r2. Defining Sk = r1k + r2k and P = r1r2. Then which one of
the following relation is true?
(i) S13 = S3 + 3PS1
(ii) S3 = S1(S12 − 3P )
(b) only (ii) is true
(d) none are true
65 The coefficient of x in the equation x 2 + bx + c = 0 was
wrongly written as 17 in place of 13 and the roots thus
found were − 2 and − 15. Find the roots of the correct
equation.
(a) − 7, − 3
(b) − 13, 1
(c) − 10, − 3 (d) 13, − 1
66 Find the possible value(s) of p if the equation
3x 2 − 2x + p = 0 and 6 x 2 − 17 x + 12 = 0 have a common
root.
(a) 7, − 12
(b) − 8 / 3, − 15 / 4
(c) − 3 / 8, − 4 / 15
(d) data insufficient
67 If a, b, c belong to R; and equations x + 2x + 9 = 0 and
2
ax 2 + bx + c = 0 have a common root, then a : b : c is
(c) 1:3:9
2
71 If (a2 − 1)x 2 + (a − 1)x + a2 − 4a + 3 = 0 be an identity in
ax + bx + c = 0, then which one of the following is
correct?
(a) ac(1 + r)2 = (rb)2
(b) ac(1 + r)2 = rb2
(b) 2:1:7
(d) 4
69 If the equations x − px + q = 0 and x + qx − p = 0 have
2
equations has a common positive real root, find the value
of m + n + p.
(a) 13
(b) 15
(c) 47
(d) 23
2
(a) 1:2:9
common root then the value of a is
(a) 5
(b) ±7
(c) ±8
x 2 − 2n + 7 = 0 and x 2 − 2p + 35 = 0. If each pair of
62 If r be the ratio of the roots of the equation
(a) only (i) is true
(c) both are true
68 If x 2 − ax − 45 = 0 and x 2 − 3ax + 27 = 0;(a > 0) has a
70 There are three quadratic equations x 2 − 2m + 5 = 0,
equal to the nth power of the other, then
1  
1 

(a) (acn)n + 1  + (can)n + 1  + bn = 0
 


 


1  
1 

(b) (acn)n + 1  + (can)n + 1  − b = 0
 


 


1  
1 

(c) (acn)n + 1  + (can)n + 1  + b = 0
 


 


1  
1 

1
(d) (acn)n + 1  + (can)n + 1  + n+ 1 = 0
 

 b
 


(c) ab(1 + r)2 = rc2
CAT
(d) 1:2:4
Then what is the most appropriate answer regarding the
above equation?
(a) It has only two real roots
(b) It has no roots
(c) It’s an identity
(d) none of the above
73 If a, b, c belong to R , a ≠ 0, and the quadratic equation
ax 2 + bx + c = 0 has no real roots. Then, which one of the
following is correct?
(a) (a + b + c) c > 0
(c) (a + b + c) c = 0
(b) (a + b + c) c < 0
(d) (a + b + c) c ≤ 0
74 Find the condition that the roots of the equation
x 3 − px 2 + qx − r = 0 are in Harmonic Progression (HP).
It is known that if any three terms a, b, c have a common
difference and are in increasing or decreasing order, they
are in Arithmetic Progression (AP) and their reciprocals
1 1 1
, , are in HP.
a b c
(a) 27 r2 − 9 pqr − 2q3 = 0
(b) 27 r2 − 9 pqr + 2q3 = 0
(c) 27 r2 − 9 pqr + 3q3 = 0
(d) 27 r2 − 9 pqr + 3p3 = 0
75 Find the equation whose roots are negative of the roots of
the equation x 3 − 3x 2 + x + 1 = 0
(a) x 3 + 3x 2 + x − 1 = 0
(b) x 3 − 3x 2 + x − 1 = 0
(c) x 3 + 3x 2 − x − 1 = 0
(d) − x 3 + 3x 2 + x + 1 = 0
Theory of Equation
831
76 Find the equation whose roots are square of the roots of the
equation x − 2x + 3x + 1 = 0
3
2
83 If the graph of the function 16 x 2 + 8(a + 5)x − 7 a − 5 = 0 is
strictly above the X-axis, then
(a) 2 < a < 15
(b) − 15 < a < − 2
(c) 17 < a < 30
(d) a > 0
(a) x 3 + 2x 2 + 13x − 1 = 0
(b) x 3 − 13x 2 + 2x − 1 = 0
(c) x 3 + 3x 2 − 12x − 1 = 0
84 Find the value of x, if
(d) − x 3 + 2x 2 + 13x + 1 = 0
x =1 +
77 Find the equation whose roots are cubes of the roots of the
equation ax 3 + bx 2 + cx + d = 0
1
3+
1
2+
(a) a3 x 3 + x 2(abd 2 + ab − cd ) −
x(3ad 2 − 3bcd − c2 ) + d 3 = 0
(b) a x + x (ab + 4cd − ad ) −
3 3
2
2
x(3ad 2 − 3bcd − c2 ) + d 3 = 0
(c) a x + x (b − 4ac + abcd ) −
3 3
2
x(3ad 2 − 3bcd + c2 ) + d 3 = 0
(d) a3 x 3 + x 2(3a2d − 3abc + b3 ) +
x(3ad 2 − 3bcd + c3 ) + d 3 = 0
78 What are the values of x, if
(7 + 4 3)x
2
−8
+ (7 − 4 3)x
2
−8
= 14?
(b) ±3, ± 4
(d) ± 4, ± 6
(a) 3, 7
(c) ±3, ± 7
79 If a + b + c = 0, then 3ax 2 + 2bx + c = 0 has always
(a) real and distinct roots
(c) rational roots
(b) imaginary roots
(d) real and equal roots
80 Find the roots of the equation 2x 4 + x 3 − 11 x 2 + x + 2 = 0
−8−
2
− 3−
(b)
2
− 3−
(c)
2
− 3−
(d)
2
(a)
−8+ 5 1
, ,2
2
2
5 − 3+ 5
,
, − 2, 2
2
7 − 3+ 7 1
,
, ,2
2
2
5 − 3+ 5 1
,
, ,2
2
2
5
,
81 Find the solution of ( x )2 = [ x]2 + 2x; where ( x ) is the
integer just less than or equal to x and[ x] is the integer just
greater than or equal to x.
1
1
(a) 0, n +
(b) 0, n −
2
2
(c) 0, 1 / 2
(d) none of these
82 For every a, b, c being real numbers the root of
ax 2 + 2bx + c = 0 are real numbers. If m, n are real
numbers such that m2 > n > 0, then the roots of the
equation ax 2 + 2mbx + nc = 0 are
(a) real
(b) complex
(c) both real and complex
(d) cannot be determined
1
3+
1
2+… ∞
(a) ± 5 / 3
(c) 5 / 3
(b) ± 5 / 2
(d) 8 / 5
85 The number of integral solutions of y 4 = 2x 4 + 1402 .
(a) 4
(c) 1
(b) 2
(d) none
86 Let a, b, c be real, if ax 2 + bx + c = 0 has two real roots α
and β, where α < − 1 and β > 1, then the value of 1 +
is
(a) less than zero
(c) equal to zero
c
b
+
a a
(b) greater than zero
(d) equal to b2 − 4ac
87 Find all the values x that satisfy
3x 3 = [( x 2 + 18 x +
32)( x 2 − 18 x − 32)] − 4 x 2
− 3 ± 7i
2
− 3 ± 17i
(b) 3 ± 17 ,
2
− 3 ± 7i
(c) 3 ± 7 ,
2
3 ± 17i
(d) 3 ± 17 ,
2
(a) 3 ± 17 ,
88 For how many values of x, the following equation is an
identity?
( x − a)( x − b) ( x − b)( x − c) ( x − c)( x − a)
+
+
=1
(c − a)(c − b) (a − b)(a − c) (b − c)(b − a)
(a) 0
(b) 6
(c) 2
(d) 3
89 The number of solutions of the following equation is
a2( x − b)( x − c) b2( x − c)( x − a)
+
(a − b)(a − c)
(b − c)(b − a)
c2( x − a)( x − b)
+
= x2
(c − a)(c − b)
(a) 0
(b) 2
(c)3
(d) 4
90 Number of positive integral solutions of the following
inequation
x 3(2x − 3)2 ( x − 4)6
≤0
( x − 3)3(3x − 8)4
(a) 0
(b) 3
(c) 5
(d) 4
832
QUANTUM
91 Find the most appropriate solution of the following
rational inequation.
x2 + 6x − 7
<0
x+3
(a) (− 3,
(b) (− 7,
(c) (− 7,
(d) (− 7,
99 Find the real a if the equation x 3 − 3x 2 + ax − 1 = 0 has
three positive solutions, then
(a) 4
(b) 10/3
(c) −10 / 3
(d) 3
100 The equation x + ax + bx − 4 x + 1 = 0 has four positive
4
− 1) ∪ (− 1, 7 )
− 3) ∪ (− 3, 1) ∪ (1, 3)
3) ∪ (3, 1)
− 3) ∪ (− 3, 1)
3
2
roots, then
(a) a > b
(c) a<0
(b) a > 0
(d) a − b > 0
101 If f ( x ) = 4 x 2 + px + ( p − 3) is negative for at least one x,
92 If x is real, then find the minimum and maximum value of
the following rational function f ( x )
x 2 − 3x + 4
f (x) = 2
x + 3x + 4
(a) 1 / 7, 7
(b) − 1 / 7, 7
(c) − 1 / 7, − 7
(d) − 1, 1 / 7
find all possible values of p.
(a) (− ∞, − 12] ∪ [ − 4, ∞ )
(b) (− ∞, 4] ∪ [12, ∞ )
(c) (4, 12)
(d) (− ∞, 4) ∪ (12, ∞ )
102 If f ( x ) = ( p2 + 3)x 2 + ( p + 2)x − 5 < 0, for at least one x,
93 If one root of the equation 24 x 3 − 14 x 2 − 63x + 45 = 0 is
double the other, then the third root of this equation is
(a) − 3 / 2
(b) 3 / 2
(c) − 5 / 3
(d) − 7 / 2
find all possible values of p.
(a) (− ∞, − 3) ∪ (8, ∞ )
(b) (− ∞, ∞ )
(c) (− 3, 5)
(d) none of these
103 Let a, b, c ∈ R and a ≠ 0 . Let α,β be the roots of equation
ax 2 + bx + c = 0, where α < − p and β > p. Then, for every
sum of the two roots of the equation
4 x 3 + 16 x 2 − 9 x − 36 = 0 is the first whole number, then
n ∈ N , which one of the following is correct?
c
1 b
c
1 b
(a) 1 + 2 +
(b) 1 + 2 +
<0
>0
p a
p a
ap
ap
the third root of this equation is:
(a) − 3 / 4
(b) 3 / 2
(c) − 3 / 2
(c) 1 +
94 If
(d) − 4
95 Find a cubic equation in which each root is greater by unity
than a root of the equation x 3 − 5x 2 + 6 x − 3 = 0. Which
of the following could be the required equation?
(a) p3 − 8 p2 + 19 p − 15 = 0
(b) p3 − 19 p2 + 8 p − 15 = 0
(c) p3 − 8 p2 − 19 p − 15 = 0
(d) p3 − 19 − 8 p + 15 = 0
96 Find a cubic equation in which each root is the cube of the
roots of x 3 + 3x 2 + 2 = 0. Which of the following could be
the required equation?
(a) p3 + 33p2 + 12p + 8 = 0
(b) p − 19 p + 8 p − 15 = 0
3
2
c
1 b
+
≤0
2
p a
ap
(d) p3 − 19 − 8 p + 15 = 0
(d) 1 +
c
1 b
+
≥0
2
p a
ap
104 Let α be the root of ax 2 + bx + c = 0 and β be the root of
− ax 2 + bx + c = 0; where (α , β ≠ 0). Let there be a
a
quadratic function f ( x ) = x 2 + bx + c = 0. Then, which
2
of the following is certainly correct?
(a) One root of f ( x ) must lie between α and β
(b) Both the roots lie between α , β
(c) Both the roots are greater than α , β
(d) Both the roots are less than α , β
105 Find
the number
x 2 − 3| x| + 2 = 0
(a) 0
(c) 4
(c) p3 − 8 p2 − 19 p − 15 = 0
of
solutions
of
the
equation
(b) 2
(d) none of these
106 Find all the values of x that satisfy 3| x 2 + 4 x + 2| = 5x + 16
97 Let
a1, a2, a3 be the roots of
Find
the
x 3 + x 2 + 3x + 1.
a1 + a2 + a3 + a1a2 + a2a3 + a3a1 + a1a2a3
(a) 5
(b) 6
(c) 3
(d) 1
the
polynomial
value
of
98 For a cubic polynomial, a3 x + a2 x + a1 x + a0, the roots
3
2
r1, r2, r3 come in three symmetric combinations :
C1 = r1 + r2 + r3, C 2 = r1r2 + r2r3 + r3r1, C 3 = r1r2r3. The sum of
the k th powers of the roots is defined as Sk = r1k + r2k + r3k .
Then
(i) a3S1 + a2 = 0
(a) only (i) is true
(c) both are true
CAT
(ii) a3S2 + a2 S1 + 2a1 = 0
(b) only (ii) is true
(d) none are true
(a) {1, − 10 / 3}
(c) {− 2, − 11 / 3}
(b) (−2, 1)
(d) { − 2, 1}
107 Find all the values of x that satisfy 2|x| −|2x − 1 − 1| = 2x − 1 + 1
(a) [ − 2, 2]
(b) [ 2, 8]
(c) [ 0, 9)
(d) [ − 1, ∞ ) − (− 1, 1)
108 Find all the values of x that satisfy
(| x − 1| − 3)(| x + 2| − 5) < 0
(a) (− 7, − 2)
(b) (3, 4)
(c) (− 7, 4) − (− 2, 3)
(d) (− 7, − 2) ∪ (3, 4)
Theory of Equation
833
109 Find all the values of x that satisfy the following inequation
x 2 − 5x + 4
≤1
x2 − 4
(a) (− 2, 8 / 5) ∪ [ 2, ∞ )
(b) [ 0, 8 / 5] ∪ [ 5 / 2, ∞]
(c) (− ∞, − 2) ∪ (0, ∞ )
(d) (0, 8 / 5) ∪ (2, 5 / 2) ∪ (5 / 2, ∞ )
110 Find all the values of x that satisfy the following inequation
x − 3x − 1
<3
x2 + x + 1
2
Directions (for Q. Nos. 115 and 116) Answer the following
questions on the basis of the information given below.
Consider the cubic polynomial f ( x ) = 4 x3 + 3x2 + 2x + 1.
115 At least one real root of f ( x ) certainly lies in the interval
(a) (− 1 / 4, 0)
(c) (− 3 / 4, − 1 / 2)
(b) (− 11, − 3 / 4)
(d) (0, 1 / 4)
116 The number of positive real roots is
(a) 2
(c) 0
(b) 1
(d) cannot be determined
117 Let p and q be real numbers such that p ≠ 0, p3 ≠ ± q. If α
and β are nonzero complex numbers satisfying α + β = − p
α
β
and α 3 + β 3 = q, then a quadratic equation having and
β
α
as its roots is
(a) ( p3 + q)x 2 − ( p3 + 2q)x + ( p3 + q) = 0
(a) (− ∞, − 1) ∪ (− 2, ∞ )
(b) (− ∞, − 2) ∪ (− 1, ∞ )
(c) (− ∞, ∞ )
(d) (− 9, 6)
111 For a given quadratic/polynomial function f ( x ) having all
(b) ( p3 + q)x 2 − ( p3 − 2q)x + ( p3 + q) = 0
the real coefficients, if f ( p) and f (q) have opposite signs,
then f ( x )= 0 have
(a) at least one real root between p and q
(b) all the real roots between p and q
(c) none of the roots between p and q
(d) none of the above
(c) ( p3 − q)x 2 − (5p3 − 2q)x + ( p3 − q) = 0
112 Consider
the
polynomial
function
f ( x ) = ( x − 2)( x + 3)(2x − 1)2 (3x − 10)5 . Then, which of
the following relations is/are incorrect regarding this
polynomial.
(i) f (− 1010 ) f (1010 ) < 0
(iii) f (− 2) f (− 4) > 0
(ii) f (− 2) f (3) < 0
(iv) f (− 2) f (1) < 0
(v) f (− 2) f (−1) < 0
(vi) f (1) f (3) f (5) < 0
(a) only (ii), (iii) and (v)
(b) only (ii), (iv), (v) and (vi)
(c) Except (i), (iii) and (v)
(d) Except (i) and (ii)
113 Let α and β be the roots of x 2 − 6 x − 2 = 0, with α > β. If
an = α n − β n for n ≥ 1, then the value of the following
rational function is:
f =
(a) 1
(c) 3
a10 − 2a8
2a9
(b) 2
(d) 4
114 If a > b, where a and b are the roots of the equation
3x 3 − 8 x − 14 = 0 and f n = an − bn, find the value of
3 f 9 − 14 f 6
.
f7
(a) 8
(c) 11
(b) 3
(d) 3/7
(d) ( p3 − q)x 2 − (5p3 + 2q)x + ( p3 − q) = 0
118 The maximum value of the function
f ( x ) = 2x 3 − 15x 2 + 36 x − 48 on the set
A = { x| x 2 + 20 ≤ 9 x} is
(a) 7
(b) 6
(c) 9
(d) none
119 Statement I (Assertion) The curve
−x2
+ x + 1 is symmetric with respect to the line x = 1
2
Statement II (Reason) Because a parabola is symmetric
about its axis.
(a) Statement I is true, statement II is true; statement II
is a correct explanation for statement I.
(b) Statement I is true, statement II is true; statement II
is not a correct explanation for statement I.
(c) Statement I is true; statement II is false.
(d) Statement I is False; statement II is2true.
x − 6x + 5
120 Consider the rational function f ( x )= x 2 − 5x + 6
Match the conditions/expressions in Column I with
statements in Column II where more than one item of
column II can match with an item of column I.
y=
Column I
Column II
(P)
If − 1 < x < 1, then f (x ) satisfies
(p)
0 < f (x ) < 1
(Q)
If 1 < x < 2, then f (x ) satisfies
(q)
f (x )< 0
(R)
If 3 < x < 5, then f (x ) satisfies
(r)
f (x )> 0
(S)
If x < 5, then f (x ) satisfies
(s)
f (x ) <1
(a) P (p,r,s); Q(q, s); R(q,s); S(p, r, s)
(b) P(p,r); Q(q); R(q, s); S(r,s)
(c) P(p,s); Q(s); R(q); S(p, r, s)
(d) P(p); Q(q ,s); R(s); S(p,s)
834
QUANTUM
121 For a, b, c and d are distinct numbers, if roots of the
equation x − 10cx − 11d = 0 are a, b and those of
x 2 − 10ax − 11b = 0 are c, d, then the value of a + b + c + d
is
(a) 1221
(b) 1100
(c) 1210
(d) cannot be determined
2
122 Let f ( x ) = ax 3 + bx 2 + cx + d, then f ′ ( x ) = 3ax 2 + 2bx + c.
Also, if f ′ (k ) = 0 then f (k ) will give maxima/minima.
Also, if there are various peak values of a graph then the
highest peak value is called the absolute or global
maximum and rest are called the local maxima. Similarly
absolute or global minimum and local minima too can be
defined.
f ( x ) is cubic polynomial which has local maximum at
x = − 1. If f (2) = 18, f (1) = − 1, and f ′ ( x ) has local
minimum at x = 0. Then
(i) f ( x ) has local minima at x = 1
(ii) Number of real roots are 3
(iii) f ( x ) is increasing for x = [1, 2 5]
(iv) Sum of the roots is zero
(a) Only (ii) and (iv) are true
(b) Only (i) and (iii) are false
(c) All of the four statements are true
(d) None of the four statements are true
then
f ′ ( x ) = 3ax + 2bx + c. Also, if f ′ (k ) = 0, then f (k )will give
maxima/minima.
Also, if there are various peak values of a graph then the
highest peak value is called the absolute or global
maximum and rest are called the local maxima. Similarly
absolute or global minimum and local minima too can be
defined.
f ( x ) is cubic polynomial which has local maximum at
x = − 1. If f (−1) = 10, f (1) = − 6, and f ′ ( x ) has local
minimum at x = 1. Find the distance between the local
maximum and local minimum of the curve given by f ( x ).
(a) 4 65
(b) 7 65
(c) 12 35
(d) 65 6
124 If α and β are the roots of ax 2 + bx + c = 0, and
α + β, a2 + β 2, α 3 + β 3 are in GP (Geometric Progression)
and ∆ = b2 − 4ac, then
(b) b ∆ = 0
(c) c ∆ = 0
+ q(bx
q −1
f ′ ( x ) = p(ax p − 1 )
) + cx . Also, if f ′ (k ) = 0, then f (k )will give local
0
maxima or local minima.
Also, if there are various peak values of a graph then the
highest peak value is called the absolute or global
maximum and rest are called the local maxima. Similarly
absolute or global minimum and local minima too can be
defined.
f ( x ) = x 3 + bx 2 + cx + d is a cubic polynomial such that
0 < b2 < c, then in the interval of all the real numbers
(− ∞, ∞ )
(a) f ( x ) is a strictly increasing function
(b) f ( x ) is a strictly decreasing function
(c) f ( x ) has a local maxima
(d) f ( x ) has a local minima
126 If f ( x ) = x 2 + 2bx + 2c2 and g( x )= − x 2 − 2cx + b2 such
that min f ( x ) > max g( x ), then the relation between b and
c is
(a) No real value of b and c
(b) 0 < c < b 2
(c)| c| < | b| 2
(d)| c| > | b| 2
and less than 3, then
(a) a < 2
(c) 3 < a ≤ 4
2
(a) ∆ ≠ 0
f ( x ) = ax p + bx q + cx + d, then
127 If the roots of the equation x 2 − 2ax + a2 + a − 3 = 0 are real
f ( x ) = ax 3 + bx 2 + cx + d,
123 Let
125 Let
CAT
(d) ∆ = 0
(b) 2 ≤ a ≤ 3
(d) a > 4
128 If 9 ≤ x ≤ 16, then
(a) ( x − 9)( x − 16) ≤ 0
(b) ( x − 9)( x − 16) ≥ 0
(c) ( x − 9)( x − 16) < 0
(d) ( x − 9)( x − 16) > 0
129 The quadratic equations f ( x )= 0 and g( x ) = 0 have a
common root such that f (2) = g(7 ) = 0 and f (4) × g(9) = 24.
Which among the following give (s) all the correct values of
the possible common root?
(i) 10, 3, − 4, − 9
(ii) −6. 5, 10, 0, − 4
(iii) −9, − 8, − 4. 5, − 4
(iv) 2 2, 3 3, 8, 27
(v) 3, 10, − 5, − 8
(a) Only I and IV
(c) Only V
(b) Only I
(d) Only II and III
LEVEL 02 > HIGHER LEVEL EXERCISE
1 The value of x satisfying the equation
| x − 1|log 3 x
2
− 2 log x 9
= ( x − 1)7 :
4
(b) 3
(d) log 4 3
(a) 3
(c) 3
2 If the roots of 10 x 3 − cx 2 − 54 x − 27 = 0 are in HP, then
find the value of c :
(a) 2
(c) 9
(a) p = 1
(b) p = 1 or 0
(c) p = − 2
(d) p = − 2 or 0
11 If p, q, r are positive and are in AP, the roots of quadratic
(b) 6
(d) none of these
3 Find the number of pairs for ( x, y ) from the following
equation px 2 + qx + r = 0 are real for :
(a)
r
−7 ≥ 4 3
p
(c) all p and r
equations :
(a) 0
(c) 2
10 If p and q are the roots of x 2 + px + q = 0, then :
1
log100 | x + y | =
2
log10 y − log10 | x | = log100 4
(b) 1
(d) none of these
4 The number of solutions for real x, which satisfy the
equation 2 log 2 log 2 x + log1/ 2 log 2 (2 2x ) = 1 :
(a) 1
(b) 2
(c) 4
(d) none of these
5 The solution set for which the equation satisfies
| x 2 + 4 x + 3| + 2x + 5 = 0 :
(a) (−1 + 3, − 1 − 3)
(c) [ −4, (−1 − 3)]
(b) [ − 2, (−1 + 3)]
(d) none of these
6 The real numbers x1, x 2, x 3 satisfying the equation
x 3 − x 2 + βx + γ = 0 are in AP. Find the intervals in
which β and γ lie, respectively :
(a) (−∞, 1 / 3][ −1 / 27, ∞ )
(b) (−∞, 3)
(c) (−1 / 3, 1 / 3)(−1 / 27, 1 / 27 )
(d) None of the above
p
−7 ≥ 4 3
r
(d) no p and r
12 The sum of all the real roots of the equation
| x − 2|2 + | x − 2| − 2 = 0 is :
(a) 2
(c) 4
(b) 3
(d) none of these
13 Let p and q be the roots of the equation x 2 − 2x + A = 0
and let r and s be the roots of the equation
x 2 − 18 x + B = 0. If p < q < r < s are in arithmetic
progression, then A, B respectively equal to :
(a) 8, 17
(b) 3, 7
(c) –3, 11
(d) none of these
14 If the roots of the equation x 2 − 2ax + a2 + a − 3 = 0 are
real and less than 3, then :
(a) a < 2
(b) 2 ≤ a ≤ 3
(c) 3 < a ≤ 4
(d) a > 4
15 If α and β (α < β), are the roots of the equation
x 2 + bx + c = 0, where c < 0 < b, then :
(a) 0 < α < β
(c) α < β < 0
(b) α < 0 < β < | α |
(d) α < 0 < | α | < β
16 If b > a, then the equation ( x − a)( x − b) − 1 = 0 has :
(a) both roots in [ a, b]
(b) both roots in (− ∞, 0)
(c) both roots in (b, + ∞ )
(d) one root in (− ∞, a) and other root in (b, + ∞ )
7 For all x ∈(0, 1) :
(a) e < 1 + x
x
(b) log e (1 + x ) < x
(c) sin x > x
(d) log e x > x
17 Let α, β be the roots of x 2 − x + p = 0 and γ, δ be the roots
8 Let p ≥ 3 be an integer and α , β be the roots of
x 2 − ( p + 1)x + 1 = 0, then the value of α n + β n, where
n ∈N
(a) is divisible by ‘ p’
(c) is a rational number
(b)
(b) is an integer
(d) both (b) and (c)
9 Let a, b, c be real, if ax 2 + bx + c = 0 has two real roots α,
c
b
is :
β, where α < − 1 and β > 1, then the value of 1 + +
a
a
(a) less than zero
(b) greater than zero
(c) equal to zero
(d) equal to b2 − 4ac
of x 2 − 4 x + q = 0. If α, β, γ, δ are in GP, then the integral
values of p and q respectively are :
(a) −2, − 32
(b) −2, 3
(c) −6, 3
(d) −6, − 32
18 If α, β are the roots of the equation x 2 − px + q = 0, then
find the quadratic equation, the roots of which are
(α 2 − β 2 )(α 3 − β 3 ) and (α 3β 2 + α 2β 3 ) :
(a) px 2 − (5p + 7 q)x − ( p6q6 + 4 p2q6 ) = 0
(b) x 2 − ( p5 − 5p3q + 5pq2 )x + ( p6q2 − 5p4q3 + 4 p2q4 )
=0
(c) x 2 − ( p3q − 5p5 + p4q) − ( p6q2 − 5p2q6 ) = 0
(d) all of the above
836
QUANTUM
19 Given that α, γ are the roots of the equation
Ax 2 − 4 x + 1 = 0 and β, δ are the roots of the equation
Bx 2 − 6 x + 1 = 0, then the values of A and B respectively
such that α, β, γ and δ are in HP :
(a) −5, 9
(b) 3/ 2, 5
(c) 3, 8
(d) none of these
20 Let α, β be the roots of the equation ( x − a)( x − b) = c;
c ≠ 0, then the roots of the equation ( x − α )( x − β ) + c = 0
are :
(a) a, c
(b) b, c
(c) a, b
(d) a + c, b + c
21 Let a,b,c be real numbers such that a + b + c = 0, then
 a5 + b5 + c5   a3 + b3 + c3   a2 + b2c2 
(a) 
 >


5
3
2

 


 a5 + b5 + c5   a3 + b3 + c3   a2 + b2 + c2 
(b) 
 <


5
3
2

 


 a5 + b5 + c5   a3 + b3 + c3   a2 + b2 + c2 
(c) 


 =
5
3
2


 

 a3 + b3 + c3   a2 + b2 + c2 
 a5 + b5 + c5 
(d) 
 = 6


5
3
2





22 Given that p and q are prime, the equation
px 2 − qx + q = 0 has
(a) No rational roots
(c) Two rational roots
(b) one rational root
(d) Either (b) or (c)
23 The equation x 4 − px 3 + q = 0 has an integral root. Find
this only integral root if p and q are prime numbers.
(a) 6
(b) 5
(c) 7
(d) 1
24 If x, y, z ar real and distinct, then
x + 4 y + pz − 2xy − 6 yz − 3zx is always
2
2
2
(a) Non-positive
(c) negative
(b) non-negative
(d) zero
25 The number of real solutions of esin x + e − sin x = 4 .
(a) 0
(c) 2
(b) 1
(d) 4
CAT
28 Let f ( x ) be a quadratic expression, which is positive for
all real x. Let
f ( x ) = ax 2 + bx + c, f ′ ( x ) = 2ax + b, f ′ ′ ( x ) = 2a; where
f ′ ( x ) and f ′ ′ ( x ) are the first and second derivatives of
f ( x ), respectively. If g( x ) = f ( x ) + f ′ ( x ) + f ′ ′ ( x ), then
for every real x
(a) g( x ) > 0
(b) g( x ) < 0
(c) g( x ) ≥ 0
(d) g( x ) ≤ 0
29 Find the values of x, if
( x − 1)3 + ( x − 2)3 + ( x − 3)3 + ( x − 4)3 + ( x − 5)3 = 0.
(a) 3, 2 ± 6 i
(c) 2, 3 ± 5 i
(b) 3, 3 ± 6 i
(d) 0, 3 ± 5 i
30 Find all the real solutions of ( x + 3)5 − ( x − 1)5 ≥ 244.
(a) (− ∞, −1] ∪ [1, ∞ )
(c) (− ∞, −2) ∪ (0, ∞ )
(b) (− ∞, −2] ∪ [ 0, ∞ )
(d) none of these
31 If α , β are the roots of x 2 + px + q = 0 and α n, β n are the
α  β
roots of x 2n + pnx n + qn = 0, and if   ,   are the roots
 β α
of x n + 1 + ( x + 1)n = 0. Then n is
(a) an integer
(c) an even integer
(b) an odd integer
(d) an irrational number
32 If Pi and pi are the real numbers, then the number of
non-real roots of the following equation is
P12
P22
Pk2
+
+… +
= x +1
( x − p1 ) ( x − p2 )
( x − pk )
(a) 2
(c) 0
(b) 1
(d) data insufficient
33 If a < b < c < d and k ∈ R , then the roots of the equation
( x − a)( x − c) + k ( x − b)( x − d ) = 0 are
(a) real
(b) non-real
(c) Either real or imaginary (d) none of these
34 For every {b, c} ∈ R ⊕ , the quadratic curve y = x 2 − bx − c
is drawn. The roots are denoted by A and B.
Y
26 If α and β are the roots of x 2 + ux + v = 0 and
x 2n + u nx n + v n = 0, where n is an even integer. What are
D
the roots of x + 1 + ( x + 1) = 0?
n
n
(a) (α n + 1) and (β n + 1)
1
1
(b) and
α
β
A
(c) (α + β ) and (α − β )
α
β
(d) and
β
α
B
X
C
27 If sum of the roots of the equation ax + bx + c = 0 is
2
equal to the sum of the squares of their reciprocals, then
bc2, ca2, ab2 are in
(a) HP
(c) AP
O
(b) GP
(d) AGP
The curve intersects Y-axis at C and another point D is
taken on the Y-axis such that A, D, B and C are concyclic.
Find the coordinates of D.
(a) (0, 1)
(b) (0, 3/2)
(c) (0, b/c)
(d) cannot be determined
Answers
Introductory Exercise 14.1
1
11
21
31
(b)
(d)
(a)
(b)
2
12
22
32
(c)
(b)
(a)
(b)
3
13
23
33
(d)
(c)
(a)
(c)
4
14
24
34
(a)
(a)
(a)
(c)
5
15
25
35
(b)
(a)
(b)
(c)
6
16
26
36
(c)
(c)
(b)
(b)
7
17
27
37
(a)
(b)
(a)
(c)
8
18
28
38
(b)
(b)
(b)
(b)
9
19
29
39
(c)
(c)
(c)
(b)
10
20
30
40
4
14
24
34
44
54
64
74
84
94
104
114
124
(c)
(c)
(a)
(a)
(d)
(c)
(c)
(b)
(c)
(d)
(a)
(a)
(c)
5
15
25
35
45
55
65
75
85
95
105
115
125
(b)
(c)
(b)
(b)
(c)
(d)
(c)
(a)
(d)
(a)
(c)
(c)
(a)
6
16
26
36
46
56
66
76
86
96
106
116
126
(b)
(c)
(c)
(a)
(b)
(d)
(b)
(a)
(a)
(a)
(d)
(c)
(d)
7
17
27
37
47
57
67
77
87
97
107
117
127
(c)
(c)
(c)
(a)
(c)
(c)
(a)
(d)
(a)
(d)
(d)
(b)
(a)
8
18
28
38
48
58
68
78
88
98
108
118
128
(c)
(b)
(c)
(a)
(b)
(c)
(d)
(c)
(d)
(c)
(d)
(a)
(a)
9
19
29
39
49
59
69
79
89
99
109
119
129
(b)
(b)
(b)
(d)
(d)
(c)
(d)
(a)
(c)
(d)
(b)
(a)
(c)
10
20
30
40
50
60
70
80
90
100
110
120
4
14
24
34
(a)
(a)
(b)
(a)
(a)
(a)
(d)
(d)
Level 01 Basic Level Exercise
1
11
21
31
41
51
61
71
81
91
101
111
121
(c)
(b)
(c)
(c)
(c)
(c)
(c)
(c)
(d)
(d)
(d)
(a)
(c)
2
12
22
32
42
52
62
72
82
92
102
112
122
(b)
(b)
(b)
(b)
(c)
(d)
(b)
(c)
(a)
(a)
(b)
(d)
(c)
3
13
23
33
43
53
63
73
83
93
103
113
123
(c)
(d)
(a)
(d)
(b)
(c)
(c)
(a)
(b)
(c)
(a)
(c)
(a)
(b)
(c)
(d)
(a)
(c)
(d)
(a)
(d )
(b)
(c)
(b)
(a)
Level 02 Higher Level Exercise
1
11
21
31
(b)
(b)
(c)
(c)
2
12
22
32
(c)
(c)
(a)
(c)
3
13
23
33
(c)
(d)
(d)
(a)
5 (c)
15 (b)
25 (a)
6 (a)
16 (d)
26 (d)
7 (b)
17 (a)
27 (c)
8 (d)
18 (b)
28 (a)
9 (a)
19 (c)
29 (b)
10 (b)
20 (c)
30 (b)
QUANTUM
CAT
Hints & Solutions
6 Check through options
Introductory Exercise
−b ±
1Q
x=
∴
x=
∴
54
x=
30
9
x=
5
7±
Alternatively, square both the sides twice.
b − 4ac
2a
2
⇒
49 − 4 × 15 × (−36)
7 ± 47
⇒x=
2 × 15
30
−40
x=
30
−4
x=
3
⇒
NOTE (i) The above problem can also be solved by factorization.
⇒
or
or
x=
x=
−b ±
6±
2
b2 − 4ac
2a
x=
2× 7
36 + 364
∴
2 7
6t − 5t + 1 = 0
…(i)
y−3
2y + 1
…(ii)
Solve the eq. (i) and then get the values of t and
subsequently y.
( x + 2)( x − 5)( x − 6)( x + 1) = 144
⇒
[( x + 2)( x − 6)][( x − 5)( x + 1)] = 144
⇒
( x 2 − 4 x − 12)( x 2 − 4 x − 5) = 144
⇒
{( x 2 − 4 x − 5) − 7} {x 2 − 4 x − 5} = 144
Let
Assume
∴
get the values of t by solving the eq. (i) and then substitute
the value of t in the relation
t =
t = x2 − 4x − 5
∴ (t − 7 )(t ) = 144
⇒ t 2 − 7t − 144 = 0
⇒ (t − 16)(t + 9) = 0
⇒ t = 16 or t = − 9
Substituting the values of t in eq. (i), we get
x = 7, − 3, 2
5
2
x=
∴
y−3
= t then, 6 t 2 + 1 = 5t
2y + 1
⇒
5
∴
8
2
1
= t)
x
−1 ± 1 − 4 −1 ± −3
=
2
2
5
1 5
1
2
and if t = then x + = ⇒ 2x − 5x + 2 = 0 ⇒ x = , 2
x 2
2
2
⇒x=
8 
5
8 5

x −  x −  = 0⇒ x = ,




3
2
3 2
4 Consider
(Substituting x +
1
= − 1.
x
x2 + 1 + x = 0 ⇒ x2 + x + 1 = 0
⇒
3 6 x 2 − 31 x + 40 = 0
⇒
2t2 − 3t − 5 = 0
Now if t = − 1, then x +
36 − 4 7 × (−13 7 )
6±
1
1


2x +  − 3x +  − 5= 0


x
x
Now solve it and you will get, t = − 1 and t =
6 ± 20
2 7
13
−7
13
or x =
or x = − 7
⇒ x=
x=
7
7
7
⇒
1
 −1 = 0
x
1
 −1 = 0
x
2


1
1

2  x +  − 2 − 3  x +  − 1 = 0



x
x


⇒
(ii) The same problem can also be solved through options.
2Q
1


2  x2 + 2 − 3  x +



x
1



2  x 2 + 2 + 2 − 2 − 3  x +

x


7
1− x 1
=
x
t
1 13
t + =
t
6
Now, solve it as in the previous problem.
2x 2 − 2x + 1 = 2x − 3
9
Square on both sides and simplify (Hint Type -3 problem)
10 Hint Type-3 problem. It requires double squaring.
Alternatively
Go through options.
11 Hint Type 6 problem.
1
1
1
1
+
=
+
x+1 x+5 x+2 x+4
12
…(i)
x
=t
1− x
⇒
 1
1   1
1 
−
−

 +
 =0
 x + 1 x + 4  x + 5 x + 2
⇒
3
3
−
=0
( x + 1)( x + 4) ( x + 5)( x + 2)
⇒
3[( x 2 + 7 x + 10) − ( x 2 + 5x + 4)]
=0
( x + 1)( x + 4)( x + 5)( x + 2)
⇒
( x 2 + 7 x + 10) − ( x 2 + 5x + 4) = 0
⇒
2x + 6 = 0 ⇒
x = −3
Theory of Equation
839
23 For equal roots, D = 0
13 For equal roots D = 0
b − 4ac = 0
2
i.e.,
b2 − 4ac = 0
i.e.,
But now
x 2 + p(4 x + p − 1) + 2 = 0
⇒
x 2 + 4 px + p2 − p + 2 = 0
14 D = b2 − 4ac = 4 − 4 × (−3) × (−8) = − 92
⇒
x 2 + 4 px + ( p2 − p + 2) = 0
15 D = b2 − 4ac = 25 − 4 × 1 × 7 = − 3
∴
(4 p)2 − 4 × 1 × ( p2 − p + 2) = 0
⇒
[ −2(1 + 3k )]2 − 4 × 1 × 7 × (3 + 2k ) = 0
Solve it and get the value of k.
Since D < 0, therefore roots are not real, i.e., roots will be
⇒ 16 p2 − 4 p2 + 4 p − 8 = 0 ⇒ 12p2 + 4 p − 8 = 0
imaginary.
⇒ 12p2 + 12p − 8 p − 8 = 0 ⇒ 12p( p + 1) − 8( p + 1) = 0
a x + abx − b = 0
2 2
16
2
⇒
D = b − 4ac = (ab) − 4 × a × (− b ) = (ab) + 4a b
2
2
2
2
2
2 2
24 αβ =
= (ab)2[1 + 4] = (ab)2(5)
∴ The roots are real and unequal
b − 4ac < 0
x 2 − x − 2x + 2 = 0
⇒
( x − 1)( x − 2) = 0 ⇒ α = 1 and β = 2
∴ (− p) − 4 × 1 × q < 0
∴
⇒
∴ The required equation is
2
p2 − 4q < 0 ⇒ p2 < 4q
4 x 2 − 3kx + 1 = 0
18
∴
⇒
⇒
k2 =
16
9
⇒ k=±
4
3
⇒
then the equation will be
( x − α )( x − β ) = 0
∴
1
b 1 c
α + β = αβ and − = ⋅
2
a 2 a
c
2(2k − 1)
⇒ − b = ⇒ (k + 6) =
⇒k =7
2
2
⇒
∴
−2k = 4 ⇒
x 2 − (α + β )x + (α ⋅ β ) = 0
k = −2
21 Since − 4 is a root of x − px − 4 = 0
∴
⇒
x 2 − ( 3 + 2 3)x + ( 3 ⋅ 2 3) = 0
⇒
x 2 − 3 3x + 6 = 0
α+β=
27
2
∴
(−4)2 − p(−4) − 4 = 0
⇒
p= −3
Since both the roots of this equation are equal.
α 2 + β 2 = 40
22
⇒
⇒
(α + β )2 − 2αβ = 40 ⇒ (8)2 − 2k = 40
k = 12
− b −1
c
2
1
=
⇒ αβ = = − = −
a
6
a
6
3
α β α 2 + β 2 (α + β )2 − 2αβ
+ =
=
β α
αβ
αβ
1
2 25
+
(−1 / 6)2 − 2(−1 / 3) 36 3 36
25
=
=
=
=−
1
1
−1 / 3
12
−
−
3
3
∴ The equation becomes x 2 + 3x + k = 0
3
∴ Sum of the roots (α + α ) = 2α = − 3 ⇒ α = −
2
9
2
2
∴ Product of the roots (α ⋅ α ) = α = (−3/ 2) = = k
4
9
∴
k=
4
( x − 3)( x − 2 3) = 0 ⇒ x 2 − 3 3x + 6 = 0
Method II. The required equation
α + β = αβ
−b c
=
⇒ −b = c
a
a
( x + 1)( x + 2) = 0 ⇒ x 2 + 3x + 2 = 0
26 Method I. If α and β be the roots of a quadratic equation,
19 Let α, β be the roots of the equation, then
20
− α = − 1 and − β = − 2
[ x − (− α )][ x − (− β )] = 0
D = b2 − 4ac = 0
9k 2 − 4 × 1 × 4 = 0
8 2
=
12 3
x 2 − 3x + 2 = 0
⇒
2
p=
c 3
= =1
a 3
25
17 For non-real roots D < 0
∴
( p + 1)(12p − 8) = 0 ⇒ p = − 1 or
Alternatively
⇒
⇒
⇒
∴
6x2 + x − 2 = 0
6 x 2 − 3x + 4 x − 2 = 0
3x(2x − 1) + 2(2x − 1) = 0
1
2
or x = −
x=
2
3
α β
1/ 2
−2/ 3
+ =
+
β α −2/ 3 1 / 2
=−
3 4
25
− =−
4 3
12
840
QUANTUM
α + β = 10 and α ⋅ β = 10
28
⇒
−
⇒
b
= 10 and
a
5
= 10
⇒
a
∴
c
= 10
1/ 2
∴
a=
1
2
c
= 10
a
1
a=
2
⇒
c=5
and
c=5
b
c
= 1 and αβ = = − 4
a
a
1 1
α+β
+ − αβ =
− αβ
α β
αβ
α+β=−
29
∴
=
1
1 15
− (−4) = 4 − =
−4
4 4
α + β = 2 and αβ = − 1
30
∴
α β + αβ 2 = αβ(α + β ) = 2(−1) = − 2
2
31 Sum of roots = (2α + 1) + (2β + 1)
= 2α + 2β + 2 = 2(α + β ) + 2
2b2 c
c
 − b
⇒ 2 ×   = ⇒ 2 = ⇒ 2b2 = 9ac
 3a
a
a
9a
35 Let the base of the triangle be x cm, then BC = x cm and
hypotenuse AC = ( x + 2) cm and the perpendicular
( x + 2) − 1 ( x + 1)
AB =
=
2
2
[Let the length of AB be k, then
AC = 2k + 1 ⇒ k =
and product of roots = 4αβ + 2(α + β ) + 1
= 4 × (−2) + 2(1) + 1 = − 8 + 2 + 1 = − 5
∴ Required equation is x 2 − 4 x − 5 = 0.
6
4
= 2 and αβ =
3
3
α β
 1 1
 +  + 2  +  + 3αβ
 β α
 α β
α+β=
32
∴
=
α +β
 α + β
+ 2
 + 3αβ
 αβ 
αβ
2
2
(α + β )2 − 2αβ
 α + β
=
+ 2
 + 3αβ
 αβ 
αβ
 2 
4 − 8/ 3
4 4/ 3
=
+ 2
+ 3+ 4 = 8
 + 3× =
4/ 3
3 4/ 3
 4 / 3
α + β = 5 ⇒ α −β =1
33
∴
∴
α = 3 and
β=2
c k
αβ = = = 3 × 2 = 6 ⇒ k = 6
a 1
34 Let the one root be α and other root be β, then β = 2α .
∴
and
−b
−b
α + β = 3α =
⇒ α=
a
3a
c
2
αβ = 2α =
a
AC − 1
]
2
∴ By Pythagorus theorem
AB 2 + BC 2 = AC 2
(see the figure)
2
 x + 1
2
2

 + x = ( x + 2)
 2 
Now since α + β = 1
∴ Sum of roots = 2(α + β ) + 2 = 2(1) + 2 = 4
C
B
2
= 4 αβ + 2α + 2β + 1 = 4 α β + 2(α + β ) + 1
αβ = − 2
c
a
A
and product of roots = (2α + 1)(2β + 1)
and
2α 2 =
∴
CAT
⇒
x 2 + 1 + 2x
+ x2 = x2 + 4 + 4x
4
⇒ x 2 + 1 + 2x + 4 x 2 = 4 x 2 + 16 + 16 x
⇒
x 2 − 14 x − 15 = 0 ⇒ ( x − 15)( x + 1) = 0
⇒
x = 15 cm
∴
base = 15 cm
(Q x = − 1 cm is inadmissible)
and perpendicular = 8 cm and hypotenuse = 17 cm
36 Let the tens digit be x and unit digit be y. Then the two digit
number = 10 x + y
but
x × y = 12
⇒
y=
12
x
12

∴ the number is 10 x + 

x
Again 10 x +
12
 12
+ 9 = 10 ×   + x
 x
x
⇒ 10 x 2 + 12 + 9 x = 120 + x 2
⇒ 9 x 2 + 9 x − 108 = 0 ⇒ x 2 + x − 12 = 0
⇒
( x + 4)( x − 3) = 0
⇒
x = − 4, 3
but x cannot be negative
∴
∴
x = 3 only.
12 12
y=
=
=4
x
3
∴ the number = 10 x + y = 10 × 3 + 4 = 34
Theory of Equation
841
⇒
37 Let the usual speed of plane be x km/hr
∴ Usual time taken =
1600
hr
x
∴
∴
2x 640000 800
+
−
3
3
x
⇒ 2x 2 − 1920000 + 800 x = 0 ⇒ x 2 − 960000 + 400 x = 0
x 2 + 400 x − 960000 = 0
⇒ x 2 + 1200 x − 800 x − 960000 = 0
1
80
1 
1
80
+1=
⇒ −
 =
( x + 4)
x
 x ( x + 4) 80
4
1
=
⇒ x 2 + 4 x − 320 = 0
x 2 + 4 x 80
⇒
( x + 20)( x − 16) = 0
⇒
( x + 1200)( x − 800) = 0
80
x
the new number of books = ( x + 4)
80
the new price =
∴
( x + 4)
⇒
⇒ x( x + 1200) − 800( x + 1200) = 0
x = 800 km/ hr
Rate of the book =
∴
⇒ 4800 x = 4800 x − 2x 2 + 1920000 − 800 x
⇒
x + 2 = 13
39 Let the shopkeeper buys x books
 1600 2
∴ Distance = speed × time, 1600 = ( x + 400) 
− 
 x
3
⇒
x = − 13, but x is positive
∴ Sum of the numbers = 11 + 13 = 24
New speed = ( x + 400) km/ hr
1600 40  1600 2
and the new time =
−
=
−  hr
x
60  x
3
1600 = 1600 −
x = 11 or
x = 16, (x = − 20, is inadmissible)
∴ He bought 16 books.
(Q x = − 1200 is inadmissible)
38 Let the two consecutive odd numbers be x and x + 2
40 x 2 + ( x + 4)2 = 656
⇒
x 2 + x 2 + 16 + 8 x = 656
x 2 + ( x + 2)2 = 290
⇒
x 2 + 4 x − 320 = 0
⇒ x 2 + x 2 + 4 + 4 x = 290 ⇒ 2x 2 + 4 x + 4 = 290
⇒
( x + 20)( x − 16) = 0
∴
⇒
⇒
x + 2x + 2 = 145
⇒
2
x = 16, (x = − 20 is inadmissible)
∴ Sides of the square are x = 16 cm
x + 2x − 143 = 0
2
and
⇒ ( x + 13)( x − 11) = 0
( x + 4) = 20 cm.
Level 01 Basic Level Exercise
y min =
1
∴
y min
− b2 + 4ac
4a
at
x=−
3 Let f ( x ) = x 2 + 2( p + 1)x + 9 p − 5. Let α, β be the roots of
b
2a
f ( x ) = 0. The equation f ( x ) = 0 will have both negative
roots, if
− p2 + 4 × 1 × q 4q − p2
=
=
4
4 ×1
(i) D ≥ 0
(ii) a < 0, β < 0, i.e., (α + β ) < 0
Now, since p and q are the roots of the equation
x 2 + px + q = 0.
(iii) f (0) > 0
Now,
D ≥ 0 ⇒ 4( p + 1)2 − 36 p + 20 ≥ 0
∴
⇒
p2 − 7 p + 6 ≥ 0
⇒
⇒
( p − 6)( p − 1) ≥ 0
p ≤ 1 or p ≥ 6
α + β < 0 ⇒ −2( p + 1) < 0
p + 1 > 0 ⇒ p > −1
f (0) > 0 ⇒ 9 p − 5 > 0
5
p>
9
⇒
and
∴
∴
p+ q=− p
q = − 2p
pq = q ⇒
p=1
(Q q ≠ 0)
q= −2
y min =
4q − p2 4 × (−2) − (1)2
9
=
=− .
4
4
4
2 Let α , β be the roots of the given equation, then,
α + β = p − 2 and αβ = − ( p + 1).
Now,
α 2 + β 2 = (α + β )2 − 2αβ = ( p − 2)2 + 2( p + 1)
= p2 − 2p + 6 = ( p − 1)2 + 5
Clearly α 2 + β 2 ≥ 5. So minimum value of α 2 + β 2 is 5,
which is attained at p = 1.
⇒
and
⇒
…(i)
…(ii)
…(iii)
from eqs. (i), (ii) and (iii), we get p ≥ 6.
4 Let α 1, β1 be the roots of ax 2 + bx + c = 0 and α 2, β 2 be the
roots of px 2 + qx + r = 0, then,
α1 α 2
=
β1 β 2
842
QUANTUM
⇒
α 1 + β1 α 2 + β 2
=
α 1 − β1 α 2 − β 2
⇒
(α 1 + β1 )2 (α 2 + β 2 )2
=
(α 1 − β1 )2 (α 2 − β 2 )2
⇒
6 By hypothesis,
α
α+1
b
α
α+1 c
and
+
=−
⋅
=
α −1
α
a
α −1
α
a
b2 / a2
(α 2 + β 2 )2
(α 1 + β1 )2
= 2
=
2
2
(α 1 + β1 ) − 4α 1β1 (α 2 + β 2 ) − 4α 2β 2 b − 4ac
a2
b2
q2
q2 / p2
D
b2
⇒ 1 = 2
= 2
=
=
D2 q
q − 4rp D1 D 2
2
p
Alternatively
Let 2, 3 be the roots of the equation
∴ the equation is x 2 − 5x + 6 = 0
∴
a = 1, b = − 5 and
∴
D1 = b2 − 4ac = 25 − 4 × 1 × 6 = 1
r = 24
∴
D1 1
= .
D
4
⇒
(a + b + c)2 = b2 − 4ac
two roots are equal.
∴
D = 0 ⇒ (2)2 − 4a ⋅ 1 = 0
⇒
a = 1.
and
αβ = α β
Now,
αβ = α β
⇒
2
2 2
…(ii)
⇒ αβ(αβ − 1) = 0
α = 0 or β = 0 or αβ = 1.
α1 α1
≠
, when solving through numerical
β1 α 2
β(1 − β ) = 0 ⇒ β = 0, β = 1.
Thus, we get two sets of values of α and β viz. α = 0, β = 0
and α = 0, β = 1. Now, if αβ = 1,
1
1
1
then
α + = α 2 + 2 [putting β = in Eq. (i)]
α
α
α
2
⇒
values.
α+
1 
1
= α +  − 2

α
α
2
⇒
5 The given equations are :
x 2 + px + qr = 0
…(i)
x + qx + rp = 0
…(ii)
⇒
x 2 + rx + pq = 0
…(iii)
⇒
2
and
…(i)
2 2
If α = 0, then β = β 2 [putting α = 0 in Eq. (i)]
(−5)
b
1 D
= = 1.
=
2
2
4 D2
(−10)
q
Caution :
(c + a)2 + 2b(a + c) + b2 = b2 − 4ac
⇒
Now, check through option, only option (c) is correct, as
2
⇒
α + β = α 2 + β2
x 2 − 10 x + 24 = 0
D 2 = q2 − 4 pr = 100 − 4 × 1 × 24 = 4
(c + a)2 + 4ac = − 2b(c + a)
the roots of another quadratic, Since the quadratic remains
the same, we have
Again, if 4, 6 be the roots of the equation
px 2 + qx + r = 0, then the equation will be
∴
⇒
c+ a
b
2α 2 − 1
= − and α =
c−a
a
α2 − α
8 Let α , β be the roots of a quadratic equation and α 2, β 2 be
c=6
p = 1, q = − 10 and
⇒
7 One double root means a single root appears two times i.e.,
ax 2 + bx + c = 0
⇒
CAT
1
1


α +  − α +  − 2 = 0


α
α
α+
1
1
= 2 or α + = − 1
α
α
α = 1 or α = ω, ω 2
Let α, β be the roots of (i), β, γ be roots of (ii) and γ, α be
the roots of (iii). Since β is a common root of (i) and (ii)
Putting α = 1, in αβ = 1, we get β = 1, and putting α = ω in
αβ = 1, we get β = ω 2
∴
β 2 + pβ + qr = 0
Putting α = ω 2 in αβ = 1, we get β = ω. Thus, we get four
and
β 2 + qβ + rp = 0
sets of values of α, β viz., α = 0, β = 0; α = 0, β = 1; α = ω,
⇒
( p − q)β + r(q − p) = 0 ⇒ β = r
β = ω 2; α = 1,β = 1.
Now,
αβ = qr ⇒ αr = qr ⇒ α = q
Thus, there are four quadratics which remain unchanged
by squaring their roots.
Since β and γ are roots of (ii). Therefore
βr = rp ⇒ γr = rp ⇒ γ = p
∴
α +β+ γ =q+ r+ p= p+ q+ r
1
NOTE α + β + γ can also be equal to 0 and − ( p + q + r).
2
9 Let x = 6 + 6 + + K ∞ ⇒ x 2 = x + 6
⇒
x2 − x − 6 = 0
⇒
x = 3, −2.
Q
x > 0, ∴ x = 3.
Theory of Equation
843
10 The best way is to consider some values of the roots α , β
α = 2, β = 3 and γ = 4, then
( x − α )( x − β )( x − γ ) = 0
Let
⇒
( x − 2)( x − 3)( x − 4) = 0
⇒
x 3 − 9 x 2 + 26 x − 24 = 0
⇒
As α is a root of x 2 − ax − 21 = 0,
2
∴
a0 = − 9
a1 = 26
Now, (1 − α 2 )(1 − β 2 )(1 − γ 2 ) = (1 − 4)(1 − 9)(1 − 16)
= − 3 × − 8 × − 15 = − 360
Now, go through options and verify.
Consider option (b)
(1 + a1 ) − (a0 + a2 ) = (1 + 26) − (−9 − 24)
2
2
 28
 28
  − a   − 21 = 0
 a
 a
a2 = 42
or
a2 = − 24
2
14 Let α be a common root of the two given equations, then
α 2 − 3aα + 35 = 0 and α 2 − aα − 21 = 0. On subtracting
28
we get −2aα + 56 = 0 or α =
.
a
and γ.
2
Q a > 0, we get a = 4
15 (α + β + γ )2 = α 2 + β 2 + γ 2 + 2(αβ + βγ + γα )
4 = 6 + 2(αβ + βγ + γα )
⇒
αβ + βγ + γα = − 1
Also, α + β 3 + γ 3 − 3αβγ
3
= (27 )2 − (33)2
= (27 + 33)(27 − 33)
= 60 × (−6) = − 360
Hence option (b) is correct.
x2 − x + 1
11 k = 2
x + x+1
kx 2 + kx + k = x 2 − x + 1
⇒
(k − 1)x 2 + (k + 1)x + k − 1 = 0
Since x is real, the discriminant
D = (k + 1)2 − 4(k − 1)2 ≥ 0
(3k − 1)(− k + 3) ≥ 0
1

 k −  (k − 3) ≤ 0

3
1
≤k≤3
3
⇒
⇒
12 Let
∴
⇒
x = 2+
2+
2+ K∞
x2 = 2 + x
⇒
Q
13 We have,
∴
2+
x = 2+ x
⇒
8 − (3αβγ ) = 2(6 + 1)
⇒
αβγ = − 2
Now, (α 2 + β 2 + γ 2 )2 = Σα 4 + 2 Σβ 2γ 2
⇒
they are real and their product is negative, i.e.,
D ≥ 0 and product of roots < 0
⇒
(a3 + 8a − 1)2 − 8(a2 − 4a) ≥ 0
and
a2 − 4a
<0
2
⇒ a2 − 4a < 0
[Q a2 − 4a < 0 ⇒ (a3 + 8a − 1)2 − 8(a2 − 4a) ≥ 0]
⇒
17 We have
and
x − x − 2= 0
1± 9 1± 3
x=
=
= 2, − 1
2
2
x > 0, ∴x = 2
α+β+γ=0
β + γ = − α , γ + α = − β and α + β = − γ.
Therefore the required equation is :
(− x )3 − 3(− x ) + 11 = 0
x 3 − 3x − 11 = 0
Σα 4 = 36 − 2[(−1)2 − 2(−2)(2)] = 18
16 The roots of the given equation will be of opposite signs if
2
∴The equation whose roots are − α, − β, − γ can be
obtained by replacing x by − x in the given equation.
or
= (α + β + γ )(α 2 + β 2 + γ 2 − αβ − βγ − γα )
= Σα 4 + 2[(Σβγ )2 − 2αβγ(Σγ )]
⇒
⇒
or a = ± 4
and
⇒
⇒
∴
0 < a < 4.
b
c
, αβ =
a
a
q
α+k+β+k=−
p
α+β=−
(α + k )(β + k ) =
r
p
α + β + 2k = −
q
p
−
b
q
+ 2k = −
a
p
k=
b

Q α + β = − 

a
1  b q
 − 
2  a p
NOTE The problem can also be solved by assuming
some appropriate values.
844
QUANTUM
18 Let f ( x ) = ax 2 + bx + c. It is given that α, β are real roots of
f ( x ) = 0. So, k lies between α and β, if
⇒
27 × 3
4( x 2 − 1) = 1 − 4 x + 4 x 2 ⇒ x =
1
9
1
1
cannot be zero. So k = − .
k
2
20 Let f ( x ) = x − 2(a − 1)x + (2a + 1) = 0.
2
a2 − 4a ≥ 0 ⇒ a ≤ 0 or a ≥ 4
Sum of the roots > 0
⇒
2(a − 1) > 0 ⇒ a > 1
and f (0) > 0
1
⇒
(2a + 1) ⇒ a > −
2
From eqs. (i), (ii) and (iii) we get a ≥ 4.
Hence, the least integral value of a is 4.
23x
21
2
−7x + 4
3x 2 − 7 x + 4 = 0
⇒
3x 2 − 3x − 4 x + 4 = 0
⇒
3x( x − 1) − 4( x − 1) = 0
4
x = 1 or x =
3
4
x = 1,
3
⇒
∴
…(i)
…(iii)
x = 2,
x = 3.
but x = 1, 2, 3 do not satisfy eq. (i).
but
LHS ≠ RHS
α2
∴  
 β
1/ 3
 β2 
+ 
α
1/ 3
α3
= 
 αβ 
=
1/ 3
 β3 
+ 
 αβ
1/ 3
3/ 8
α+β
3/ 8 1
=
=
=
(αβ )1/ 3 (27 / 8)1/ 3 3/ 2 4
26 The quadratic equation 3x 2 + 2(a2 + 1)x + (a2 − 3a + 2) = 0
and
∴
i.e., if
a2 − 3a + 2
<0
3
a2 − 3a + 2 < 0,
(a − 1)(a − 2) < 0 or 1 < a < 2
( x − a)( x − b)
27 Let y =
⇒ x 2 − (a + b + y )x + ab + cy = 0
( x − c)
Since x is real, then D ≥ 0
⇒
(a + b + y )2 − 4(ab + cy ) ≥ 0
⇒
y 2 + 2(a + b − 2c)y + (a + b)2 − 4ab ≥ 0
⇒
y 2 + 2(a + b − 2c)y + (a − b)2 ≥ 0 …(i)
Since y takes all real values, (i) is possible iff
…(i)
It is possible only when x − 1 = 0, x − 2 = 0, x − 3 = 0
x = 1,
 5
RHS = 4   − 1 = 4 = 2
 4
will have two roots of opposite sign if it has real roots and
the product of the roots is negative, i.e., if
D ≥ 0 and α ⋅ β < 0
⇒
4(a2 + 1)2 − 12(a2 − 3a + 2) ≥ 0
⇒ (2| x | − 1)(| x | − 2) = 0
1
| x| = , 2
∴
2
1
∴
x =± ,± 2
2
( x − 1)2 + ( x − 2)2 + ( x − 3)2 = 0
and
…(ii)
22 2| x |2 − 5| x | + 2 = 0
23
5
−1 =1
4
5
is not a root of the equation.
4
3
27
25 α + β = , αβ =
8
8
= 1 = 20
⇒
5
+1−
4
LHS =
5
4
∴x=
Then, f ( x ) = 0 will have both roots positive, if
(i) D > 0 (ii) Sum of the roots > 0 and (iii) f (0) > 0
Now,
D ≥ 0 ⇒ 4 (a − 1)2 − 4 (2a + 1) ≥ 0
⇒
5
,
4
For x =
1
= 0 or −2
k
⇒
But
Squaring again, we get
= 27 or 3
31/ k = 1 or 31/ k =
⇒
2 x 2 − 1 = 1 − 2x
⇒
27 × 31/ k − 15 = ± 12
1/ k
4x − 1 ≥ 0
( x + 1) + ( x − 1) − 2 x 2 − 1 = 4 x − 1
(27 × 31/ k − 15)2 − 144 = 0
⇒
x − 1 ≥ 0,
All these inequalities are satisfied when x ≥ 1. Squaring
both the sides of the equation we get
a2k 2 + abk + ac < 0
19 The given equation will have equal roots iff D = 0
⇒
x − 1 = 4x − 1
∴ x + 1 ≥ 0,
af (k ) < 0 ⇒ a(ak 2 + bk + c) < 0
⇒
x+1−
24
CAT
4(a + b − 2c)2 − 4(a − b)2 < 0
⇒
⇒
(a + b − 2c + a − b)(a + b − 2c − a + b) < 0
4(a − c)(b − c) < 0
⇒ (a − c)(b − c) < 0
This is possible if c lies between a and b, that is if
a < c < b or a > c > b.
Theory of Equation
845
32 Let f ( x ) = x 2 − 2(4k − 1)x + 15k 2 − 2k − 7. Then f ( x ) > 0
28 Q D = b2 − 4ac
⇒
and, if D ≥ 0, then the roots of the equation are given by
x=
As
−b± D
2a
D = b2 − 4ac < b2
(Q a > 0 and c > 0)
⇒
4(4k − 1)2 − 4(15k 2 − 2k − 7 ) < 0
⇒
k 2 − 6k + 8 < 0 ⇒ 2 < k < 4
33 According to the given condition a lies between the roots.
It follows that the roots of the quadratic equation are
negative.
Let
−b±i −D
2a
which have negative real parts.
29 It is very obvious that at x = 3 the given expression
Now,
D ≥ 0 ⇒ 4(2a + 1)2 − 8a(a + 1) ≥ 0
⇒
1

8 a2 + a +  ≥ 0,
2

∴
satisfies.
Now, y = 3x −1 and y = 5x −1 are both increasing functions
of x (exponential functions with base greater than 1).
Therefore their sum y = 3x −1 + 5x −1 is also an increasing
f ( x ) = 2x 2 − 2(2a + 1)x + a(a + 1).
For a to lie between the roots, we must have D ≥ 0 and
f (a) < 0
In case D < 0, then the roots of the equation are given by
x=
D < 0 [Q coefficient of x 2 > 0]
f (a) < 0 ⇒ 2a2 − 2a(2a + 1) + a(a + 1) < 0
⇒
− a2 − a < 0 ⇒ a2 + a > 0
⇒
a > 0 or a < − 1
34 Let α, β and γ be the roots of the given equation such that
function of x. It means for x < 3, y = 3x −1 + 5x −1 < 34 and
α = − β. Then,
for x > 3, y = 3x −1 + 5x −1 > 34.
α+β+γ=a ⇒ γ=a
Since γ is a root of the given equation, so it satisfies the
equation, i.e.,
Y
34
γ 3 − aγ 2 + bγ − c = 0
y = 34
⇒
y = 3x –1 + 5x – 1
O
f ( x ) = 0. So,
∴the given equation can be expressed as
2
1
(2 + 3)x − 2x + 1 +
= 2(2 +
2
(2 + 3)x − 2x − 1
3)x
2
− 2x + 1
af (1) > 0 ⇒
y+
⇒
y 2 − 2(2 +
⇒
y
3)y + (2 +
[ y − (2 +
…(i)
3)
= 2(2 +
x − 2x + 1
(2 +
⇒
x 2 − 2x + 1 = 1 ⇒
3)
= (2 +
x ≠ ± 2 ⇒ ( x − 2) = 0
⇒ x = 2, which is inadmissible.
3)
2x / 2 + ( 2 + 1)x = (5 + 2 2)x / 2
x
x
 2+1 

2 
 =1
⇒ 
+
 5 + 2 2
 5 + 2 2




3
which is of the form
cosx α + sin x α = 1
3)
∴
x( x − 2) = 0
x = 0 or 2.
1
one of the quadratic equations becomes
2
1
linear, so a ≠ 0, a ≠ .
2
2
Hence the only answer is a =
9
31 For a = 0 or a =
x2 − 4 ≠ 0
⇒
3)2 = 0
∴
⇒
Here
37
3)]2 = 0 ⇒ y = 2 +
2
(Q a > 0)
a+ b+ c>0
1
1
36 x − 2
= 2− 2
x −4
x −4
expressed as
3)2
f (1) > 0
⇒
, then the eq. (i) can be
(2 +
a3 − a3 + ab − c = 0 ⇒ ab = c
35 Let f ( x ) = ax 2 + bx + c. Since 1 lies outside the roots of
X
x=3
Thus, the equation has no other solution.
1
30 Q
2− 3 =
2+ 3
Let us put y = (2 +
which is always true
38
x = 2.
xy = 2( x + y ) ⇒
∴ y=
y( x − 2) = 2x
2x
but x, y ∈ N by trial, we get x = 3, 4, 6
x−2
∴ y = 6, 4, 3 But x ≤ y
∴ x = 3, 4 and
y = 6, 4
Thus two solutions are possible.
846
QUANTUM
39 Let the roots be α and
1
α

As  − x −

1 c
=
⇒ c=a
α a
∴
α⋅
Q
(| a| − | b|)2 ≥ 0 or| a| ≥ | b|
c = a ∴| c| ≥ | b|
but
2
1 
 > 0, we must have,
−x 
p− 2+
Since
⇒
i.e., iff
| x − 1| = a − 2a − 3
⇒
2
x 2 − 1 = a2 − 2a − 3
⇒
x 2 = a2 − 2a − 2, − a2 + 2a + 4 ≥ 0
For real solutions,
a2 − 2a − 2 ≥ 0 and
⇒
q = 0 and
p = 0 and
3
2
2)a.
∴
1
1

+ 2p  x +  + 1 = 0

x
x2
…(i)
1
Put x + = y. Then we can write eq. (i) as
x
y 2 − 2py − 1 = 0
Since D > 0, i.e., 4 p2 + 4 > 0, hence eq. (ii) has two
distinct real roots. These roots are
p2 + 1
Since − p − p2 + 1 < 0, both the roots of the equation
1
x + = − p − p2 + 1 are negative.
x
6 )a < a
6 )a = ( 6 − 1)a.
⇒
2x + 1 − 1 = | 2x + 1 − 1|
⇒
2x + 1 − 1 ≥ 0 or
x + 1 ≥ 0 or
or
x ≥ −1
Let x + 2 < 0.
| x + 2| = − ( x + 2) and the equation becomes
2−( x + 2) − |2x + 1 − 1| = 2x + 1 + 1
…(i)
1
−| y − 1| = y + 1
= y. Then eq. (i) becomes
2y
2y 2 + 2y + | y − 1| 2y = 1
If y ≥ 1, then LHS of eq. (ii) becomes
2y 2 + 2y − 2y( y − 1) − 1 = 0
We can write Eq. (iii) as
p +1
6 > 1 and (−1 +
x = (−1 +
2x + 1(2 − 1) − 1 − | 2x + 1 − 1| = 0
Put 2x + 1
2
(Q x < a)
⇒
…(iii)
2
x ≠ (−1 − 6 )a
2x + 2 − 2x + 1 − 1 − | 2x + 1 − 1| = 0
has either positive roots or non-real complex roots.
1 
 = p− 2+
−x 
6 )a
44 Case 1. Let x + 2 ≥ 0.
∴
p + 1 > 0, the equation
p2 + 1
∴
Case 2.
2
1
=− p+
x
−2a ± 24a
= (−1 ±
2
∴| x + 2| = x + 2 and the equation becomes
…(ii)
4 p2 + 4
=− p±
2
x=
Again, as a < 0, −1 +
2
1
1


 x +  + 2p  x +  − 1 = 0



x
x
−2 p ±
⇒
x 2 + 2ax − 5a2 = 0
As a < 0, −1 − 6 < 0 < 1, so (−1 − 6 )a > a,
as a root. Dividing by x 2, we can write as

 −x −

2)a < a.
x = (1 − 2)a.
⇒
r=0
42 The equation x + 2px + x + 2px + 1 = 0 cannot have 0
x+
2)a
x 2 − 2a(a − x ) − 3a2 = 0
p=q=r=0
4
Since − p +
2 > 1, (1 +
2a ± 2 2a
= (1 ±
2
Case 2. Suppose x < a, then the given equation becomes
rk 2/ 3 = 0
y=
x=
∴
k≠0
⇒
⇒
Again as 1 − 2 < 1, (1 − 2)a > a.
Again p = 0, qk1/ 3 = 0 and
x2 +
x 2 − 2ax − a2 = 0
Therefore, x ≠ 1(1 +
41 Since k is a prime number
∴
⇒
As a < 0 and 1 +
− a2 + 2a + 4 ≥ 0
Thus we have a ∈ (−∞, − 1] ∪ [ 3, ∞ )
∴
3
.
4
x 2 − 2a( x − a) − 3a2 = 0
a2 − 2a − 3 ≥ 0
∴
p>
43 Case 1. Suppose x ≥ a. Thus the given equation becomes
x 2 − 1 = ± (a2 − 2a − 3)
but
p2 + 1 > | 2 − p|.
p2 + 1 > | 2 − p| iff 4 p − 3 > 0.
| x + 1|| x − 1| = a − 2a − 3
2
p2 + 1 > 0, i.e.,
p2 + 1 = ( p − 2)2 + 4 p − 3,
2
40
CAT
⇒
4 y = 1 or
Also
2x + 1 = 2−2
y=
or
1
4
x = −3
…(ii)
Theory of Equation
847
1
= 5 − 2 6. Thus the given
y
45 Let y = 5 + 2 6, then
equation becomes y x
Again let y x
2
t +
Thus,
∴
−3
2
x2 − 3
= 10
…(i)
= t . Then eq. (i) becomes
1
= 10 ⇒ t 2 − 10t + 1 = 0
t
t = 5± 2 6
(5 + 2 6 )x
⇒ x − 3=1
2
−3
= 5 ± 2 6 = (5 + 2 6 )±1
2
2
x=±
x = ± 2 or
2
b2 ≥ 4a (Q c = 1)
b2
1
2
3
4
1
4
9
16
1
1/2
1/3
1/4
1/5
2
3
4
×
4
→1
4, 8
→2 7
4, 8, 12, 16 → 4
Thus, there are 7 possible equations of the given form.
1
…(i)
47 Let y = a +
a
a +
2
2.5
3.33
4.25
5.20
2.5
3.33
4.25
1
Let us consider y = a + ; (a > 0)
a
For a to be a real number, (− y )2 − (4 × 1 × 1) ≥ 0
y 2 ≥ 4 ⇒ y ≤ − 2 or y ≥ 2
Given that a > 0, so y > 0.
⇒
y≥2
It implies that the minimum value of y = 2.
1
Thus the minimum value of a = = 2.
a
48 (a + b + c)2 ≥ 0
∴ (a + b + c)2
= a2 + b2 + c2 + 2(ab + bc + ca) ≥ 0
…(i)
Differentiating above eq. (i), we get
1
1 − 2 = 0 ⇒ a2 = 1 ⇒ a = ± 1
a
∴ 1 + 2(ab + bc + ca) ≥ 0
But, since a > 0, therefore, only a = 1 is acceptable. Now
substituting a = 1 in the original Eq. (i), we get the
required minimum value of the expression.
1
Hence, minimum of y = 1 + = 2
1
(a2 + b2 + b2 + c2 + c2 + a2 ) ≥ 2ab + 2bc + 2ac
Alternatively
following graph of y = x +
1
2
Also, a2 + b2 ≥ 2ab, b2 + c2 ≥ 2bc, c2 + a2 ≥ 2ac
⇒
a2 + b2 + c2 ≥ ab + bc + ac
⇒
∴
1
.
x
50 Let
4
∴
3
x>0
ab + bc + ac ≤ 1
1
− ≤ ab + bc + ac ≤ 1.
2
(15 + 4 14 )t = k
(15 − 4 14 )t =
2
1
x<0
−2
−3
∴
3
4
5
6
∴The given equation can be written as
1
k + = 30
k
⇒
k 2 − 30k + 1 = 0 ⇒ k = 15 ± 4 14
Now, the equation will hold only at t = ± 1.
−5
∴
x 2 − 2| x | = ± 1
Again if
x 2 − 2| x | = 1
and if
x 2 − 2| x | = − 1 ⇒ x = ± 1
x=a
y=x+
2
1
k
−4
−6
Here,
(Q a2 + b2 + c2 = 1)
integer.
5
1
1
1
=a+
x
a
…(ii)
49 Best way is to go through options. It satisfies for every even
6
−6 −5 −4 −3 −2 −1 0
−1
(Q a2 + b2 + c2 = 1)
(ab + bc + ca) ≥ −
⇒
See the
1
a
a2 + 1
⇒ a2 − ya + 1 = 0
a
y=
⇒
4a
a
×
1
1, 2
1, 2, 3, 4
1
2
3
4
5
1/2
1/3
1/4
⇒
2
46 For the real roots we must have D ≥ 0. i.e., b2 − 4ac ≥ 0 ⇒
b
1
a
Alternatively
or x − 3 = − 1 ⇒ x = 4 or x = 2
2
⇒
 1
+ 
 y
−3
a
Hence
⇒ x = ± (1 +
x = ± 1, ± (1 +
2)
2)
848
QUANTUM
51 We have| 2x − 3| < 1 ⇔ −1 < 2x − 3 < 1
⇔
1< x<2 ⇔
2
Alternatively For real values of x, x +
x ∈(1, 2)
α =ω
Let
β = ω 2. Then
and
Hence α 19 and β7 are roots of the same equation.
Therefore, we have 0 ≤
53 Let α, β be the roots of the equation x 2 + px + 8 = 0.
Then α + β = − p and αβ = 8
p2 − 32 = 4 ⇒
⇒
(Q xy = 1)
For the minimum value of the expression, x = y
∴
x = 1 and y = 1
( x + y )min =
( x + y )min = 2
x2 + y2 + 2 = 4 = 2
(Q x > 0 and xy > 0∴y > 0)
x 2 + y2 + 2 = (x + y)
xy
x
y
1
1
1
1
1
1
1
1/10
1/5
1/2
1
2
5
10
10
5
2
1
1/2
1/5
1/10
101/10
26/5
5/2
2
5/2
26/5
101/10
Alternatively
We know that for any real number, AM ≥ GM .
x+ y
≥ xy ⇒ ( x + y ) ≥ 2 xy ⇒ ( x + y ) ≥ 2 (Q xy = 1)
∴
2
x =7 + 4 3
y=
∴
∴
1
=7 − 4 3
7+4 3
1
1
x 2 + y 2 (7 + 4 3)2 + (7 − 4 3)2
=
+ 2 =
2
[(7 + 4 3)(7 − 4 3)]2
( xy )2
x
y
=
2(49 + 48)
= 194
1
56 For real x, the ever power of x is always a non-negative
number.
Therefore, we have x 2 /(1 + x 4 ) ≥ 0
Also 1 + x 4 − 2x 2 = (1 − x 2 )2 ≥ 0 ⇒ 1 + x 4 ≥ 2x 2
∴
 x2  1
 ≤ ,

4
1 + x  2
1
x2
≤
1 + x4 2
⇒
x3 − x2 − x + 1 ≥ 0
⇒
x ≥ −1
Therefore a + b = 5 and ab = 3.
a b 19
a b
Now + =
and × = 1.
b a 3
b a
19
x+1=0
3
3x 2 − 19 x + 3 = 0
So the required equation is x 2 −
⇒
Hence, choice (c) is the answer.
Alternatively
55
4
and b are the roots of x 2 = 5x − 3 or x 2 − 5x + 3 = 0.
( x + y )2 = x 2 + y 2 + 2
Hence,
1
x2
≤
x +1 2
4
58 Since a2 = 5a − 3 and b2 = 5b − 3, (a ≠ b). It implies that a
p = ± 6.
( x + y )2 = x 2 + y 2 + 2xy
54
⇒ ( x − 1) ( x + 1) ≥ 0
2
1
≥2
x2
x2
≥0
x +1
x3 + 1 ≥ x2 + x
57
α − β = 2 ⇒ (α + β )2 − 4αβ = (2)2
⇒
⇒
And, for real values of x,
α 19 = ω19 = ω and β7 = ω14 = ω 2
But
x4 + 1
≥2
x2
⇒
52 The roots of the equation are ω and ω 2.
CAT
So 0 ≤
1
x2
≤
1 + x4 2
59 Let D1 and D 2 be the discriminants of f ( x ) and g( x )
respectively.
Then D1 = b2 − 4ac and D 2 = b2 + 4ac
Since ac ≠ 0 ⇒ ac > 0 or ac < 0.
If ac > 0, D 2 must be positive; D1 may or may not be.
And if ac < 0, D1 must be positive; D 2 may or may not be.
It implies that when ac ≠ 0, at least one of D1 and D 2
must be positive. That means at least one of the two
equations f ( x ) and g( x ) must have real roots.
Hence choice (c) is the answer.
NOTE In any quadratic equation either both the roots
are real or both the roots are non-real.
Alternatively The four roots of f ( x )⋅ g( x ) = 0 are the
two roots of f ( x ) and two roots of g( x ). It implies that the
number of real roots of f ( x )⋅ g( x ) = 0 will be either 4 or 2
or 0.
Now we have to find the minimum number of real roots
of f ( x )⋅ g( x ) = 0
So let us consider that none of the equations have real
roots. That means, both the discriminants D1 and D 2 are
negative. That is D1 < 0 and D 2 < 0
It implies that b2 − 4ac < 0 and b2 + 4ac < 0
But, we observe that, whether ac > 0 or ac < 0, at least
one of the above discriminants is certainly invalid. That
means one of the two discriminants has to be positive.
Thus, it is certain that one equation must have real roots.
This, in turn, implies that at least 2 roots have to be real.
Theory of Equation
849
Alternatively
Let D1 and D 2 be the discriminants of
and
respectively.
f (x)
g( x )
Now, D1 + D 2 = (b2 − 4ac) + (b2 + 4ac)
⇒
2
D1 + D 2 = 2b ⇒ D1 + D 2 > 0
It implies that at least one of D1 and D 2 must be positive.
So, at least one of the two equations f ( x ) and g( x ) must
have real roots.
Therefore, the equation f ( x )⋅ (g( x ) = 0 has at least two
real roots.
60 Let D1 and D 2 be the discriminants of f ( x ) and g( x )
respectively.
⇒
D1 + D 2 = u 2 + j2 − 4(v + k )
⇒
D1 + D 2 = u 2 + j2 − 2uj ⇒ D1 + D 2 = (u − j)2
⇒
D1 + D 2 ≥ 0
So at least one of D1 and D 2 must be positive. Therefore
it can be inferred that at least one of f ( x ) and g( x ) must
have real roots.
Hence choice (d) is the answer.
61 Let α and α n be the roots of the given equation, then
 c
α= 
 a
⇒
1
n+ 1
⇒
⇒
⇒
 c
a 
 a
 c
a 
 a
1
n+ 1
1
n n+1
(ca )
 c
+ a 
 a
 cn 
+ a b 
a 
1
n+ 1
1
n n+ 1
+ (ac )
( x + 2)( x − 2) − 5 = 0 ⇒ ( x + 3)( x − 3) = 0
c = − 3, d = 3
Now substituting the values of c and d in the given
choices, we try to find out the correct choice whose roots
are a, b. That is −2, 2 .
(a) x 2 − 14, which is not true.
(b) x 2 − 14, which is not true.
(c) x 2 − 4, which is true, since it gives the required roots
− 2, 2.
(d) x 2 + 1, which is not true.
Alternatively ( x − a)( x − b) − k = 0
...(i)
⇒
x 2 − (a + b)x + ab − k = 0
⇒
cd = ab − k
The new quadratic equation whose roots are a, b is
Now from eqs. (i) and (ii), we get
x 2 − (a + b)x + ab = 0
+ b=0
⇒
x 2 − (c + d )x + cd + k = 0 ⇒ ( x − c)( x − d ) + k = 0
Hence choice (c) is the answer.
+ b=0
64 The best way to solve it is by considering arbitray values of
roots then framing a cublic equation as follows.
Let’s assume that r1 = 1, r2 = 2, then we will have a
corresponding equation to these roots as
+ b=0
( x − 1)( x − 2) = x 2 − 3x − 2
62 Let α and β be the roots of this equation, then
α +β =
and
Also,
...(i)
...(ii)
...(iii)
Now go through the given choices and check their
validity.
Let’s consider choice (b), then
ac(1 + r)2 = rb2
...(ii)
x 2 − (a + b)x + ab = 0
Hence choice (c) is the answer.
−b
a
c
αβ =
a
α
=r
β
...(i)
c+d =a+ b
⇒
...(ii)
n
n+ 1
63 Let’s assume that a = −2, b = 2, and k = 5, then
Hence choice (c) is the answer.
From the eq. (i) we have, aα + aα n + b = 0
1
n+ 1
1 α
 b
= × b2 ⇒
ac −  = αβ × b2
2
 a
β
β
c
c
× b2 = × b2
⇒
a
a
Since both the sides are equal. Therefore the chosen
relationship is true. Hence choice (b) is the answer.
⇒
Therefore D1 + D 2 = (u 2 − 4v ) + ( j2 − 4k )
−b
a
c
c
α ⋅α n = ⇒ α n + 1 =
a
a
2
α
 α + β
2
ac
 = ×b
 β 
β
⇒ ac(α + β )2
2
α + αn =
2
α
α

ac1 +  = × b2 ⇒

β
β
⇒
Then S1 = r1 + r2 + r2 = 3, P = r1r2 = 2, and S3 = r13 + r23 = 9
∴
S13 = S3 + 3PS1 ⇒ 27 = 9 + 3(2 × 3), which is valid.
Similarly S3 = S1(S12 − 3P ) ⇒ 9 = 3(9 − 6), which is valid
Therefore (i) and (ii) both are true
Hence choice (c) is the answer
Alternatively Use the following formulae
a + b3 = (a + b)(a2 + b2 − ab) = (a + b)[(a + b)2 − 3ab]
3
and
(a + b)3 = a3 + b3 + 3ab (a + b)
850
QUANTUM
65 If the two roots are α and β, then
So, one root alone cannot be common, as the conjugate
root of the common root be same for both the equations.
It implies that both the roots of the equations will be
common.
a b c
= = ⇒ a: b : c = 1 : 2: 9
∴
1 2 9
x 2 + bx + c = 0
⇒ x 2 − (α + β )x + αβ = 0
It implies that α + β = − 17 and αβ = 30
Thus the wrong equation is x 2 + 17 x + 30 = 0
Hence choice (a) is the answer.
And the correct equation is x 2 + 13x + 30 = 0
Therefore the roots of correct equation can be computed
by using any of the known methods as −10 and −3.
Hence choice (c) is the answer.
NOTE Since, only the coefficient of x is wrong, it means
the constant term c (or αβ) is correct.
66 The two roots of 6 x 2 − 17 x + 12 = 0 are
3
4
and .
2
3
3
is a common root in both the equations then it
2
will satisfy 3x 2 − 2x + p = 0 also.
15
4
3x 2 − 2x + p = 0
 4
 16
3  − 2  + p = 0
 3
 9
but the sum of roots is unknown. However, it is obvious
that the roots will be odd as sum of roots is an even
number.
equations.
Alternatively Consider two quadratic equations,
ax 2 + bx + c = 0; (where a ≠ 0)
...(i)
a′ x 2 + b′ x + c′ = 0; (where a′ ≠ 0)
...(ii)
If one root is common, then the following condition must
be satisfied.
(ab′ − a′ b)(bc′ − b′ c) = (ca′ − c′ a)2
Therefore [(3 × −17 ) − (6 × −2)][(−2 × 12) − (−17 × p)]
= [( p × 6) − (12 × 3)]2
⇒ 12p2 + 77 p + 120 = 0
⇒
70. Since in each equation product of roots is clearly known,
x 2 − 8 x + 7 = 0 and x 2 − 12x + 15 = 0 be the possible
Hence choice (b) is the answer.
And
Hence choice (d) is the answer.
With this little known information, we can assume, with
some confidence, that (1, 5), (1, 7) and (5, 7) are the
roots of respective equations. Then x 2 − 6 x + 5 = 0,
8
p=− .
3
⇒
Hence choice (d) is the answer.
⇒ x =1
4
is a common factor in both the equations then
3
it will satisfy 3x 2 − 2x + p = 0 also.
⇒
It shows that when a = 4, there is one common root (9)
between the two given equations.
It implies that 1 is the common root, then substituting
x = 1 in any of the given equations you will get, p − q = 1.
Again, if
Therefore,
x 2 − 4 x − 45 = 0 ⇒ ( x − 9)( x + 5)
x 2 − px + q = x 2 + qx − p ⇒ ( p + q) = x( p + q) ⇒ ( p + q)
 3
 9
3  − 2  + p = 0
 2
 4
p=−
contain the minus sign while it is given that a > 0.
Now try the following approach and see if it works.
Consider
a = 4, from the last choice.
69. Comparing the given equations you will get,
Therefore, 3x 2 − 2x + p = 0
⇒
68 Choices (b) and (c) can’t be the required answer as they
x 2 − 12x + 17 = 0 ⇒ ( x − 9)( x − 3)
Now if
⇒
CAT
8
15
p=− , −
3
4
67 The roots of the equation x 2 + 2x + 9 = 0 are imaginary as
the discriminant D = − 32, which is negative. That means
there is a pair of conjugate roots.
Now comparing these equations with the given ones, we
find that m = 3, n = 4 and p = 6. Therefore m + n + p = 13.
Hence choice (a) is the correct one.
Alternatively Let (a, b), (b, c) and (c, a) be the roots of
the given equations, respectively. Then a × b = 5, b × c = 7
and c × a = 35 ⇒ a × b × c = 35
But since, a, b, c are positive real numbers, therefore
a = 5, b = 1, c = 7.
Now, 2m = a + b = 6, 2n = b + c = 8, 2p = c + a = 12.
Therefore, 2m + 2n + 2p = 26 ⇒ m + n + p = 13.
71 As it’s an identity, so it must be satisfied by any value of x.
So the coefficients and the constant term must be zero.
Therefore, a2 − 1 = 0 ⇒ a = − 1, 1
And a − 1 = 0 ⇒ a = 1 and a2 − 4a + 3 = 0 ⇒ a = 1, 3
Theory of Equation
Thus you can see that when a = 1, the constant term and
the coefficients of x and x 2 are zero. So, for a = 1, the
given relation will be an identity.
Hence choice (c) is the answer.
Hint Identity equations are the equations which are
always true no matter what value is plugged in for the
variable. If you simplify an identity equation, you'll
always get a true statement.
When an equation becomes an identity in x, then the
equation gets satisfied by every real number. And, this
can happen only when all the 3 terms of the given
equation become equal to zero. So, you can see when
a = 1, each of the three terms given in the equation
becomes zero. And, then whatever value you substitute
for x, you’ll always get LHS = RHS.
72 The given equation is satisfied for three distinct values of
851
Hint Please recall that the product of two numbers is
always positive if both of them have the same sign. So, if
f (0) and f (1) have the same sign, then their product will
certainly be positive.
Here, f (0) = c and f (1) = a + b + c
74 Let us consider any three arbitrary roots for a cubic
equation, which are in HP.
As 1, 1, 1 are in HP, so we can consider them the valid
roots. Then we will have the cubic equation as following
( x − 1)( x − 1)( x − 1) = 0 or ( x − 1)3 = 0
Or
x 3 − 3x 2 + 3x − 1 = 0
...(i)
Now, comparing the above eq. (i) with given equation
x 3 − px 2 + px − r = 0, we get p = 3, q = 3, r = 1
Now, if we substitute these values of p, q and r in the
given choices, then only choice (b) will be satisfied.
x; x = − a, x = − b, x = − c.
Hence choice (b) is the answer.
The highest power of x in the given equation is 2 but this
equation is satisfied by three distinct values of x. So it is
an identity.
NOTE Even if you consider the three roots as 2, 3, 6 (or
any three roots which are in HP), the answer will always be
choice (b).
Hence choice (c) is the answer.
Alternatively The equation whose roots are
reciprocals of the roots of the given equation is given by
1
p
q
−
+ − r=0
x
x3 x2
Hint Actually, when you simplify the given equation you
will find that it’s a quadratic equation, which should not
have more than 2 roots. However, this equation has more
than 2 roots (a, b and c). So, it is obvious that the given
equation is an identity.
73 The best way is to consider any quadratic equation whose
roots are not real. So, I’m assuming such an equation:
x2 − x + 2 = 0
Now, comparing this equation (x 2 − x + 2 = 0) with the
given equation (ax 2 + bx + c = 0), we get a = 1, b = − 1
and c = 2
Now, if you plug in the values of a, b and c in the given
options, you will see that options (b), (c) and (d) do not
satisfy.
Hence, choice (a) is the valid one.
Hint How can you find a quadratic equation whose roots
are non-real?
(i) Have b2 < 4ac
(ii) Directly consider any two non-real roots, say
α = (1 + i ) and β = 1 − i and then
form the equation x − (α + β )x + (αβ ) = 0
⇒
Let f ( x ) = ax 2 + bx + c. Since, the
equation has no real roots, the graph of f ( x )will lie either
above the X-axis or below the X-axis. It clearly suggests
that for all the arbitrary values of x, the value of f ( x ) will
have the same sign. So, you can claim that f (1) and f (0)
will have the same sign.
Therefore f (1)⋅ f (0) > 0 ⇒ (a + b + c)c > 0
Hence choice (a) is the answer.
...(i)
Since the roots of the given equation are in HP, the roots
of the eq. (i) are in AP. Let the roots of eq. (i) be
(b − d ), b, (b + d ).
(− q)
Then, sum of the roots = (b − d ) + b + (b + d ) = −
r
q
b=
⇒
3r
Since b is the root of eq. (i), so
rb3 − qb2 + pb − 1 = 0
3
2
⇒
 q
 q
 q
r   − q   + p  −1 = 0
 3r
 3r
 3r
⇒
27 r2 − 9 pqr + 2q3 = 0
Hence choice (b) is the answer.
75 Replacing x by − x, we can get the desired equation.
(− x )3 − 3(− x )2 + (− x ) + 1 = 0
2
Alternatively
rx 3 − qx 2 + px − 1 = 0
⇒
− x 3 − 3x 2 − x + 1 = 0
⇒
x 3 + 3x 2 + x − 1 = 0
Hence choice (a) is the answer.
76 Replacing x by x , we can get the desired equation.
( x )3 − 2( x )2 + 3( x ) + 1 = 0
⇒
x x − 2x + 3 x + 1 = 0
852
QUANTUM
⇒
x ( x + 3) = (2x − 1)
79 The best way is to assume some relevant numerical values
for a, b, c such that a + b + c = 0. So, for the sake of
simplicity, we can assume
Squaring both sides and simplifying, we get
x + 2x + 13x − 1 = 0
3
2
a = 1, b = 0, c = − 1
Hence choice (a) is the answer.
77 Replacing x by ( x )1/ 3, we can get the desired equation.
2
3
 1
 1
 1
a ( x )3  + b ( x )3  + c ( x )3  + d = 0






2
1

ax + d = − b ( x )3 + c ( x )3 


⇒
Cubing both the sides of the above equation and
simplifying it we get,
Now, substituting the values of a, b and c in the given
equation, 3ax 2 + 2bx + c = 0, it becomes as 3x 2 − 1 = 0,
1
and its roots are ±
3
Thus you can see that the roots are real and distinct.
Hence choice (a) is the answer.
Alternatively If you assume a = 1, b = 2, c = − 3, you
will have 3x 2 + 4 x − 3 = 0, and then its discriminant will
be a positive value, which ensures that the roots are real
and distinct.
a3 x 3 + x 2(3a2d − 3abc + b3 ) +
NOTE For any quadratic equation, the coefficient of x 2
x(3ad 2 − 3bcd + c3 ) + d 3 = 0
Hence choice (d) is the answer.
Alternatively Since the equation given in the
problem and the equations given in the choices are purely
in variables, so you have the flexibility to solve this
problem by assuming any convenient numbers.
Now, let us assume that three roots of the given equation
be 1, 2, 3. Then the original equation would be
( x − 1)( x − 2)( x − 3) = 0
⇒
x 3 − 6 x 2 + 11 x − 6 = 0
Then the required equation would be
⇒
cannot be zero, so you can assume any real value for x, but
not zero.
Remember that roots can be rational only when the
discriminant is a perfect square. It means whether roots are
rational or not, but they must be real and distinct (unequal)
when discriminant is a positive value. That’s why choice (a) is
always correct but choice (c) may or may not be true.
Please try to understand that choice (c) represents the
subset (or a special case) of choice (a). So as per the given
condition you can tell the general answer, but not the very
specific one.
...(i)
Now, form another equation whose roots are 1, 8, 27
(i.e., cube of 1, 2, 3).
( x − 1)( x − 8)( x − 27 ) = 0
x 3 − 36 x 2 + 251 − 216 = 0
CAT
...(ii)
Alternatively
The discriminant of this equation is
D = 4b2 − 12ac = 4(b2 − 3ac)
⇒
D = 4[{− (a + c)}2 − 3ac]
⇒
D = 4[(a2 + c2 + 2ac) − 3ac]
⇒ D = 4(a2 + c2 − ac)
Now, if we put in the values of a, b, c, d in the given
choices, only choice (d) will give the Eq. (ii).
⇒
Hence choice (d) is the answer.
78 The best way is to go through the given choices. Since 3 is
But, as we can see that D is a not a perfect square so the
roots are irrational.
there in each of the choices so it’s a sure shot value for x.
And you can easily notice that whether you have x = 3 or
x = − 3 in both the case x 2 − 8 = 1
Hint We know that for any two real numbers, AM ≥ GM .
x+y
that is
≥ xy
2
So, for x = 3 or −3, (7 + 4 3) + (7 − 4 3) = 14.
⇒
That is valid.
⇒
Now there are basically two values to choose from the
given choices: ±4 and ± 7 . If we consider x = 4, it does not
lead to any potential possibility. So let’s consider x = 7
So, for x = ± 7 , x 2 − 8 = − 1. Therefore,
(7 + 4 3)−1 + (7 − 4 3)−1
=
1
1
+
= 14 that is valid
7+4 3 7−4 3
Hence choice (c) is the answer.
D>0
It implies that the roots are real and distinct.
⇒
x + y ≥ 2 xy ⇒ a2 + c2 ≥ 2 a2c2
a2 + c2 ≥ 2ac
⇒ a2 + c2 > ac
a2 + c2 − ac > 0
NOTE
Since, a2 + c2 − ac > 0, therefore D > 0.
Otherwise, the square of any real number is always
non-negative. So, we have ( a − c)2 ≥ 0
⇒
a2 + c2 − 2ac ≥ 0 ⇒ a2 + c2 ≥ 2ac
⇒
a2 + c2 ≥ 2ac > ac ⇒a2 + c2 > ac ⇒a2 + c2 − ac > 0
Theory of Equation
853
80 Since x = 0 is not a solution of given equation. Dividing by
2
84 x = 1 +
x , we get
1
3+
1 
1 

2  x 2 + 2  +  x +  − 11 = 0



x 
x
2

1
1 
1
⇒ 2  x +  − 2 × x ×  +  x +  − 11 = 0



x
x
x


1
Putting x + = y in eq. (i), we get
x
...(i)
1
2+
⇒x+1 = 2 +
1
3+
1
3+
2( y 2 − 2) + y − 11 = 0 ⇒ 2y 2 + y − 15 = 0
⇒
5
1
5
1
y = − , 2 ⇒ x + = − or x + = 2
x
x
3
3
⇒
x=
⇒ x +1 =2+
−3 − 5 −3 + 5 1
,
, ,2
2
2
2
⇒ x +1 = 2+
Hence choice (d) is the answer.
⇒
81 Case 1: If x ∈ I, then ( x ) = [ x]
Therefore, ( x )2 = [ x]2 + 2x ⇒ x = 0
⇒
Case 2 : x ∉ I, then ( x ) = [ x] − 1
( x )2 = [ x]2 + 2x
Therefore,
⇒
x = [ x] +
⇒
x =n +
1
2 + .... ∞
1
3+
1
( x + 1)
x +1
x +1
⇒ x =1+
3x + 4
3x + 4
4x + 5
3x + 4
3x 2 = 5
⇒
82 Since the roots of the equation ax 2 + 2bx + c = 0 are real.
Therefore,
4b2 − 4ac ≥ 0
And given that m > n
3+
⇒ x(3x + 4) = 4 x + 5
⇒
x=±
5
3
y 4 = 2x 4 + 1402
1
; n ∈I
2
2
1
85 Let us assume that x 2 = a and y 2 = b. Then,
1
2
b2 ≥ ac
2+
Hence choice (c) is the answer.
Hence choice (a) is the answer.
⇒
1
But, x cannot be a negative value.
[ x]2 + 1 + 2[ x] = [ x]2 + 2x
⇒
x=
1
2 + .... ∞
...(i)
...(ii)
Now the discriminant of the equation
ax 2 + 2mbx + nc = 0 is 4m2b2 − 4anc
From the eqs. (i) and (ii), we get m2b2 > anc
∴ 4m2b2 > 4anc
b2 = 2a2 + 1402
As the problem asks for integral solutions, then we can
assume that x and y are integers, so a and b are also
integers. Since perfect squares can never be negative, so
a2 ≥ 0 and b2 ≥ 0. As we can see that a2 and b2 are perfect
square integers, so the minimum possible value of b2 must
be a perfect square number just greater than 1402 in order
to satisfy the eq. (i).
Therefore the minimum possible value of b2 = 1444 and
the minimum possible value of b = 38.
From eq. (i) we know that b is an even number, so b can
be expressed as following. b = 38 + 2k; Where k is any
whole number.
Hence choice (a) is the answer.
83 16 x 2 + 8(a + 5)x − 7 a − 5 = 0
∴
b2 = 2a2 + 1402
⇒
(38 + 2k )2 = 2a2 + 1402
Since the whole graph is strictly above the X-axis, so it
implies that the roots of the graph are non-real. That
means the discriminant of this equation D is negative.
⇒
Therefore,
discriminant D should be zero.
[ 8(a + 5)] − 4[16 × (−7 a − 5)] < 0
That is
⇒
a2 + 17 a + 30 < 0
It shows that 2k 2 + 76k + 21 is not a perfect square.
⇒
(a + 15)(a + 2) < 0
Therefore a, b cannot be positive integers.
2
⇒
−15 < a < −2
Hence choice (b) is the answer.
...(i)
a2 = 2k 2 + 76k + 21
Since 2k 2 + `76k + 21 is a perfect square, so its
D = 762 − 4 × 2 × 21 ⇒ D > 0
So, x, y cannot have integral solutions.
Hence choice (d) is the answer.
854
QUANTUM
86 ax 2 + bx + c = 0
⇒
x2 +
b
c
x + =0
a
a
CAT
Now testing each of the 6 intervals we get the results as
shown on the above number line.
...(i)
Here we need non-positive intervals so we exclude all the
positive intervals and combine the others keeping in
mind the restriction on limits of the intervals.
Let’s draw a quadratic graph such that α < −1 and β > 1
Therefore the valid values of x are
a –1
1

8  8 
 0, 3 ∪  3 , 3 ∪ {4}.


b
From the graph, we observe that f (−1) < 0
b c
1+ + <0
⇒
a a
The critical points 8/3, 3 are not included in the domain
since the function is not defined there and 3/2, 4 are
included since they satisfy the given relation. Since the
problem asks only integral solutions so we have to look
back in the domain of this function. Now looking in the
domain we find that 1, 2, 4 are the three integral
solutions to this problem.
...(ii)
Similarly from the graph, we observe that f (1) < 0
b c
...(iii)
⇒
1− + <0
a a
Hence choice (b) is the answer.
From eqs. (ii) and (iii), we can conclude that
+ c < 0
b
1 +
a a
Alternatively The least critical number is 0 and the
greatest critical number is 4. So first of all by putting 1, 2, 3
and 4 in the above rational function you can check whether
a particular integer gives the value of function f ( x ) ≤ 0.
Hence choice (a) is the answer.
Now test the same by selecting any random and
convenient integer number before 1 and after 4.
87 3x 3 = [( x 2 + 18 x + 32)( x 2 − 18 x − 32)] − 4 x 2
⇒ 3x 3 = x 4 − ( 18 x +
32)2 − 4 x 2
The result is that only 1, 2 and 4 make the function
f ( x ) ≤ 0.
⇒ 3x 3 = x 4 − 2(3x + 4)2 − 4 x 2
⇒ x 4 − 2(3x + 4)2 − x 2(3x + 4) = 0
91
Now putting 3x + 4 = y, the above equation becomes
⇒ ( x 2 − 2y )( x 2 + y ) = 0 ⇒ ( x 2 − 2y ) = 0
⇒
( x 2 + y ) = 0 ⇒ x 2 − (3x + 4) = 0
( x + 7 )( x − 1)
<0
| x + 3|
(−∞, −7 ) ∪ (−7, − 3) ∪ (−3, 1) ∪ (1, ∞ )
−3 ± 7 i
x = 3 ± 17 or x =
2
Now you can test each interval and then you see that
only two intervals (−7, − 3) and (−3, 1) give negative
regions, which is the requirement of the problem.
Hence choice (a) is the answer.
88 For x = a, b, c the given equation is an identity. Actually the
given equation is a quadratic equation of degree two. But,
since this equation has more than two solutions, therefore
it is an identity.
Hence choice (d) is the answer.
89 Since at x = a, b, c both the sides are equal. It means there
are 3 solutions.
Hence choice (d) is the answer.
Hint While testing for the signs in each of the intervals
you can test the intervals by testing the numerator only.
You need not consider the denominator as for every real
number denominator will give positive results only. Also
keep in mind that at x = − 3, domain is broken (or
undefined).
92 In each of the numerator and denominator, the
Hence choice (c) is the answer.
NOTE Please bear in mind that the given equation is a
quadratic equation which should have only two roots, but
this equation has more than two roots, so it’s an identity.
90 The critical points of this rational function are 0, 3/2 , 8/3,
discriminant D < 0; and the coefficient of x 2 is positive. So
both the numerator and denominator are positive. It
implies that the ratio of numerator and denominator is also
positive. Therefore choices (b), (c) and (d) are eliminated.
Hence choice (a) is the answer.
93 The best way is to go through the given choices.
3 and 4.
+
–∞
⇒
There are three critical points −7, − 3 and 1. So there will
be four different regions on the number line. That means
there will be four intervals as following.
x 4 − yx 2 − 2y 2 = 0
or
x2 + 6x − 7
<0
| x + 3|
–
0
–
3/2
–
8/3
+
3
4
−3
in place of x does not satisfy the equation,
2
so it’s not a root of this equation at all.
(a) Putting
+
+∞
Theory of Equation
855
3
3
(b) Putting x = , the equation gets satisfied, so is one
2
2
3
of the roots. Now put double of that is 3 in the
2
equation, but this time it is not satisfied. So what you can
3
3
do is put half of that is in the equation. This time it is
2
4
3
satisfied, so is also the root of this equation.
4
3
3
So and are the roots of this equation and one root is
2
4
double the other one. But we have to find the third root.
−5
in place of x, we see that the equation is
(c) Putting
3
−5
satisfied. So clearly
is the third root.
3
Hence choice (c) is the answer.
94 Let α, β, γ
be the roots of the given equation
4 x 3 + 16 x 2 − 9 x − 36 = 0, then sum of the roots
=α +β + γ = −
⇒
Therefore the required equation is
p3 − S1 p2 + S2 p − S3 = 0
⇒
96 Let p be one of the roots of the required equation,
then p = x 3
1
⇒
1
( p1/ 3 )3 + 3( p1/ 3 )2 + 2 = 0
⇒
⇒
p3 + 8 + 6 p ( p + 2) = − 27 p2
p3 + 33p2 + 12p + 8 = 0
Hence choice (a) is the answer.
let α, β, γ be the roots of the given
Alternatively
equation x + 3x + 2 = 0, then
2
α +β+ γ = −
3
=−3
1
0
=0
1
2
αβγ = − = − 2
1
αβ + βγ + γ α =
b
a
95 Let p be one of the roots of the required equation, then
⇒ x = p−1
Now the roots of required equation are α 3, β 3, γ 3 then
S1 = Σα = α 3 + β 3 + γ 3 = − 33
Now substituting p − 1 in the given equation, we get
S2 = Σαβ = α 3β 3 + β 3γ 3 + γ 3α 3 = 12
( p − 1)3 − 5( p − 1)2 + 6( p − 1) − 3 = 0
S3 = Σαβγ = α 3β 3 γ 3 = (αβγ)3 = − 8
p3 − 8 p2 + 19 p − 15 = 0
Therefore the required equation is
Hence choice (a) is the answer.
Let α, β, γ be the roots of the given
equation
x 3 − 5x 2 + 6 x − 3 = 0, then α + β + γ = −
p + 3p2/ 3 + 2 = 0 ⇒ ( p + 2)3 = (−3p2/ 3)
3
16
=−4
4
Hint Zero is the first whole number. And α + β + γ = −
Alternatively
3
⇒
Hence choice (d) is the answer.
⇒
x = p3 .
Now substituting p 3 in the given equation, we get
γ = − 4 since α + β = 0
p= x +1
p3 − 8 p2 + 19 p − 15 = 0
−5
=5
1
p3 − S1 p2 + S2 p − S3 = 0
⇒
p3 + 33p2 + 12p + 8 = 0
Hint (α 3 + β 3 + γ 3 ) − 3(αβγ )
= (α + β + γ)[(α 2 + β 2 + γ 2 ) − (αβ + βγ + γα )]
6
−3
αβ + βγ + γ α = = 6, αβγ = −
=3
1
1
⇒ (α 3 + β 3 + γ 3 ) − 3(−2)
Now the roots of required equation are
(α + 1), (β + 1), (γ + 1), then
But (α + β + γ)2 = (α 2 + β 2 + γ 2 ) + 2(αβ + βγ + γα )
= (−3)[(α 2 + β 2 + γ 2 ) − (0)]
S1 = Σα = (α + 1) + (β + 1) + (γ + 1)
= (α + β + γ) + 3 = 8
⇒ (−3)2 = (α 2 + β 2 + γ 2 ) + 2(0) ⇒ (α 2 + β 2 + γ 2 ) = 9
∴(α 3 + β 3 + γ 3 ) = − 3(9) − 6 = − 33
S2 = Σαβ = (α + 1)(β + 1) + (β + 1)(γ + 1)
Similarly, (α 3β 3 + β 3 γ 3 + γ 3α 3 )
+ (γ + 1)(α + 1)
= (αβ + βγ + γα ) + 2(α + β + γ ) + 3
= 6 + 2(5) + 3 = 19
S3 = Σαβγ = (α + 1)(β + 1)(γ + 1)
= αβγ + (αβ + βγ + γα ) + (α + β + γ ) + 3
= 3 + 6 + 3 + 3 = 15
= (αβ )3 + (βγ )3 + (γα )3
∴(αβ )3 + (βγ )3 + (γα )3 − 3[(αβ )(βγ )(γα )]
= [(αβ ) + (βγ ) + (γα )]
[(αβ ) + (βγ )2 + (γα )2 − (αββγ ) + (βγγα ) + (γααβ )]
2
= [ 0][(αβ )2 + (βγ )2 + (γα )2 − (αββγ ) + (βγγα ) + (γααβ )] = 0
856
CAT
QUANTUM
⇒ (αβ )3 + (βγ )3 + (γα )3 = 3[(αβ )(βγ )(γα )]
⇒ (αβ )3 + (βγ )3 + (γα )3 = 3(αβγ )2
⇒ (αβ )3 + (βγ )3 + (γα )3 = 3(−2)2 ⇒ α 3β 3 + β 3γ 3 + γ 3α 3 = 12
97 α 1 + α 2 + α 3 = − 1, α 1α 2 + α 2α 3 + α 3α 1 = 3, α 1α 2α 3 = − 1
∴α 1 + α 2 + α 3 + α 1α 2 + α 2α 3 + α 3α 1 + α 1α 2α 3 = 1
Hence choice (d) is the answer.
98. The best way to solve it is by considering arbitrary values of
roots then framing a cubic equation as follows.
Let’s assume that r1 = 1, r2 = 2, r3 = 3, then we will have a
corresponding equation to these roots as
100 Let f ( x ) = x 4 + ax 3 + bx 2 − 4 x + 1.
Since there are four positive roots, it implies that there
will be four changes in the sign among the various terms
of the function, when the terms are written in
increasing/decreasing order of degree. So a must be
negative and b must be positive.
Hence choice (c) is the answer.
101 If you can find the real values of p so that f ( x ) ≥ 0 for all x,
then for the remaining real values of p the function
f ( x ) < 0 for at least one x.
Now, since f ( x ) ≥ 0 for all x and a > 0, then D ≤ 0.
( x − 1)( x − 2)( x − 3) = x − 6 x + 11 x − 6
Therefore p2 − 4 × 4( p − 3) ≤ 0
It implies that a3 = 1, a2 = − 6,
⇒
a1 = 11, a0 = − 6
⇒
3
2
p2 − 16 p + 48 ≤ 0 ⇒ ( p − 4)( p − 12) ≤ 0
p ∈[ 4, 12]
It implies that when p ∈[ 4, 12], the given function f ( x ) is
not negative for any value of x.
Then S1 = r1 + r2 + r3 = 6
And S2 = r12 + r22 + r32 = 14
∴a3S1 + a2 = 1 × 6 + (− 6) = 0
It means when p ∈ (−∞, 4) ∪ (12, ∞ ), the given function
f ( x ) is negative for at least one x.
Similarly a3S2 + a2S1 + 2a1 = 1 × 14 + (− 6)
× 6 + 2 × 11 = 0
Hence choice (d) is the answer.
Therefore (i) and (ii) both are true.
Since f ( x ) < 0 for at least one value of x, and a > 0 , then
D > 0. Therefore p2 − 16 p + 48 > 0
Hence choice (c) is the answer.
99. Comparing the given equation with
x 3 − S1 x 2 + S2 x − S3 = 0, you will find that a is positive.
Let α, β and γ be the roots of the given equation, then
α + β + γ = 3 and αβ + βγ + γα = a and αβγ = 1
From A.M. − G.M. inequality,
1
(α + β + γ )
≥ (αβγ )3
3
Substituting the value of α + β + γ and αβγ, you will find
1
that
(α + β + γ )
= (αβγ )3
3
⇒ A.M. = G.M., when A.M = G.M then α = β = γ
Therefore αβ + βγ + γα = 3
Hence, choice (d) is the answer.
Alternatively
Since α + β + γ = 3, αβγ = 1 and α > 0, β > 0, γ > 0
Alternatively
⇒
( p − 4)( p − 12) > 0
⇒
p ∈ (−∞, 4) ∪ (12, ∞ )
102 f ( x ) = ( p + 3)x + ( p + 2)x − 5 < 0
2
2
Since f ( x ) < 0 for at least one value of x, and a > 0, then
D > 0.
Therefore ( p + 2)2 − 4( p2 + 3) × (−5) > 0
⇒
21 p2 + 4 p + 64 > 0
Thus the given function f ( x ) < 0 for at least one x, if
p ∈ (−∞, ∞ )
Hence choice (b) is the answer.
100. Let f ( x ) = x + ax + bx − 4 x + 1.
4
3
...(ii)
Now you have to find out the values of p for which the
above expression (ii) is positive. For this you would have
to again find out the discriminant of expression (ii)
which is negative. Since in quadratic expression (ii), the
coefficient of x 2 is positive and discriminant is negative,
the quadratic expression (ii) will be positive for all real
numbers. That is for all real values of p, the discriminant,
D > 0.
It is possible only when α = β = γ = 1.
Therefore αβ + βγ + γα = 3
...(i)
2
Since there are four positive roots, it implies that there
will be four changes in the sign among the various terms
of the function, when the terms are written in
increasing/decreasing order of degree. So a must be
negative and b must be positive.
Hint p2 + 3 > 0.
Alternatively
If you can find the real values of p so
that f ( x ) ≥ 0 for all x, then for the remaining real values
of p the function f ( x ) < 0 for at least one x.
f ( x ) = ( p2 + 3)x 2 + ( p + 2)x − 5 ≥ 0
Now, since f ( x ) ≥ 0 for all x and a > 0, then D ≤ 0.
...(i)
Theory of Equation
857
Therefore ( p + 2)2 − 4( p2 + 3) × (−5) ≤ 0.
⇒
21 p + 4 p + 64 ≤ 0
2
105 Case (i) When x > 0, the given equation will be
...(ii)
Since the coefficient of p2 is positive and the discriminant
of quadratic expression (ii) is negative, then the above
quadratic inequality (ii) is not valid for any real value of p
It implies that when p ∈ (−∞, ∞ ), the given function f ( x )
is not positive for any value of x.
It means when p ∈ (−∞, ∞ ), the given function f ( x )is
negative for at least one x. Hence choice (b) is the answer.
103 let f ( x ) = ax 2 + bx + c.
⇒ f (x) = x 2 +
Since α < − p and β > p. Therefore f (− p) < 0 and f ( p) < 0.
And when
⇒
f ( p) < 0
b
c
p + p+ <0
a
a
2
...(ii)
...(i)
Similarly since β is a root of − ax 2 + bx + c = 0, therefore β
Case (i) When ( x + 4 x + 2) is positive, the given equation
will be 3x 2 + 12x + 6 = 5x + 16 ⇒ 3x 2 + 7 x − 10 = 0
⇒
⇒
⇒
⇒
Now
⇒
...(ii)
a 2
a
x + bx + c = 0 ⇒ f (α ) = α 2 + bα + c = 0
2
2
a 2
2
[from eq. (ii)]
f (α ) = α − aα
2
a
...(iii)
f (α ) = − a2
2
a
[from eq. (ii)]
f (β ) = β 2 + bβ + c = 0
2
a
...(iv)
f (β ) = β 2 + aβ 2
2
3a 2
f (β ) =
β
2
3
f (α)⋅ f (β ) = − a2α 2β 2
4
f (α )⋅ f (β ) < 0
−10
.
3
Case (ii) When ( x 2 + 4 x + 2) is negative, the given
equation will be
The roots are −2 and
−11
.
3
−10
−11
and
do not satisfy the given Eq. (i), so
3
3
you have to discard them. Thus the valid values of x are
1 and −2. Hence choice (d) is the answer.
Since
(Qα , β ≠ 0)
It implies that f (α ) and f (β ), have opposite signs.
Therefore equation f ( x ) = 0 has exactly one root between
α and β. Hence choice (a) is the answer.
f (x) = 0
Now if
⇒
(| x − 1| − 3) = 0
Case (i) When x − 1 > 0, then ( x − 1) − 3 = 0
⇒
x=4
Case (ii) When x − 1 < 0, then
−( x − 1) − 3 = 0
Let f ( x ) =
⇒
...(i)
2
108 Let f ( x ) = (| x − 1| − 3)
104 Since α is a root of ax 2 + bx + c = 0, therefore α will satisfy
− aβ 2 + bβ + c = 0
106 The given equation is 3| x 2 + 4 x + 2| = 5x + 16
All the values of choice (d) satisfy the equation, so it’s
the answer. Hence choice (d) is the answer.
Hence choice (a) is the answer.
will satisfy it as
Hence choice (c) is the answer.
107 The best way is to go through the given choices.
b1
c
+ 2 <0
a p ap
it as aα 2 + bα + c = 0
Thus there are total four roots, viz. −2, − 1, 1, 2.
−3x 2 − 12x − 6 = 5x + 16 ⇒ 3x 2 + 17 x + 22 = 0
...(i)
From eqs. (i) and (ii), we can conclude that
p + c < 0
b
f ( x ) = p2 + 
a
a
f (x) = 1 +
Case (ii) When x < 0, the given equation will be
x 2 + 3x + 2 = 0 and then the roots will be −1, −2
The roots are 1 and
b
c
x+
a
a
When f (− p) < 0
b
c
p2 − p + < 0
⇒
a
a
x 2 − 3x + 2 = 0 and then the roots will be 1, 2.
⇒
x =−2
After testing for the signs you will get
f ( x ) < 0, if x ∈ (−2, 4) and f ( x ) > 0, if
x ∈ (−∞, −2) ∪ (4, ∞ )
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 1
–1
–2
–3
–4
–5
–6
–7
++++++++++++++
...(i)
2
3
––––––––––
–2
4
5
6
7
8
9
+++++++++++
4
858
QUANTUM
Let
g( x ) = (| x + 2| − 5)
Now if
g( x ) = 0
109 Case (i) When both the numerator and denominator have
the same sign. That is
x 2 − 5x + 4
−( x 2 − 5x + 4)
or
2
x −4
−( x 2 − 4)
⇒ (| x + 2| − 5) = 0
Case (i) When x + 2 > 0, then
f ( x ) = ( x + 2) − 5 = 0
⇒
Therefore the given inequation is
x=3
Case (ii) When x + 2 < 0, then
⇒
f ( x ) = − ( x + 2) − 5 = 0
⇒
x = −7
After testing for the signs you will get
g( x ) < 0, if x ∈ (−7, 3) and g( x ) > 0
x ∈ (−∞, − 7 ) ∪ (3, ∞ )
if
...(ii)
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 1
–1
–2
–3
–4
–5
–6
–7
+++
8
The critical points are −2, 2 and . So there are four
5
8  8 

intervals (−∞, − 2),  −2,  ,  , 2 (2, ∞ )

5  5 
––––––––––
++++++
––––––
8/5
–2
2
Since the rational function is non positive, therefore you
8

have to consider only two intervals:  −2,  , and (2, ∞ )

5
2
3
4
5
6
7
8
9
Case (ii) when the signs of numerator and denominator
are different. That is
−( x 2 − 5x + 4) ( x 2 − 5x + 4)
or
−( x 2 − 4)
( x 2 − 4)
Therefore the given inequation is
− ( x 2 − 5x + 4)
≤1
x2 − 4
++++ ++++ ++++ +
3
–7
Now since you know that f ( x )g( x ) < 0, only when f ( x )
and g( x ) are of opposite signs.
⇒
That means f ( x )g( x ) < 0, if f ( x ) > 0 but g( x ) < 0 or
f ( x ) < 0 but g( x ) > 0.
− ( x 2 − 5x + 4)
−1 ≤ 0
x2 − 4
⇒
− x(2x − 5)
≤0
( x + `2)( x − 2)
So combining the results of (i) and (ii), we can conclude
that f ( x )g( x ) < 0, if x ∈ (−7, − 2) ∪ (3, 4).
The same result can be easily seen by the following graph.
5
The critical points are −2, 2, 0 and . So there are five
2
 5  5 
intervals (−∞, − 2), (−2, 0), (0, 2),  2,  ,  , ∞
 2  2 
––––––––
–2
2
x 2 − 5x + 4
≤1
x2 − 4
x 2 − 5x + 4
−5 x + 8
≤0
−1 ≤ 0 ⇒
( x + 2)( x − 2)
x2 − 4
+++++++++
–––––––––––––––––
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 1
–1
–2
–3
–4
–5
–6
–7
CAT
3
4
5
6
7
8
9
++++++
––––––
0
++++
2
––––––
5/2
Since the rational function is non-positive, therefore you
have to consider only three
5 
intervals: (−∞, − 2), (0, 2) and  , ∞
2 
Therefore the intersection of both the cases will be your
desired result.
That is [ 0, 8 / 5] ∪ [ 5/ 2, ∞] .
Hence choice (d) is the answer.
Alternatively The best way is to go through the given
choices. All the values of x that lie between (−7, − 2) and
(3, 4) satisfy the given equation.
Hence choice (d) is the answer.
Hence choice (b) is the answer.
NOTE Since the given inequality is not a strict
inequality, so the boundary values of the chosen intervals
will have to be included. The best way is to cross check by the
substitution of boundary values in the given (original)
expression, in case of any uncertainty.
Theory of Equation
859
110 Case (i) when both the numerator and denominator have
the same sign. That is
graph of the given polynomial or determine the sign of
each interval on the number line (or X-axis).
1
The real roots of the given polynomial are 2, − 3,
2
10
and .
3
x 2 − 3x − 1
−( x 2 − 3x − 1)
or
2
x + x+1
−( x 2 + x + 1)
Therefore the given inequation is
x 2 − 3x − 1
x 2 − 3x − 1
<3 ⇒ 2
− 3< 0
2
x + x+1
x + x+1
⇒
– – – – – – ++++++ ++++++++
−2( x + 2)( x + 1)
<0
x2 + x + 1
–3
The critical points are −2, and −1, since denominator
does not yield any real root. So, there are three intervals
(−∞, −2), (−2, − 1), (−1, ∞ )
++++++++++++
–––––––
–2
––––––––––
–1
Case (ii) when the signs of numerator and denominator
are different. That is
−( x 2 − 3x − 1)
x 2 − 3x − 1
or
2
−( x 2 + x + 1)
x + x+1
Therefore the given inequation is
−( x 2 − 3x − 1)
<3
x2 + x + 1
−( x 2 − 3x − 1)
−2(2x 2 + 1)
− 3< 0 ⇒ 2
<0
2
x +x +1
x + x+1
Since both the numerator and denominator are unable to
give any real root, so there is no critical point for this
rational function. Further since the coefficient of x 2 of
the numerator is negative so the whole graph will be
negative as roots are non-real.
–––––––––––––––––––––––––––––––––––––––
–∞
∞
Since the rational function is negative, therefore all the
real values satisfy the inequality. That is (−∞, ∞ )
Therefore the intersection of both the cases will be your
desired result. That is (−∞, − 2) ∪ (−1, ∞ ).
10/3
Now there are two methods to determine the sign of the
interval. First, either test each interval individually by
substituting any suitable numerical value from that
interval in the given polynomial or follow the ensuing
technique.
So the first interval (that is the left most) will be negative.
After that −3 is a root of odd multiplicity so after −3, the
sign of that interval will change to positive. But after that
1
− is a root of even multiplicity so there won’t be any
2
1
change in sign after − . Further 2 is a root of odd
2
multiplicity so the sign will change after 2 from positive to
10
negative. Finally,
is again a root of odd multiplicity so
3
10
sign will change from negative to positive. Now
after
3
since x = − 1010 lies in first interval which is negative and
x = 1010 lies in the last interval which is positive, so the
product of negative and positive is negative, that is
f (−1010) f (1010) < 0. Therefore (i) is correct.
Similarly, f (−2) > 0 but f (3) < 0, so f (−2) f (3) < 0.
Therefore (ii) is also correct.
It means choices (a), (b) and (c) are wrong, as they say
that (i) and (ii) are incorrect, which is not.
Hence choice (d) is the answer.
113 Since α and β are the roots of x 2 − 6 x − 2 = 0. therefore
α 2 − 6α − 2 = 0 ⇒ α 2 − 2 = 6α
Hence choice (b) is the answer.
must be a real root between p and q through which graph
intercepts the X-axis and so it changes the sign of the graph
from positive to negative or negative to positive.
As you know that a real root (with odd multiplicity)
changes the sign of graph – that is the two points, one
lying in the left side and the other one lying in the right
side of the real root, have different signs.
...(i)
β − 6β − 2 = 0 ⇒ β − 2 = 6β
2
111 Since the sign of f ( p)and f (q)are opposite, it means there
Hence choice (a) is the answer.
– – – +++++ +++
2
1/2
After simplification, the highest power (or degree) of this
equation is 9, that is odd. The coefficient of the highest
power of x is a positive integral.
Since the rational function is negative, therefore you
have to consider only negative values. That is
x ∈ (−∞, − 2) ∪ (−1, ∞ ).
⇒
112 The best way to solve this problem is to sketch a rough
Now we have,
2
a10 − 2a8 α
=
2a9
10
...(ii)
− β − 2(α − β )
2(α 9 − β 9 )
10
8
8
=
(α 10 − 2α 8 ) − (β10 − 2β 8 ) α 8(α 2 − 2) − β 8(β 2 − 2)
=
2(α 9 − β 9 )
2(α 9 − β 9 )
=
α 8(6α ) − β 8(6β ) 6(α 9 − β 9 )
=3
=
2(α 9 − β 9 )
2(α 9 − β 9 )
Hence choice (c) is the answer
860
114
QUANTUM
3 f 9 − 14 f 6 3(a9 − b9 ) − 14(a6 − b6 )
=
f7
a7 − b7
=
116 The maximum number of positive real roots of a
polynomial equation f ( x ) = 0 is the number of changes of
signs − from positive to negative and / or negative to
positive − in the coefficients of f ( x ).
a (3a − 14) − b (3b − 14)
a7 − b7
6
3
6
3
Since a and b are the roots of 3x 3 − 8 x − 14 = 0, therefore
Coefficient /constants + 4 + 3
And, 3b3 − 8b − 14 = 0 ⇒ 3b3 − 14 = 8b
+
Sign
3 f − 14 f 6 a6 (8a) − b6(8b)
Therefore, 9
=
f7
a7 − b7
8a7 − 8b7 8(a7 − b7 )
=8
= 7
a − b7
a7 − b7
Hence choice (a) is the correct one.
115 The best way is to solve through the given choices.
Choice (a)
3
4 x 3 + 3x 2 + 2x + 1
Equation
you will have 3a3 − 8a − 14 = 0 ⇒ 3a3 − 14 = 8a
=
CAT
2
10
 1
 1
 1
 1
f  −  = 4 −  + 3 −  + 2  −  + 1 =
 4
 4
 4
 4
16
 1
⇒ f −  > 0
 4
Since f (−11)⋅ f (−3 / 4) > 0, therefore it implies that
between −11 amd −3/4 either there are two real roots or
there is not real root at all.
Cloice (C) f (−3 / 4) < 0 and f (−1 / 2) > 0
 −3  −1
Since f   . f   < 0, therefore it implies that
 4  2
−3
−1
between
and
either there are three real roots or
4
2
there is one real root. In essence, there is certainly at
least one real root. Hence choice (c) is the answer.
Hint When f ( p). f (q) < 0, then there is at least one real
root between p and q.
Choice
f ( p)
f (q)
f ( p).
f (q)
Possible
number of
roots
(a)
>0
>0
>0
0 or 2
May or may not be
(b)
<0
<0
>0
0 or 2
May or may not be
(c)
>0
<0
<0
1 ro 3
At least one root is
there
(d)
>0
>0
>0
0 or 2
May or may not be
Conclusion
+1
+ +
Since there is no change in sign of coefficients/constant
values, so there won’t be any positive real root.
Hence choice (c) is the answer.
α β
117 Product of roots = × = 1
β α
Now since α 3 + β 3 = (α + β )(α 2 + β 2 − αβ )
⇒
q = (− p)[(α + β )2 − 3αβ] ⇒ q = (− p)[(− p)2 − 3αβ]
⇒ αβ =
1  2 q
p + 
3
p
Therefore sum of roots =
Similarly f (0) > 0
 1
Since f  −  ⋅ f (0) > 0 therefore it implies that between
 4
1
− and 0 either there are two real roots or there is no
4
real root at all.
 −3
Choice (b) f (−11) < 0 and f   < 0
 4
+
+2
=
(α + β ) − 2αβ
=
αβ
2
α β α 2 + β2
+ =
β α
αβ
2  2 q
p + 
p p3 − 2q
3
= 3
p +q
1  2 q
p + 
3
p
p2 −
 p3 − 2q
So the required equation is x 2 −  3
 x +1 = 0
 p + q
⇒
( p3 + q)x 2 − ( p3 − 2q)x + ( p3 + q) = 0
Hence choice (b) is the answer.
118 x 2 + 20 ≤ 9 x
⇒ x 2 − 9 x + 20 ≤ 0 ⇒ ( x − 4)( x − 5) ≤ 0 ⇒ 4 ≤ x ≤ 5
f (4) = − 4 ⇒ f (4) < 0 and f (5) = 7 ⇒ f (5) > 0
It implies that there will be either only one real root or
all the three roots between 4 and 5.
Using rule of signs you get to know that there is no
negative root so all the roots are positive. If all the three
roots lie between 4 and 5, then their sum will lie between
12 and 15. But since the sum of three roots is 15/2 (or
7.5), it means only one root lies between 4 and 5.
Since f (4) < f (5), it means the highest value of f ( x ) will
be given by f (5) , which is 7.
Hence choice (a) is the answer.
119 y = −
x2
3
1
+ x + 1 ⇒ y − = − ( x − 1)2
2
2
2
It implies that the curve is symmetric about x = 1.
Hence choice (a) is the answer.
Theory of Equation
861
x 2 − 6 x + 5 ( x − 1)( x − 5)
⇒
x 2 − 5x + 6 ( x − 2)( x − 3)
120 f ( x ) =
Since f ( x ) is a cubic polynomial which either have 1 real
root or 3 real roots. It means all 3 roots are real.
Therefore statement (ii) is true.
The critical points are 1, 2, 3 and 5.
Using the given information you can sketch the graph of
f ( x ) similar to the one as shown below.
The vertical asymptotes are x = 2, x = 3
Horizontal asymptote is y = 1.
Partial graph of the given function is shown below.
Otherwise, test the nature of the function in various
intervals.
(P) When −1 < x < 1 ⇒ 0 < f ( x ) < 1
(Q) When 1 < x < 2 ⇒ f ( x ) < 0
–1
1
2
(R) When 3 < x < 5 ⇒ f ( x ) < 0
(S) When x > 5 ⇒ 0 < f ( x ) < 1
Therefore (P)→ p, r, s; (Q) → q, s ; (R)→ q, s; (S)
→ p, r, s
f ′ ( x ) = 3ax 2 + 2bx + c is a quadratic function. It gives
−b
minimum / maximum when x = . But it is given that
2a
f ′ ( x ) has local minimum at x = 0. It implies that b = 0
4
Therefore f ′ ( x ) = 3ax 2 + c
3
Since at x = − 1, f ( x ) has local maxima, it means at
2
1
x = −1, f ′ ( x ) = 0
0
That is 3a(−1) + c = 0 ⇒ c = − 3a
2
–3
–2
–1
1
2
3
4
5
6
7
8
Therefore f ′ ( x ) = 3ax 2 − 3a = 3a( x 2 − 1)
–1
–2
Since a ≠ 0, x 2 − 1 = 0 ⇒ x = ± 1.
–3
Since at x = ± 1, f ′ ( x ) = 0, therefore at x = ± 1, f ( x ) will
have its local maxima / minima.
–4
Hence choice (a) is the answer.
121 x 2 − 10cx − 11d = 0
...(i)
a + b = 10c and ab = − 11d
x 2 − 10ax − 11b = 0
...(ii)
c + d = 10a and cd = − 11b
So (a + b) + (c + d ) = 10(a + c) ⇒ b + d = 9(a + c)
...(iii)
And (ab)(cd ) = 121bd
⇒
ac = 121
...(iv)
So
2
⇒
(a + c) = 121 or (a + c) = − 22
⇒
a + b + c + d = 10(a + c) = 1210
Therefore statement (iv) is true.
Hint As you have determined b = 0 and c = − 3a, now you
can easily find out the relevant polynomial as following
a2 + c2 − 20ac − 11(b + d ) = 0
⇒ (a + c)2 − 22(121) − 99(a + c) = 0
It means f (1) = − 1 is the local minima. So f ( x ) is
increasing for [1, ∞ ). Therefore statement (i) and (iii) are
true.
−b
As you have just found that b = 0, it means
= 0 ⇒ Sum
a
of roots is zero.
Hence choice (c) is the answer.
Now a − 10ac − 11d = 0 and c − 10ac − 11b = 0
2
As, f (−1) already has local maxima, so f (1) will have
local minima.
...(v)
Hence choice (c) is the answer.
NOTE Equation (v) is a quadratic Equation. The value of
( a + c) = − 22 has to be discarded otherwise a = − 11 and
c = − 11, which is not possible as a and c have to be distinct.
122 Since f (1) f (2) < 0, it implies that at least one real root of
f ( x ) lies between 1 and 2.
Also f (−1) > 0 and f (1) < 0 ⇒ f (−1) f (1) < 0, it implies at
least one real root lies between −1 and 1.
f ( x ) = ax 3 + bx 2 + cx + d ⇒ f ( x ) = ax 3 − 3ax + d
∴
f (2) = 18 ⇒ 8a − 6a + d = 18
⇒
2a + d = 18 and f (1) = − 1 ⇒ a − 3a + d = − 1
19
17
57
, c =−
⇒
−2a + d = − 1
⇒ a= ,d =
4
2
4
1
3
Thus f ( x ) = (19 x − 57 x + 34)
4
123 f ( x ) = ax 3 + bx 2 + cx + d
...(i)
f (−1) = − a + b − c + d = 10
f (1) = a + b + c + d = − 6
...(ii)
...(iii)
From eqs. (ii) and (iii), b + d = 2 and a + c = − 8
...(iv)
862
QUANTUM
Now, f ′ ( x ) = 3ax 2 + 2bx + c is a quadratic function. The
But it is given that f ′ ( x ) has local minimum at x = 1.
−(2b)
...(v)
That is
1=
⇒ b = − 3a
2(3a)
Therefore f ′ ( x ) = 3ax 2 − 6ax + c
Now, since f ( x ) has local maxima at x = − 1, it implies
that at x = − 1, f ′ ( x ) = 0. That is f ′ (−1) = 0
3a + 6a + c = 0 ⇒ c = −9a
Now the discriminant of f ′ ( x ) is
D = 4b2 − 12c
quadratic function has max / min at
−(coefficient of x )
.
x=
2(coefficient of x 2 )
⇒
...(vi)
From eqs. (vi), (iv) and (v), we have
⇒
D = 4(b2 − c) − 8c < 0
⇒
D<0
Since the quadratic function f ′( x ) has no real root, so
the cubic function f ( x ) will have no local
maxima/minima. It implies that either f ( x ) is strictly
increasing or strictly decreasing.
But since the coefficient of x 3 in f ( x ) is positive so the
function f ( x ) is strictly increasing in (−∞, ∞ ).
Hence choice (a) is the answer.
126 min f ( x ) = 2c2 − b2 and max g( x ) = b2 + c2
a = 1, b = − 3, c = − 9, d = 5
Therefore, 2c2 − b2 > b2 + c2 ⇒ c2 > 2b2 ⇒ | c| > 2| b|
Therefore f ( x ) = x 3 − 3x 2 − 9 x + 5 and
Hence choice (d) is the answer.
f ′ ( x ) = 3x 2 − 6 x − 9.
Since f ′ ( x ) = 0 ⇒ 3x 2 − 6 x − 9 = 0 ⇒ 3( x + 1)( x − 3) = 0
As f ′ ( x ) = 0 at x = − 1. It means at x = − 1, f ( x ) is
maximum, as already given.
Similarly, f ′ ( x ) = 0 at x = 3. It means at x = 3, f ( x ) is
minimum, as required.
f (−1) = 10 and f (3) = − 22. Therefore distance between
(−1, 10) and (3, − 22) is 4 65.
Hint The quadratic function has max/min at
−(coefficient of x )
.
x=
2(coefficient of x 2 )
Otherwise,
f ( x ) = x 2 + 2bx + 2c2 = ( x + b)2 + 2c2 − b2
f ( x ) is minimum when ( x + b)2 = 0 at x = − b
and g( x ) = − x 2 − 2cx + b2
Hence choice (a) is the answer.
b2 + c2 − ( x + c)2
Hint The distance between two points ( x1, y1 ) and
g( x ) is maximum when ( x + c)2 = 0 at x = − c.
( x 2, y 2 ) is given by
127 Since both the roots are less than the arbitrary point k then
d = ( x 2 − x1 )2 + ( y 2 − y1 )2
you must have to satisfy the following three conditions.
124 What if a root, say β, is zero! Then
(i) D ≥ 0 ⇒ 4a2 − 4(a2 + a − 3) ≥ a ≤ 3
α + β, α 2 + β 2, α 3 + β 3 ⇒ α, α 2 , α 3 , which are naturally in
GP. So, α + β, α 2 + β 2, α 3 + β 3 will also be in GP.
∆ = b2
⇒ c∆ = 0(b2 ) = 0
128 ( x − 9)( x − 16) = x 2 − 25x + 144
If x, y, z are in GP, then y 2 = xz.
Since the coefficient of x 2 is positive, it means the graph
will open upwards. So the values between roots will be
included only when, x 2 − 25x + 144 ≤ 0
(α 2 + β 2 )2 = (α + β )[α 3 + β 3]
⇒ α 4 + β 4 + 2α 2β 2 = α 4 + β 4 + αβ(α 2 + β 2 )
Hence choice (a) is the answer.
⇒ 2(αβ )2 = αβ(α 2 + β 2 ) ⇒ αβ(α 2 + β 2 − 2αβ ) = 0
129 Let the common root be r, then
2
4c 
c  − b
⇒ αβ[(α + β ) − 4αβ] = 0 ⇒   −  = 0
a  a 
a 
f = ( x − r)( x − 2) = 0 ⇒ f = m[( x − r)( x − 2)] = 0
2
and g = ( x − r)( x − 7 ) = 0 ⇒ g = n[( x − r)( x − 7 )] = 0
c  b2 − 4ac 
2
 = 0 ⇒ c(b − 4ac) = 0 ⇒ c∆ = 0

a  a2 
125 Q f ( x )= x 3 + bx 2 + cx + d
∴ f ′ ( x ) = 3x 2 + 2bx + c
b
−2a
⇒3> −
⇒a< 3
2a
2
Hence choice (a) is the answer.
Therefore,
⇒
(iii) k > −
The intersection of (i), (ii) and (iii) is a ∈ (−∞, −2)
Hence choice (c) is the answer.
Alternatively
(ii) af (k ) > 0 ⇒ 1. f (3) > 0
⇒ 9 − 6a + a2 + a − 3 ⇒ a ∈ (−∞, 2) ∪ (3, ∞ )
And when one root is zero, product of roots will be zero.
That is c = 0.
So
CAT
When, m and n are non-zero real numbers.
Since,
...(i)
...(ii)
f (4) × g(9) = 24
⇒ m(4 − r)(4 − 2) × n [(9 − r)(9 − 7 )] = 24
⇒ mn(4 − r) × (9 − r) = 6
⇒
mn(r2 − 13r + 36) = 6
Theory of Equation
⇒
(r2 − 13r + 36) =
863
6 
6

⇒ r2 − 13r +  36 −
 =0

mn
mn
A root is rational only when discriminant D is a perfect
square.
That is D = 0, 1, 4, 9, K
As you can see that r depends on m and n, so there is no
unique solution.
 24 
D = 25 + 

 mn
⇒ mn = −
But since one root in each of the equations f ( x ) = 0 and
g( x ) = 0 is real and rational, so the other root, in each
equation, must be real and rational.
Q
A root is real only when discriminant,
Q
D ≥ 0. That is b2 − 4ac ≥ 0
It means, (i), (ii) and (iii) are invalid.
It implies that D ≠ 25.
6 

∴ 169 − 4 36 −
 ≥0

mn
r=
−13 ± D
⇒ r ≠ −4, − 9
2
Further, since D is a perfect square, so D will be a
non-negative integer. Therefore the values of r would be
rational numbers not the irrational ones. Therefore (iv) is
also invalid. Hence choice (c) is the correct one.
24 

 ≥0
 25 +

mn
⇒
24
(25 − D )
Level 02 Higher Level Exercise
1 The best way is to go through options.
Consider option (b)
| 34 − 1|log 3 ( 3
4 2
) − 2 log 81 9
| 80|log 3 3
⇒
8
− log 81 81
Hence option (b) is correct.
1
2 Putting x = , we get 27 y 3 + 54 y 2 + cy − 10 = 0
y
Σα = α − β + α + α + β = 3α
−2
−54
⇒
3α =
⇒ α=
27
3
−2
will satisfy the Eq. (i) and we get
Now α =
3
4 2c
−8
27 ×
+ 54 × −
− 10 = 0 ⇒ c = 9
27
9 3
1
3
log100 | x + y | =
⇒ (100)1/ 2 = | x + y |
2
2 log 2 log 2 x − log 2 log 2 (2 2x ) = log 2 2
⇒
log 2 (log 2 x )2 − log 2 [log 2 2 2x] = log 2 2
⇒
log 2
(log 2 x )2
(log 2 x )2
= log 2 2 ⇒
=2
[log 2 2 2x]
log 2 (2 2x )
log 2 x = 3 ⇒ x =
1
or
2
x=8
1
But for x = , log 2 log 2 (1 / 2) is undefined
2
∴ Only possible value of x = 8.
then
…(ii)
(Q x > 0)
10
3
⇒
5 Consider x 2 + 4 x + 3 ≥ 0
| x + y | = 10 ⇒ | x + 2| x || = 10
x + 2x = 10 ⇒ x =
10
20
and y =
.
3
3
2 log 2 log 2 x + log1/ 2 log 2 (2 2x ) = 1
⇒
From eq. (ii) we can conclude that y is always positive.
Now, when x > 0 and y > 0 (always)
⇒
x=
⇒ (log 2 x )2 − 2 log 2 x − 3 = 0 or log 2 x = − 1
log10 y − log10 | x | = log100 4
y
= log10 2
y − log10 | x | = log10 2 ⇒ log10
| x|
x + 2| x | = 10
y = 20 and
(Q x < 0)
⇒ (log 2 x )2 = 2[ 3/ 2 log 2 x + log 2 x] = 3 + 2 log 2 x
Again,
⇒
x = − 10
⇒ (log 2 x )2 = 2 log 2 (2 2x ) = 2 log 2 (23/ 2 x )
…(i)
y = 2| x |
y = 20
4
∴
⇒
| x | = 10 ⇒
…(i)
(Q roots are in AP)
| x + y | = 10
⇒
∴
Hence, x = − 10,
This above eq. (i) must be in AP.
Let the roots of equation in y be
log10
| − x + 2| − x || = 10 ⇒| − x + 2x | = 10
= (80)7 ⇒ log 3 38 − log 81 81 = 7
8 − 1 = 7 and 7 = 7
⇒
20
3
Again, x < 0 and y > 0 (always positive)
= (34 − 1)7
α − β, α, α + β
y=
∴
…(i)
( x + 4 x + 3) + 2x + 5 = 0
2
⇒
x = − 2 and
x = −4
but x = − 2 does not satisfy eq. (i)
Now, if
x2 + 4x + 3 < 0
then
⇒
or
− ( x + 4 x + 3) + 2x + 5 = 0
2
x = −1 − 3
x = −1 + 3
…(ii)
864
QUANTUM
8 Let us consider some value of p = 3 (say), then
but only x = − 1 − 3 satisfies the eq. (ii).
Hence the solution set of x is (−4, − 1 − 3).
x2 − 4x + 1
Alternatively, check the options by substituting the values
from the options given in the question.
6 Q x1, x 2, x 3 are in AP.
⇒
x1 = a − d,
Σα = x1 + x 2 + x 3 = 1
…(i)
(a − d ) + (a) + (a + d ) = 1
Σαβ = x1 x 2 + x 2 x 3 + x1 x 3
⇒
β = (a − d )a + a(a + d ) + (a − d )(a + d ) …(ii)
and
Σ = αβγ = x1 x 2 x 3 = − γ = (a − d )(a)(a + d ) …(iii)
1
hence from eq. (i) we get a =
3
and from eq. (ii) we get
1
1

β = 3a2 − d 2 ⇒ β = − d 2
Q a = 

3
3
1
1
β = − d2 ≤
3
3
1
1

or β ∈  −∞, 
β≤

3
3
[Q d ≥ 0]
(Q d 2 ≥ 0)
Hence option (a) is correct.
Similarly for p = 4, 5, 6, K etc. we can conclude the same
results.
NOTE In this question, the discriminant D is always
positive, i.e., b2 − 4 ac > 0. So the roots will always be real,
unequal and irrational.
But the fact is that α n + β n , for n ∈N , always yields integral
value. This can be easily proved by mathematical induction
method. So if the answer is an integer, then it must be a
rational number, hence option (d) is correct.
9 Just assume some values of α, β conforming the basic
constraints of the problem.
e.g., α = − 2, β = 8, then the equation becomes
x 2 − 6 x − 16 ⇒ b = − 6 and c = − 16
∴
(c) → sin x > x
(d) → ex < x ⇔ log e x > x
α < 1 and β > 1
c
<1
∴
αβ < 1 ⇒
a
Further the product of any two numbers (n1 , n2 ≠ 0) is less
than the sum of the number if any one of them is negative.
αβ < α + β
⇒
NOTE e ≈ 2.714 and e2 ≈ 7.366
(QHere αβis negative)
1+
c
b
+
< 0.
a
a
10 Since p and q are the roots of given equation
x 2 + px + q = 0 then p + q = − p ⇒ q = − 2p
8
and
e2
pq = q ⇒
p=1
So, when p = 1, then q = − 2.
6
Again, when q = 0, then p = 0 hence,
5
p = 1, 0 and q = − 2, 0
4
Thus, option (b) is most appropriate.
3
e
2
1
11 Q p, q, r are in AP.
∴
–1
c
b
+
= 1 − 16 + 6 = − 9
a
a
c
b
c
b
; but is numerically greater than .
∴ <
a
a
a
a
(b) → ex > (1 + x ) ⇔ log e (1 + x ) < x
–2
1+
So,
7 (a) → ex < 1 + x
–3
…(i) (n ∈ N )
NOTE Since
1
d3
−1
⇒γ ≥
−
3 27
27
7
3) + (2 − 3)n
n
∴The value of the expression is negative, hence choice
(a) is correct.
2
 1  d 
a(a2 − d 2 ) = − γ ⇒   +  −  = − γ
 27   3 
γ=
(α, β are the roots)
3
n
2
Again from eq. (iii), we get
⇒
Now,
n
statement you can put n = 1, 2, 3, K etc. in Eq. (i).
So,
⇒
⇒
α + β = (2 +
n
Then, α + β will always be an integer, for the validity of
x3 = a + d
Where d is the common difference.
Now, since x1, x 2, x 3 are the roots of the given equation
x 3 − x 2 + βx + γ = c
Thus
(α , β ) = 2 ±
and
n
x 2 = a,
CAT
0
1
2
3
4
X
From the graph it is clear that when 0 < x < 1, e x > (1 + x )
Option (b) is clearly wrong.
q=
p+ r
2
[Q p + r = 2q]
For the real roots q2 − 4 pr ≥ 0
2
⇒
 p + r
2
2

 − 4 pr ≥ 0 ⇒ p + r − 14 pr ≥ 0
 2 
Theory of Equation
865
2
 p
 p
  − 14   + 1 ≥ 0
 r
 r
⇒
NOTE You should not worry about this technique. It
must be clear to you that while assuming the values of α , β,
you have to strictly follow the basic given constraints, i.e., α
must be less than β and after forming the similar equation
you should have c < 0 < b. So I think it is even very intelligent
method to solve the problems rather than conventional
methods.
2

p
 − 7 ≥ 48

r
⇒
p
−7 ≥ 4 3
r
⇒
16 Let us assume a = 3, b = 4, given that a < b then the given
12 The given equation is| x − 2|2 + | x − 2| − 2 = 0.
equation becomes
Let us assume| x − 2| = m
Then,
( x − 3)( x − 4) − 1 = 0 or x 2 − 7 x + 11 = 0
m2 + m − 2 = 0 or (m − 1)(m + 2) = 0
49 − 44
7± 5
⇒ x=
2
2
7− 5
7+ 5
<3
x=
> 4 and
⇒
3
2
Hence only option (d) is satisfied, hence correct.
∴
Only admissible value is
m =1
(Q m ≠ − 2 as m ≥ 0]
∴
| x − 2| = 1
⇒ x − 2 = 1 ⇒ x = 3 or − ( x − 2) = 1 ⇒ x = 1
Hence
x = 1, 3
x=
Alternatively
∴ Sum of the roots of equation = 1 + 3 = 4.
⇒
13 Just consider an option, and then substitute the values of A
7±
Since f ( x ) = ( x − a)( x − b) − 1 = 0
( x − a)( x − b) = 1
Y
and B from assumed option, if the roots p, q, r, s are in AP,
then the assumed option is correct, else not.
Thus we find that options a, b and c are incorrect.
a<b
Hence (d) is the right choice.
Alternatively
Let us consider p = a − 3d, q = a − d,
α
r = a + d. and s = a + 3d
∴
2a − 4d = 2
…(i)
and
2a + 4d = 18
…(ii)
p = − 1,
A = pq = − 3 and B = rs = 77
q = 3,
r = 7, s = 11
Thus (d) is the right choice
14 Let
f ( x ) = x 2 − 2ax + a2 + a − 3
Since f ( x ) has real roots, both less than 3.
Therefore, D > 0 and f (3) > 0
⇒ a2 − (a2 + a − 3) > 0 and a2 − 5a + 6 > 0
⇒ a < 3 and (a − 2)(a − 3) > 0
⇒ a < 3 and a < 2 or a > 3 ⇒ a < 2
and
c = −6
Now, we check for the given choices, which satisfy the
aforesaid conditions
(a) It is clearly wrong
(b) It is correct
(c) It is also wrong
(d) It is also wrong
Hence option (b) is appropriate.
X
αβ = p and γδ = q
17
Now since, α , β, γ, δ are in GP and integral values. So the
options (b) and (c) are ruled out, as they have no required
proper integral factors. Now let us look for option (a). We
see that
αβ = − 2 = − 1 × 2
γδ = − 32 = − 4 × 8
So, −1, 2, − 4, 8 are in GP satisfying the above conditions.
Again in option (d) the two values don’t have the factors
with common ratio, hence it is wrong. Finally option (a)
is correct.
15 Considering the given constraints in the problem.
Let us consider α, β = (−3, 2)
Then the given equation becomes
x2 + x − 6 = 0 ⇒ b = 1
β
It means X-axis is shifted to –1 unit below the original
position. So it is clear from the graph that α < a and β > b.
a = 5, d = 2
Hence
b
α < a and β > b
Solving equations (i) and (ii), we get
∴
a
Alternatively
The sum of roots and product of roots is
as follows :
α + β = 1,
αβ = p
γ + δ = 4,
γδ = q
Let r be the common ratio of the GP α , β, γ, δ.
Then,
⇒
α + β = 1 and
γ+δ=4
α + αr = 1 and αr2 + αr3 = 4
α(1 + r) = 1 and αr2(1 + r) = 4
So 1 + r =
1
and
2
 1
αr2   = 4 ⇒ r = ± 2
α
866
QUANTUM
1
which is inadmissible.
3
Hence, r = − 2 is to be considered, then α = − 1
Thus
αβ = p = − 2 and rδ = q = − 32.
20 Assume some convenient and appropriate values of a, b, c
When r = 2, then α =
as
a = 3, b = 4, c = 6.
18 When this problem will be solved by algebraic methods, it
will take too much time to solve beyond the normal
required time. So, the best way to get the correct and quick
answer is to assume some simple roots (i.e., α and β) then
Then, ( x − 3)( x − 4) − 6 = 0
[Q ( x − a)( x − b) = c, c ≠ 0]
⇒
x 2 − 7 x + 6 = 0 ⇒ α = 6, β = 1
Again ( x − 6)( x − 1) + 6 (Q( x − α )( x − β ) + c = 0)
x 2 − 7 x + 6 + 6 = 0 or x 2 − 7 x + 12 = 0
go through options.
Let us take two arbitrary values α = − 1, β = 2, then the
equation will be x 2 − x − 2 = 0
Comparing with the equation x − px + q = 0
2
⇒
∴The roots k1 = 3 and k2 = 4
which are same as a and b.
Hence, option (c) is correct.
Alternatively x 2 − (a + b)x + (ab − c) = 0
p = 1, q = − 2
∴
Now, the sum of the roots of the required equation
= [(α − β )(α − β )] + [α β + α β ]
2
2
3
3
3 2
2 3
α +β=a+ b
and
αβ = ab − c
Again if k1 and k2 be the roots of the other equation, then
( x − α )( x − β ) + c = 0
= 27 + 4 = 31
and product of roots [(α 2 − β 2 )(α 3 − β 3 )[α 3β 2 + α 2β 3]
x − (α + β )x + (αβ + c) = 0
2
k1 + k2 = α + β = a + b
…(i)
and k1 ⋅ k2 = αβ + c = (ab − c) + c = ab
Hence equation is x − 31 x + 108 = 0
x 2 − [1 − {5 × 1 (−2)} + 5 × 1 × 4] + [1 × 4 − {5 × 1 × (−8)}
+ 4 × 1 × 16] = 0
= x 2 − 31 x + 108 = 0
19 Let us consider choice (a). When we put the values of A
and B respectively, we get the values of α , β, γ and δ as –1,
1/3, 1/5, 1/3, which are not in HP. So this option is not
correct.
Now for our convenience we consider choice (c), then by
substituting the values of A and B, we get the values of
α , β, γ and δ as 1, 1/2, 1/3 and 1/4 which are in HP.
Hence this could be the correct choice.
Alternatively Ax 2 − 4 x + 1 = 0
…(i)
4
1
α + γ = or αγ =
A
A
Bx − 6 x + 1 = 0
2
…(ii)
Thus, from eqs. (i) and (ii) it is clear that the roots are a
and b. Hence correct choice is c.
Now putting the values of p and q in the equation
options a, b, and c we get option (b) is correct. as :
21 The best way is to assume any convenient arbitrary
numerical values. And then verify the options.
Let us assume a = 1, b = 2, c = − 3, as a + b = c = 0. Then we
will see that only choice (c) has a valid relation as both the
sides have same value.
Hence, choice (c) is the answer.
22 Given that px 2 − qx + q = 0 ⇒ px 2 + q = qx
As p and q are prime, so they are positive. It implies x > 0.
a
Let us consider now x = such that a and b are co-prime.
b
∴
px 2 − qx + q = 0 ⇒ p
a
a2
−q + q=0
b
b2
and pa2 − qab + qb2 = 0
Now dividing eq. (i) by a we get pa − qb +
…(ii)
6
1
β + δ = or βδ =
B
B
Since it is given that α , β, γ, δ are in HP.
2αγ
1
2βδ
1
β=
= and γ =
=
∴
α+γ 2
β+δ 3
Again since β and γ are the roots of the given equation
hence they must satisfy the equation. So
Bβ 2 − 6β + 1 = 0 and Aγ 2 − 4γ + 1 = 0 ⇒ B = 8 and A = 3
Hence option (c) is correct.
i.e.,
∴
= 27 × 4 = 108
2
and
CAT
…(i)
2
qb
= 0; as
a
qb2
too must be an integer. But,
a
since a and b are co-prime, so a cannot divide b. It implies
that a = 1 or q.
pa − qb is integer, so
Similarly by dividing eqn. (i) we get to know that b = 1 or p.
a 1 1 q q
Therefore x = = , , ,
b 1 p 1 p
Case 1: If x = 1, px 2 − qx + q = 0 ⇒ p = 0.
That’s impossible as p being a prime number cannot be
zero.
Theory of Equation
867
1
1 q
Case 2: If x = , px 2 − qx + q = 0 ⇒ − + q = 0 ⇒
p
p p
pq + 1 = q.
or
esin x = 2 + 5.
Now taking log on both the sides, we get
sin x = log(2 − 5) or sin x = log(2 +
That’s impossible as pq > q.
Case 3: If x = q, px 2 − qx + q = 0 ⇒ pq2 + q = q2.
That’s impossible as pq2 > q2.
Here, 2 − 5 is a negative number. Since log of negative
number is not defined, so sin x = log(2 − 5) does not
give the solution.
Again, sin x = log(2 +
5) > log e
q
Case 4: If x = , px 2 − qx + q = 0 ⇒ q = 0.
p
⇒ sin x > 1, which is not possible.
That’s impossible as q being a prime number cannot be zero.
Hence choice (a) is the answer.
Therefore no rational solutions to the quadratic equation
px 2 − qx + q = 0 exists.
Hence, choice (a) is the correct answer.
Hint log e = 1 and e = 2.714. Also, −1 ≤ sin x ≤ 1.
26 Since α and β are the roots of x 2 + ux + v = 0.
Therefore
23. Let r be an integral root of the given equation, then
r − pr + q = 0 ⇒ r (r − p) + q = 0 ⇒ q = r ( p − r)
4
3
3
3
Since q itself is a prime (which cannot be the product of
two distinct primes), so either p − r = ± 1 or r3 = ± 1,
If p − r = ± 1, then q = ± r which is not possible as a
3
And
If r = − 1, then q = − 1( p + 1)
⇒ q + p = − 1, which is impossible as prime numbers are
always positive.
Further, if r = 1, then q = 1( p − 1) ⇒ p − q = 1.
α + β = −u
...(i)
αβ = v
...(ii)
Again since α and β are the roots of x 2n + u nx n + v n = 0.
Therefore α and β will satisfy this equation too, which is
shown below.
prime number cannot be a perfect cube. So r3 = ± 1.
(α n)2 + u nα n + v n = 0
...(iii)
(β ) + u β + v = 0
...(iv)
n 2
n n
n
From eqs. (iii) and (iv) we can infer that α n and β n are
the roots of p2 + u np + v n = 0.
So
α n + β n = − pn
...(v)
αβ =q
...(vi)
So we can infer that p = 3 and q = 2 when the root r = 1.
And
Therefore the required integral root is 1.
From the eq. (i), we have
Hence choice (d) is the answer.
(α + β )n = (− p)n = pn, as n is even.
24. x + 4 y + 9z − 2xy − 6 yz − 3zx
2
2
2
⇒
(α n + β n) + (α + β )n = 0
...(vii)
Now dividing eq. (vii) by β , we get
n
n
That means the given expression is always non-negative.
Hence choice (b) is the correct one.
n
⇒
25 Let p = esin x , then the given equation becomes as shown
below.
n
α
α

  + 1 +  + 1 = 0
 β
β

n
n
n
n
 α + β
 β
  +1+ 
 =0
α
 β 
⇒

β
 β
  + 1 +  + 1 = 0

α
α
From eqs. (viii) and (ix), we can infer that
⇒ p2 + p − 4 = 0
p = 2 − 5 or p = 2 + 5 ⇒ esin x = 2 − 5
...(viii)
Again dividing eq. (vii) by α n, we get
Hint Since a perfect square number is always a
non-negative number.
∴( x − 2y )2 ≥ 0, ( x − 3z )2 ≥ 0, (2y − 3z )2 ≥ 0.
n
α
 α + β
  +1+ 
 =0
 β
 β 
2
1
[( x − 2y )2 + ( x − 3z )2 + (2y − 3z )2] ≥ 0
2
⇒
n
(α + β )n = − (− pn) = − (α n + β n)
+ (4 y + 9z − 12yz )
2
1
p+ =4
p
n n
Now, from the eq. (v) , we have
1
= [ 2x 2 + 8 y 2 + 18z 2 − 4 xy − 12yz − 6zx]
2
1
= [( x 2 + 4 y 2 − 4 xy ) + ( x 2 + 9z 2 − 6zx )
2
=
5)
the roots of x n + 1 + ( x + 1)n = 0
Hence choice (d) is the answer.
...(ix)
α
β
and are
β
α
868
QUANTUM
 1
α
2
 1
 β
2
27 Let α and β be the two roots, such that α + β =   +  
That is α + β =
1
1
+
(α )2 (β )2
α 2 + β 2 (α + β )2 − 2αβ
α +β =
=
(αβ )2
(αβ )2
⇒
−b
a
c
αβ =
a
And
...(iii)
Therefore using eqs. (ii) and (iii), the Eq. (i) becomes as
following.
2
⇒
−
 c
 − b
  − 2 
 a
 a
2
b b2 − 2ac
⇒ bc2 + ab2 = 2ca2
=
a
c2
2
2
b2 − 4ac < 0
a>0
...(i)
Now g( x ) = ax + (2a + b)x + (2a + b + c)
2
The discriminant of g( x ) is
...(ii)
 c
 
 a
Alternatively Since f ( x ) is positive for all real x.
and
Now since, α + β =
−b
=
a
Hence, choice (a) is the valid one.
Then,
...(i)
CAT
2
D = (2a + b)2 − 4a(2a + b + c) ⇒ D = (b2 − 4ac) − 4a2
But using eq. (i) we can conclude that (b2 − 4ac) − 4a2 < 0
∴D<0
And from eq. (i), we already know that a > 0. Therefore
g( x ) > 0, for all real x.
Hence choice (a) is the answer.
29 Since AM of ( x − 1) and ( x − 5) = AM of ( x − 2) and
( x − 4) = ( x − 3). So, for our convenience, we can substitute
y in place of x − 3. Therefore
( x − 1)3 + ( x − 2)3 + ( x − 3)3 + ( x − 4)3 + ( x − 5)3 = 0
⇒ ( y + 2)3 + ( y + 1)3 + y 3 + ( y − 1)3 + ( y − 2)3 = 0
Therefore bc , ca , ab are in A.P.
⇒[( y + 2)3 + ( y − 2)3] + y 3 + [( y + 1)3 + ( y − 1)3] = 0
Hence choice (c) is the answer.
⇒ 2( y 3 + 12y ) + 2( y 3 + 3y ) + y 3 = 0
Alternatively
Let us consider the two roots 1 and 1,
2
2
 1
 1
as they satisfy the given condition 1 + 1 =   +  
 1
 1
Therefore, you will have the following quadratic
equation,
( x − 1)( x − 1) = 0
⇒
⇒
x − 2x + 1 = 0
2
a = 1, b = −2, c = 1
⇒ 5y 3 + 30 y = 0 ⇒ 5y( y 2 + 6) = 0
⇒ y = 0, ± −6
Hence choice (b) is the answer.
30 The AM of ( x + 3) and ( x − 1) is ( x + 1).
So, now putting x + 1 = y, we get the given equation in
the following form.
( y + 2)5 − ( y − 2)5 ≥ 244
Then, bc2 = − 2, ca2 = 1, ab2 = 2
As, −2, 1, 2 are in AP, therefore choice (c) is the valid
one.
28 Since f ( x ) is positive for all real values of x, it means the
graph of f ( x )is strictly above the X-axis. That means roots
are not real.
So, the best way is to consider any quadratic equation
whose roots are not real. So,
I’m assuming a desired equation: x 2 − x + 2 = 0
Now, comparing this equation ( x 2 − x + 2 = 0) with the
given equation (ax 2 + bx + c = 0), we get
a = 1, b = − 1 and c = 2
⇒ y = 0, ± 6i
∴x = 3, 3 ± 6i
⇒ 2(10 y 4 + 80 y 2 + 32) ≥ 244 ⇒ ( y 2 + 9)( y 2 − 1) ≥ 0
⇒ ( y 2 − 1) ≥ 0, since y 2 + 9 > 0 for every real value of y.
The corresponding roots for y 2 − 1 = 0 are −1 and 1.
So the inequation y 2 − 1 ≥ 0 is valid only when y ≤ −1 or
y ≥ 1.
+
–
–1
+
1
Now if y ≤ −1 ⇒ x ≤ −2 And if y ≥ 1 ⇒ x ≥ 0
It implies that the real solutions of the given equation is
x ≤ −2 or x ≥ 0
Therefore the solution set is (−∞, − 2] ∪ [ 0, ∞ )
But, g( x ) = ax 2 + (2a + b)x + (2a + b + c)
Hence choice (b) is the answer.
Now, if you plug in the values of a, b and c in g( x ), you
will get g( x ) = x 2 + x − 3. And, you see, as the
Hint ( y + 2)5 = y 5 + 5C1 y 4 × (2) + 5C 2 y 3 × (22 )
discriminant of g( x ) is strictly negative and coefficient of
x 2 is positive, so the roots of g( x ) will be non-real and the
and ( y − 2)5 = y 5 − 5C1 y 4 × (2) + 5C 2 y 3 × (22 )
graph will be strictly above the X-axis. That is g( x ) > 0.
+ 5C 3 y 2 × (23 ) + 5C 4 y × (24 ) + 5C 5 y × (25 )
− 5C 3 y 2 × (23 ) + 5C 4 y × (24 ) − 5C 5 × (25 )
Theory of Equation
869
Therefore ( y + 2)5 − ( y − 2)5
= 2[ 5C1 y 4 × (2) + 5C 3 y 2 × (23 ) + 5C 5 × (2)5]
⇒ ( y + 2)5 − ( y − 2)5
= 2[ 5y 4 × (2) + 10 y 2 × (8) + 1 × (32)]
31 From the given information we will have the following
results.
If α + β = − p and αβ = q
...(i)
And α n + β n = − pn and α nβ n = qn
...(ii)
α
is root of x n + 1 + ( x + 1)n = 0, so it must
β
satisfy this equation.
Now since
n



P12
P22
+
+


2
2
2
2 
(a − p2 ) + b 
(a − p1 ) + b
2ib1 + 
 = 0

2
Pk

. . . +



(a − pk )2 + b2



Since all the values inside the bracket are positive, so the
expression in the bracket cannot be zero. Therefore
2ib = 0
⇒ b = 0 since i ≠ 0
a + ib = a and a − ib = a
Therefore all the roots are real.
Hence choice (c) is the answer.
33 ( x − a)( x − c) + k( x − b)( x − d ) = 0
⇒ x 2 − (a + c)x + ac + k[ x 2 − (b + d )x + bd] = 0
n
∴
α
α

  + 1 +  + 1 = 0
 β
β

⇒
αn
 α + β
+1+ 
 =0
 β 
βn
Let’s assume that the eq. (i) will have real root, then the
discriminant D ≥ 0.
⇒
α n + β n (α + β )n
=0
+
βn
βn
⇒[(b + d )2 − 4bd]k 2 + [ 2(a + c)(b + d ) − 4ac − 4bd]k +
⇒ (1 + k )x 2 − [(a + c) + k(b + d )] x + (ac + kbd ) = 0
n
⇒
− pn (− p)n
− pn + (− p)n
=0⇒
+
=0
n
n
β
β
βn
⇒
− pn + (− p)n = 0
That is [(a + c) + k(b + d )]2 − 4[(1 + k )(ac + kbd )] ≥ 0
[(a + c)2 − 4ac] ≥ 0
⇒ (b − d )2 k 2 + 2(ab + ad + bc + cd − 2ac − 2bd )k
+ (a − c)2 ≥ 0
It is valid only when n is an even integer.
32 Let’s assume that a + ib is an imaginary root of the given
equation, then the conjugate of this root a − ib is also a root
of this equation. Putting x = a + ib and x = a − ib in the
given equation we get
P12
P22
Pk2
+
+ ... +
(a + ib − p1 ) (a + ib − p2 )
(a + ib − pk )
...(i)
P12
P22
Pk2
+
+ ... +
(a − ib − p1 ) (a − ib − p2 )
(a − ib − pk )
= a − ib + 1
Subtracting eq. (i) from eq. (ii), we get


P12
p22
+
+

2
2
2
2 
(a − p1 ) + b
(a − p2 ) + b 
= − 2ib
2ib 
2


Pk

. . . +
(a − pk )2 + b2




P12
P22
+

2
2
2
2
(a − p1 ) + b
(a − p2 ) + b 
⇒ 2ib
+ 2ib = 0
2


Pk
 +. . . +

(a − pk )2 + b2


...(ii)
Now, the expression (ii) looks like a quadratic inequality.
In the above inequation (ii) the coefficient of k 2
Hence choice (c) is the answer.
= a + ib + 1
...(i)
...(ii)
[i.e., (b − d )2] is positive.
Let’s assume that the discriminant of the expression (ii)
is J. Now,
(a) If J > 0, for certain values of k, expression will be
positive and for certain values of k, expression will
be negative. So it won’t be valid for all the real
values of k.
(b) If J =`0, for all the values of k, expression will be
non-negative, so it will be valid.
(c) If J < 0, for all the values of k, expression will be
positive, so it will be valid.
Therefore the need arises to determine the discriminant
J. Then,
J = 4(ab + ad + bc + cd − 2ac − 2bd )2 − 4(b − d )2(a − c)2
⇒ J = 4(ab + ad + bc + cd − 2ac − 2bd )2
− 4[(b − d )(a − c)]2
⇒ J = 4[(ab + ad + bc + cd − 2ac − 2bd )2
− {(b − d )(a − c)}2]
⇒ J = 4[{(ab + ad + bc + cd − 2ac − 2bd + (b − d )(a − c)} ×
{ab + ad + bc + cd − 2ac − 2bd + (b − d )(a − c)}]
⇒ J = 16(b − c)(a − d )(a − b)(d − c)
870
QUANTUM
⇒ J = 16(− ve)(− ve)(− ve)(+ ve); since a < b < c < d. ⇒ J < 0
Since the discriminant is negative and the coefficient of k 2
is positive, it means the given inequation is valid for all the
real values of k. This in turn implies that the roots of the
given equation are real. Hence choice (a) is the answer.
Hint Since b < d, so (b − d )2 is positive; as the square of
any negative number is always positive.
Alternatively
Let f ( x ) = ( x − a)( x − c) + k( x − b)( x − d ) = 0
Also it is given that a < b < c < d.
Now, f (b) = (b − a)(b − c) ⇒ f (b) < 0
(Q b − a > 0 and b − c < 0)
And f (d ) = (d − a)(d − c)
(Q d − a > 0 and d − c > 0)
⇒ f (d ) > 0
Since f (b) and f (d ) have opposite signs therefore the
given equation has one real root between b and d.
Since one root is real and a, b, c, d, k are also real, therefore
the other root of the given equation will also be real.
Thus the given quadratic equation
( x − a)( x − c) + k( x − b)( x − d ) = 0 has both the real roots.
CAT
34 Since the discriminant of the quadratic equation
x 2 − bx − c = 0 is b2 + 4c, which is positive, so there will be
two real roots, say α and β.
The product of the roots = αβ = − c, so one root will be
negative and one root will be positive.
Then the three coordinates will be
A(α , 0), B(β, 0) and C(0, c).
Therefore the line segment AO = − α
(as the root is negative), BO = β and CO = c.
As we know that in a circle, with two intersecting chords
the product of the segments is
equal. That is AO × BO = CO × DO .
∴
⇒
⇒
−α × β = c × DO
c = c × DO
DO = 1
Since D lies on the Y-axis, so the coordinates of D are
(0, 1), which is coincidently independent of both b and c.
Hint When you put x = 0 in y = x 2 − bx − c, you will get
y = OC