QUADRATIC EQUATIONS
Objectives
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Identify a quadratic equation.
Distinguish between a pure and an adfected quadratic
equation.
Solve simple problems on pure and adfected
quadratic equations.
Identify the standard form of a quadratic equation.
Solve the quadratic equations by factorization.
Solve the quadratic equations by using formula.
Find the value of the discriminant and know the
nature of the roots.
Frame the quadratic equation for the given roots.
Solve the quadratic equation graphically.
Definition and explanation
The name Quadratic comes from “quad(in
Latin)” meaning square, because the
variable gets squared(X²).
 Also an equation of second order or
degree 2 is quadratic equation.
Eg: X²–3X+2
(This makes it quadratic)
Similarly equation with degree one is a linear
equation. Eg: 3X=9
General form: ax²+bx+c=0
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Example 1
Consider a square of side ‘a’ units and its
a
area 25 square units.
a
Area of the square = (side)²
25 = a²
or
a²= 25
∴ a =±5 i.e. a=+5 or a=-5
A Quadratic equation has only two
roots.
Example 2
Take a simple example of a
ball dropped from a height.
According to physics any
particle moving with uniform
acceleration the equation is
given as ½at²+ut-s=0
This is a quadratic equation
having the form
ax²+bx+c=0
Types of quadratic equations
(Pure and Adfected)
Quadratic equation involving a
variable only in second degree is a
“Pure Quadratic Equation’’.
From the general form when b=0 we
get ax² + c = 0, where a and c are real
numbers and a ≠ 0 is a pure
quadratic equation.
Eg: (1) x² – 9 = 0
(2) 2a² – 18 = 0
Quadratic equation involving a
variable in second degree as well as
in first degree is an “Adfected
Quadratic Equation”.
ax2 + bx + c = 0 is the standard form of a
quadratic equation where a, b and c
are variables and a ≠ 0.
Eg: (1) x2 + 3x – 10 = 0
(2) 3a2 – a – 2 = 0
Thus a easy way to differentiate between the 2
types is when b = 0 and b ≠ 0.
Problems (Pure)
Example 1 : Solve the equation 3X²-27=0
Solution : 3X² – 27 = 0
∴ 3X² = 27
get 27 to RHS
X² = 27/3
divide it by 3
X² = 9
result is 9
∴X² = 9
Find the square root
X = ±3
Ans
X=+3 or X= – 3
Are the roots of the equation
Example 2 : Solve the equation
(m + 8)² –5 = 31
Solution : (m + 8)² –5 = 31
(m + 8)² = 31 + 5
get 5 to RHS
(m + 8)² = 36
add it to 31
∴ (m + 8) = √36
square root of 36
(m + 8) = ±6
roots
∴ m = –8 ± 6
simplify
m = –8 + 6 or m = –8 – 6
m = -2 or m= -14
ans
Are the roots of equation
Example 3 : If l2 = r2 + h2. Solve for h and
find the value of ‘h’ if l = 15 and r =9
Solution : l2 = r2 + h2
r² + h² = l²
h² = l² – r²
solving for h
h = √(l² - r²)
h = √(15² - 9²)
substitute the values
h = √(255 - 81)
solving
h = √(144)
taking the roots
h = ±12
ans
h = +12 or h = -12
Are the roots
Problems(Adfected)
Example 1 : Solve the quadratic equation
a2 – 3a + 2 = 0
Solution : a2 – 3a + 2 = 0
a² – 2a – 1a + 2 = 0 Resolve the expression
a(a – 2) –1(a – 2) = 0 Factorize
(a – 2) (a – 1) = 0
Take common factor
a – 2= 0 or a – 1 = 0 Equate each factor to 0
i.e. a = 2 or a = 1
Are the roots
Example 2 : Solve the quadratic equation
2x² – 3x + 1 = 0
Solution : 2x² – 3x + 1 = 0
2x² – 2x – 1x + 1= 0 Resolve the expression
2x (x – 1) –1(x – 1)=0 Factorize
(x – 1) (2x – 1) = 0 Take common factor
(x – 1) = 0 or
(2x – 1) = 0
Equate each factor to 0
x = 1 or x = ½
Are the roots
Example 3 : Solve the quadratic equation
4k (3k – 1) = 5.
Solution : 4k (3k – 1) = 5
12k² – 4k – 5 = 0
12k² – 10k + 6k – 5 = 0
2k (6k – 5) + 1(6k – 5) = 0
(6k – 5) (2k + 1) = 0
(6k – 5) = 0 or (2k + 1) = 0
k = 5/6 or k = -½
Solving the equation
Consider the equation x² + 3x + 1 = 0
It cannot be factorised by splitting the middle term.
How do you solve such an equation ?
It can be solved by using Formula.
Derivation is as follows
Derivation
Problems(Formula)
Problems(Factorization)
Example 1 : Solve the equation (x + 6) (x + 2) = x
Solution :(x + 6) (x + 2) = x
x² + 6x + 2x + 12 = x
Resolve the expression
x² + 8x + 12 – x = 0
Factorize
x² + 7x + 12 = 0
x² + 4x + 3x + 12 = 0
x(x + 4) + 3 (x + 4) = 0
Take common factor
(x + 4) (x + 3) = 0
(x + 4) = 0 or (x + 3) = 0 Equate each factor to 0
x = -4 or x = -3
Are the roots
Problems based on Quadratic
Equation
Example 1 : If the square of a number is added to 3
times the number, the sum is 28. Find the number.
Solution : Let the number be = x
Square of the number = x²
3 times the number = 3x
Square of a number + 3 times the number = 28
x² + 3x = 28
x²+ 3x – 28 = 0
x²+ 7x – 4x – 28 = 0
x(x + 7) –4 (x + 7) = 0
(x + 7) (x – 4) = 0
x + 7 = 0 or x – 4 = 0
x = –7 or x = 4
∴ The required number is 4 or –7
Nature of the roots and
Discriminant
After solving the equations we get two roots
and there can be three nature of roots:
(1)Equal
Eg: X = 2 and X = 2
(2)Distinct
Eg: X = 1 and X = 2
Eg: X = 1 and X = -2
(3)Imaginary Eg: X = 1 +√(-2) and
X = 1- √(-2)
Thus for all the problems the roots can be
classified into these 3 categories.
It is clear that,
1) Nature of the roots of quadratic equation
depends upon the value of (b2 – 4ac)
2) The Expression (b2 – 4ac) is denoted by Δ
(delta) which determines the nature of the
roots.
3) In the equation ax2 + bx + c = 0 the
expression (b2 – 4ac) is called the discriminant.
Discriminant (b2 – 4ac) Nature of the roots
Δ=0
Roots are real and
equal
Δ > 0 (Positive)
Roots are real and
distinct
Δ < 0 (negative)
Roots are imaginary
Example 1 : Determine the nature of the roots of the equation 2x2 – 5x –
1 = 0.
Consider the equation 2x2 – 5x – 1 = 0
This is in form of ax2 + bx + c = 0
The co-efficient are a = 2, b = –5, c = –1
Δ = b2 – 4ac
Δ = (–5)2 –4(2) (–1)
Δ = 25 + 8
Δ = 33
∴Δ>0
Roots are real and distinct
Example 2 : Determine the nature of the roots of the equation 4x2 – 4x +
1=0
Consider the equation 4x2 – 4x + 1 = 0
This is in the form of ax2 + bx + c = 0
The co-efficient are a = 4, b = –4, c = 1
Δ = b2 – 4ac
Δ = (–4)2 –4 (4) (1)
Δ = 16 – 16
∴Δ=0
Roots are real and equal
Example 3 : For what values of ‘m’ roots of the equation x² + mx + 4
= 0 are
(i) equal (ii) distinct
Consider the equation
x² + mx + 4 = 0
This is in the form
ax² + bx + c = 0
the co-efficients are
a = 1, b = m, c = 4
Δ = b² – 4ac
Δ = m² – 4(1) (4)
Δ = m² – 16
(1) If roots are equal Δ = 0
m² – 16 = 0
m² = 16
m = √16 ∴ m = ±4
(2) If roots are distinct Δ > 0
m² – 16 > 0
m² > 16
m > √16
m > ±4
Sum and product of the roots
Formation of quadratic equation
If ‘m’ and ‘n’ are the roots then the
Standard form of the equation is
x² – (Sum of the roots) x + Product of the
roots = 0
x² – (m+ n) x + mn = 0
Let ‘m’ and ‘n’ are the roots of the equation
∴ x = ‘m’ or x = ‘n’
i.e., x – m = 0, x – n = 0
(x – m) (x – n) = 0
∴ x² – mx – nx + mn = 0
x² – (m + n) x+ mn = 0
Example 1 : Form the quadratic equation
whose roots are 2 and 3
Let ‘m’ and ‘n’ are the roots
∴m = 2, n = 3
Sum of the roots = m + n = 2 + 3
∴m+n=5
Product of the roots = mn
= (2) (3)
∴ mn = 6
Standard form x² – (m+ n) x+ mn = 0
x² – (5)x + (6) = 0
Example 2 : Form the quadratic equation
whose roots are 3 + 2√5 and 3 – 2√5
Let ‘m’ and ‘n’ are the roots
∴ m = 3 + 2√5 and n = 3 – 2√5
Sum of the roots = m + n
= 3 + 2√5 + 3 – 2√5
∴m+n=6
Product of the roots = mn
= (3 + 2√5 ) (3 – 2√5 )
= (3)² –( 2√5 )²
= 9 – 20
∴ mn = – 11
x² – (m + n) x + mn = 0
∴ x² – 6x – 11 = 0
Example 3 : If ‘m’ and ‘n’ are the roots of equation x² – 3x + 4 = 0 form the equation whose
roots are m² and n².
Solution:
Consider the equation x² – 3x + 4 = 0
The coefficients are a = 1, b = –3, c = 4
Let ‘m’ and ‘n’ are the roots
Sum of the roots = m + n = − b/a
= − (−3)/1
∴m+n=3
ii) Product of the roots = mn = c/a
=4/1
∴ mn = 4
If the roots are ‘m²’ and ‘n²’
Sum of the roots m² + n² = (m + n)² – 2mn
= (3)² – 2(4)
=9–8
∴ m² + n² = 1
Product of the roots m²n² = (mn)²
= 4²
∴ m²n² = 16
x² – (m² + n²) x + m²n² = 0
∴ x² – (1)x + (16) = 0
∴ x² – x + 16 = 0
Graphical method of solving a
Quadratic Equation
Graphical method of solving is another way
to solve a quadratic equation
The graph of a
quadratic
polynomial is
a curve called
‘parabola’
Eg: X² - 3X – 10
Roots are X = -2
and X = 5