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QUADRATIC EQUATIONS
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Name
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Quadratic Equations
CHAPTER 2: QUADRATIC EQUATIONS
1.1 Recognize Quadratic Equations and express it in general form
General form
ax 2  bx + c = 0
, where a , b and c are constants , a  0
Properties
1. Equation must be in one unknown only
2. The highest power of the unknown is 2
Examples
1. 2x 2 + 3x – 1 = 0 is a quadratic equation
2. 4x 2 – 9 = 0 is a quadratic equation
3. 8x 3 – 4x2 = 0 is not a quadratic equation
Activity 1
1. Determine whether each of the following equation is a quadratic equation or not .
Equations
Answer
(a) x 2 – x = 0
Yes
(b) 2x 2 – y = 0
(c) 3x + 2 = 0
(d) 2m 2 – 7m – 3 = 0
(e) k 2 – 4k = 0
(f) y 2 – 2 = 0
2. Rewrite each of the following quadratic equation in the general form. State the value
of a , b and c .
Quadratic equations
(a) 1 + 2x = x(x + 3)
1 + 2x = x2 + 3x
x2 + x – 1 = 0
(b) m 2 = 21 – 4m
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Value of a , b and c
a =1
b =1
c = -1
Quadratic Equations
(c) (y + 6)(y – 2) = - 7
(d) x 2 =
7x  3
2
(e) (x + 1) 2 = 16
1.2 Roots of Quadratic Equations
Notes
1. The root of a quadratic equation is the value(number) of the unknown(variable)
that satisfy the equation .
2. A quadratic equation has at most two roots only
Exercises
1. Determine which of the values of the variable x given are roots of the respective
quadratic equation.
1
(a) x 2 – x – 2 = 0 ; x = - 1 , 1 , 2
(b) 2x 2 + 7x + 3 = 0 ; x = - 3, - , 1 , 3
2
2. Determine by inspection which of the values of x are roots of the following quadratic
equations .
(a) (x + 3)(x – 2) = 0 ; x = 3 , 2 , - 3
(b) x(x + 4) = 0 ; x = 4 , 0 , - 4
3. If x = 2 is the root of the quadratic equation x 2 – 3kx -10 = 0 , find the value of k .
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Quadratic Equations
2
. SOLVING QUADRATIC EQUATION
2.1 Solving Quadratic Equations
A. By Factorization
If a quadratic equation can be factorized into a product of two factors such that
(x – p)(x – q) = 0 ,
Hence
x – p = 0 or x – q = 0
x = p or
x=q
 p and q are the roots of the equation .
Notes
1. If p  q  the equation have two different roots
2. If p = q  the equation have two equal roots (one root only)
3. The equation must be written in general form ax 2 + bx + c = 0 before
factorization.
Activity 2
Solve the following quadratic equations by factorization .
1. x 2 – 7x – 8 = 0
2. x 2 – 4x + 4 = 0
(x–8)(x+1)=0
x – 8 = 0 or x + 1 = 0
x = 8 or x = -1
3. x 2 – 8x = 0
4. 4x 2 – 9 = 0
5. 6x 2 + 13x – 5 = 0
6. (3x + 1)(x - 1) = 7
7.
40  3x
x
5  2x
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8. (x + 1)(x – 5) = 16
x2 – 4x – 5 = 16
x2 – 4x – 21 = 0
(x–7)(x–3)=0
x = 7 or x = 3
Quadratic Equations
9.
10. (2p + 1)(p + 1) = 0
t
t

16 9t
Exercise 1
Solve the following quadratic equation by factorisation.
1. x 2 – 5x – 6 = 0
[6,-1]
9. x 2 – 9x + 20 = 0
2
2. m + 5m – 24 = 0
[-8,3]
10. 4x 2 – 13x + 3 = 0
[5,4]
[ 1 ,3 ]
4
2
3. y + 10y + 24 = 0
2
4. 2x + 3x – 5 = 0
[-6,-4]
2
11. 2x – 3 = 5x
2
[  1 ,3 ]
2
[ 1 , 7 ]
2 3
[ 5,-2]
12. 6x – 11x = 7
5. 16x 2 – 6x – 7 = 0
[ 1,  5 ]
2
[ 7 , 1 ]
8 2
6. 2a 2 + 4a = 0
[0.-2]
14. (3m + 1)(m – 1) = 7
[ 2, 4 ]
7. 100 – 9n 2 = 0
[ 10 ,  10 ]
15. 10x 2 + 4 = 13x
[
[  1 ,3 ]
2
16. x(x + 4) = 21
[ -7,3]
13. (2x – 3) 2 = 49
3
3
8. (2x + 1)(x + 3 ) = 0
3
B. By Completing the Square
Notes
1. The expression x 2 – 2x + 1 can be written in the form (x – 1) 2
This is called “perfect square”.
Example
Solve each of the following quadratic equation
(a) (x + 1) 2 = 9
(b) x 2 = 49
x+1=3
x + 1 = 3 , x + 1 = -3
x=2 , x=-4
(c) (x + 2) 2 = 36
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1 4]
,
2 5
Quadratic Equations
2. From the example , note that, if the algebraic expression on the LHS of the quadratic
equation are perfect squares , the roots can be easily obtained by finding the square
roots.
3. To make any quadratic expression x2 + hx into a perfect square , we add the term
h
( ) 2 to the expression .
2
2
h
h

And this will make x  hx  x  hx      x  
2
2

2
2
2
4. To solve the equation by using completing the square method for quadratic equation
ax 2 + hx + k = 0 , follow this steps ;
Step 1 : Rewrite the equation in the form ax 2 + hx = - k
Step 2 : If the coefficient of x2 is  1 , reduce the coefficient to 1 (by dividing) .
h
Step 3 : Add ( )2 to both sides of the equation.
2
Step 4 : Write the expression on the LHS as perfect square.
Step 5 : Solve the equation
Examples
1. x2 + 6x – 9 = 0
x2 + 6x = 9
2
2. 2x2 – 5x – 8 = 0
2
6
6
x + 6x +   = 9 +  
2
2
2
( x + 3 ) = 18
x+3
= 18
x+3
=  4.243
x = 4.243 – 3 , x = -4.243 – 3
x = 1.243 , x = -7.243
2
Exercise 2
Solve the following equations by completing the square. (Give your answers correct to
four significant figures)
1.
2.
3.
4.
5.
6.
7.
x 2 – 8x + 14 = 0
2x 2 – 7x – 1 = 0
x 2 + 5x + 1 = 0
– x 2 – 3x + 5 = 0
x 2 = 5(x + 4)
-4x 2 – 12x + 3 = 0
2x 2 – 3x – 4 = 0
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[5.41 , 2.59]
[3.64 , -0.14]
[-0.209,-4.79]
[-4.19,1.19]
[7.62 , -2.62]
[-3.23,0.232]
[2.35,-0.85]
Quadratic Equations
C. By Using the quadratic formula
The quadratic equation ax 2 + bx + c can be solved by using the quadratic formula
b  b 2  4ac
x =
, where a  0
2a
Example
2x 2 – 7x – 3 = 0
a = 2 , b = -7 , c = -3
x
 (7)  (7) 2  4(2)(3)
2(2)
x
7  73 7  8.5440

4
4
x = 3.886 , -0.386
Exercise 3
Use the quadratic formula to find the solutions of the following equations. Give your
answers correct to three decimal places .
1. x 2 – 3x – 5 = 0
[4.193 , -1.193]
2
2. 9x = 24x – 16
[1.333 ]
2
3. 2x + 5x – 1 = 0
[0.186 , -2.686]
4. 3x 2 + 14x – 9 = 0
[2.899 , -6.899]
2
5. 7 + 5x – x = 0
[0.768 , -0.434]
2
6. m = 20 – 4m
[0.573 , -5.239]
k 1
7.
[-1.140 , 6.140]
 k2
3
8. x(x + 4) = 3
[0.646 , -4.646]
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Quadratic Equations
2.2 Forming a quadratic equation from given roots
A. If the roots of a quadratic equation are known, such as
then, the quadratic equation is (x – p)(x – q) = 0
x 2 – px – qx + pq = 0
x 2 – (p + q)x + pq = 0
x=p
and
x=q
Notice that p + q = sum of roots ( SOR )
and
pq = product of roots ( POR )
Hence, the quadratic equation with two given roots can be obtained as follows :x 2 – (SOR)x + (POR) = 0
Examples
Form the quadratic equations from the given roots.
1. x = 1 , x = 2
Method 1
(x – 1)(x – 2) = 0
x2 - 2x – x + 2 = 0
x2 - 3x + 2 = 0
Method 2
SOR = 1 + 2 = 3
POR = 1 x 2 = 2
x2 – 3x +2 = 0
2. x = - 2 , x = 3
Exercise 4
Form the quadratic equations with the given roots.
1. x = 3 , x = 2
1
2. x = - 6 ,
3
3. x = - 4 , x = - 6
4
4. x = -3 , x =
5
5. x = -7 , 3
[x2 - 5x + 6 = 0]
6. x = 5 only
[x2 - 10x + 25 = 0]
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[3x2 +17x - 6 = 0 ]
[x2 + 10x + 24 = 0]
[5x2 + 11x - 12=0 ]
[x2 + 4x - 21 = 0]
Quadratic Equations
1
3
1
1
8. x  , x 
2
3
[3x2 - x = 0]
7. x = 0 , x =
[6x2 - 5x + 1 = 0]
B. To find the S.O.R and P.O.R from the quadratic equation in general form
ax 2 + bx + c = 0
b
c
a , x 2 + x  = 0
a
a
Compare with
Then ,
x 2 – (SOR)x + (POR) = 0
b
a
c
POR =
a
SOR =
If  and  are the roots of the
quadratic equation ax2 + bx + c = 0,
b
then  +  =
a
c
 =
a
Activity 3
1. The roots for each of the following quadratic equations are  and  . Find the value
of  +  and  for the following equation
Quadratic Equations
a. x 2 – 12x + 4 = 0
b. x 2 = 4x + 8
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

12
4
Quadratic Equations
c. 3 – 2x 2 = 10x
d. 3x 2 + 8x = 10
e. 2x 2 + 3x + 4 = 0
C. Solving problems involving SOR and POR
Activity 4
1. Given that  and  are the roots of the quadratic equation 2x 2 + 3x + 4 = 0 . Form a
quadratic equation with roots 2 and 2.
2x 2 + 3x + 4 = 0
3
x2  x  2  0
2
3
 = 
2
 =2
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New roots
 3
SOR = 2  2 = 2 (    ) = 2   = -3
 2
POR = 2 ( 2 )  4 = 4(2) = 8
x 2 – (SOR)x + (POR) = 0
x 2 – (-3)x + 8 = 0
x 2 + 3x + 8 = 0
Quadratic Equations
2. If  and  are the roots of the quadratic equation 2x 2 – 5x – 1 = 0 , form a
quadratic equation with roots 3 and 3.
3. Given that  and  are the roots of the quadratic equation 2x 2 – 3x + 4 = 0 . Form a
1
1
quadratic equation with roots
and .


4. Given that m and n are roots of the quadratic equation 2x2 – 3x – 5 = 0 , form a quadratic
equation which has the roots 2m
n
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and 2n
m
.
Quadratic Equations
Exercise 5
1.
If  and  are roots of the quadratic equation 2x2 + 3x + 1 = 0, form a quadratic
equation for the following roots
a. 2 and 2
[x2 + 3x + 2 = 0]
b.
c. 
2
d.
2.
[x2 - 3x + 2 = 0 ]
2 + 3 and 2 + 3
and 
[8x2 + 6x + 1 = 0 ]
2
[x2 - 6x - 5 = 0]
2 - 1 and 2 - 1
If  and
 are the roots of equation 2x


equation with roots
2
and
2
.
2
– 5x – 6 = 0 , form a quadratic
[ 4 x 2  5x  3  0 ]
3. Given that  and  are the roots of the equation 3x 2 = 4 – 9x , form a quadratic
equation with roots  2 and  2 .
[ 9 x 2  105x  16  0 ]
4. Given m and n are the roots of the equation x 2 + 10x – 2 = 0 , form a quadratic
equation with roots;
(a) 2m + 1 and 2n + 1
[ x 2  18x  27  0 ]
3
3
(b)
and
[ 2 x 2  30 x  9  0 ]
m
n
5. Given that  and 3 are the roots of the equation x 2 + 2bx + 3a = 0 , prove that
4a = b 2 .
6. Given one of the root of the quadratic equation x 2 – 5kx + k = 0 is four times the
other root, find the value of k .
[k  1 ]
4
7. One of the roots of the quadratic equation 2x2 + 6x = 2k – 1 is twice the value of
the other root whereby k is a constant. Find the roots and the value of k.
[-1, -2 ; k =  3 ]
2
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Quadratic Equations
3. DISCRIMINANT OF A QUADRATIC EQUATIONS
3.1 Determining the types of roots of quadratic equations
For the quadratic equation ax2 + bx + c = 0 , the value of b2 – 4ac will determine the
types of roots.
b2 – 4ac is called the “discriminant”
Condition
b – 4ac > 0
b2 – 4ac = 0
b2 – 4ac < 0
2
Type of roots
Two different roots
Two equal roots
No roots
Example
Determine the type of roots for each of the following quadratic equations .
(a) 2x2 – 7x + 9 = 0
(b) 2x2 – 3x – 9 = 0
a = 2 , b = -7 , c = 9
b2 – 4ac = (-7)2 – 4(2)(9)
= 49 – 72
= -23
<0
 no roots
Exercise 6
Calculate the discriminant for each of the following quadratic equation and then state the
type of roots for each equation .
1. x2 – 8x + 14 = 0
5. x(3x – 5) = 2x- 5
2. 2x2 – 7x – 1 = 0
6. 5(5 – 4x) = 4x2
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Quadratic Equations
3. 4 + x2 = 4x
7. x2 = 2 – 4x
4. (x – 2)2 = 3
8. 2x2 + 3x = 0
3.2 Solving problems involving the use of the discriminant
Activity 5
1. The quadratic equation 2kx2 + 4x – 3 = 0 has two equal roots , find the value of k .
2. The quadratic equation x2 + 2kx + (k + 1)2 = 0 has real roots , find the range of values
of k.
3. Show that the equation x2 + m + 1 = 8x has two different roots if m < 15 .
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Quadratic Equations
4. The straight line y = tx – 2 is a tangent to the graph of a curve y = 2x2 + 4x , find
the value of t (t > 0) .
5. Given that the quadratic equation p(x2 + 9) = - 5qx has two equal roots , find the ratio
of p : q . Hence, solve those quadratic equation .
6. Show that the quadratic equation x2 + kx = 9 – 3k has real roots for all the value of k .
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Quadratic Equations
Exercise 7
1. Find the possible values of m if the quadratic equation (4 – 2m)x2 – 2m = 1 – 3mx has
two equal roots .
k
2. The equation x2 – 2x + = 0 has two different roots , find the range of values of k .
3
3. Given that the equation (p + 1)x2 – 2x + 5 = 0 has no roots , find the range of values
of p .
4. Find the range value of k if the quadratic equation x2 + 1 = k – 4x has real roots .
5. The quadratic equation 2x(x – 3) = k – 2x has two distinct roots. Find the range of
values of k.
6. The quadratic equation (m – 2)x 2 + 2x + 3 = 0 has two distinct roots. Find the range
of values of m.
7. A quadratic equation 4x(x + 1) = 4x – 5mx – 1 has two equal roots. Find the possible
values of m.
8. The straight line y = 2x – 1 does not intersect the curve y = 2x 2 + 3x + p.
Find the range of values of p.
9. The straight line y = 6x + m does not intersect the curve y = 5 + 4x – x 2 . Find the
range of values of m.
10. The straight line y = 2x + c intersect the curve y = x2 – x + 1 at two different points,
find the range of values of c.
11. Find the range values of m if the straight line y = mx + 1 does not meet the curve
y2 = 4x .
12. Show that the quadratic equation kx2 + 2(x + 1) = k has real roots for all the values
of k.
Answers for Exercise 7
1. m   4 , 4
7
4. k  3
7. m =  4 or m = 4
5
5
9. m > 6
k<3
3. p >  4
5. k > -2
6. m < 7
3
2.
5
8. p > – 3
8
10. . c >  5
4
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11. m > 1
Quadratic Equations
Enrichment Exercise – Quadratic Equations
1. The quadratic equation kx2 + 4x + 3= 0 has two different roots, find the range of
values of k .
2. Find the possible values of k if the quadratic equation x2 + (2 + k)x + 2(2 + k) = 0 has
two equal roots.
3. Show that the quadratic equation x2 + (2k – 1)x + k2 = 0 has real roots if k 
1
.
4
4. Find the possible values of k if the straight line y = 2x + k is a tangent to the curve
y = x2 + x + 1 .
5. Given that  and  are the roots of the quadratic equation 2x2 – 8x + 1 = 0 . Form
the quadratic equation with roots   2 and 2   .
6. Solve each of the following quadratic equation :a. 6x2 + 5x – 4 = 0
b. y(y + 1) = 10
c. 2x(x + 5) = 7x + 2
d. 16x2 + 8x + 1 = 0
7. The roots of the equation 2ax2 + x + 3b = 0 are
3
4
and . Find the value of a and b.
2
3
8. If  and  are the roots of quadratic equation 2x2 – 3x – 6 = 0 , form the quadratic


equation with roots
and .
3
3
1
9. Given
and – 5 are the roots of the quadratic equation . Write the quadratic equation
2
in the form of ax2 + bx + c = 0 .
10. Given that m + 2 and n – 1 are the roots of the equation x2 + 5x = - 4 . Find the
possible values of m and n .
11. Given that 2 and m are the roots of the equation (2x – 1)(x + 3) = k(x – 1) such that
k is a constant . Find the value of m and k .
12. Given one of the root of the equation 2x2 + 6x = 2k – 1 is twice the other root,
such that k is a constant . Find the value of the roots and the value of k .
13. One of the root of the quadratic equation h + 2x – x2 = 0 is - 1 . Find the value of h.
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Quadratic Equations
1
. Give your answer in the
2
form ax2 + bx + c = 0 , where a , b and c are constants. (SPM 2004)
14. Form the quadratic equation which has the roots -3 and
15. Solve the quadratic equation x(2x - 5) = 2x – 1 . Give your answer correct to three
decimal places .(SPM 2005)
16. The straight line y = 5x – 1 does not intersect the curve y = 2x2 + x + p . Find the
range of the values of p .(SPM 2005)
17. A quadratic equation x2 + px + 9 = 2x has two equal roots. Find the possible values
of p.(SPM 2006)
Answers on Enrichment Exercises
4
3
3
4. k =
4
1. k <
6. (a)
4 1
x ,
3 2
7. a = 3 , b = -4
9.
2x 2  9x  5  0
2. k = 6 , - 2
5.
(b) y = 2.702 , - 3.702
8.
(c)
2 x 2  24 x  65  0
x
1
,2
2
(d)
6 x 2  3x  2  0
10. n = 0 , - 3 ; m = - 6 , - 3
11. m = 3 , k = 15
12. roots = - 1 , -2 and k =
13. h = 3
14. 2x2 + 5x – 3 = 0
15. x = 3.35 , 0.15
16. p > 1
17. p = -8 , 4
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x

3
2
1
4