Quadratic Equations  Solutions by Completing the Square Method
Consider the following identities
x^{2}+2ax+a^{2}=(x+a)^{2}
x^{2}2ax+a^{2}=(xa)^{2}
These trinomials are perfect squares. The highest power x^{2} has unity (i.e., 1) as its coefficient. In this case, the term without an x (the constant term) must equal the square of half the coefficient of x. Therefore, if the terms x^{2} and x are given, the square may be completed by adding the square of half the coefficient of x.
If ax^{2}+bx+c=0
Transposing ax^{2}+bx=c
Dividing by a to make the coefficient of x^{2} = 1 we get
To complete the square, we add the square of half the coefficient of x, or
= , to both sides of the equation
We obtain
Taking the square roots on both sides
which is, as you know, the formula.
Example 1
Solve using the completing the square method
x^{2}+14x=32
a=1, b=14, c= 32
(x + 7)^{2} = 32 + 49
(x + 7)^{2} = 81
x + 7 = ± 9
i.e., x = 9  7 or x = 9  7
x = 2 or x = 16
Example 2
Solve the equation x^{2}2x=8 by completing the square,
comparing x^{2}  2x = 8 with x^{2}+ b / a x = c / a
a = 1, b = 2, c = 8
x^{2}  2x + (1)^{2} = 8 + (1)^{2}
(x  1)^{2} = 8 + 1
(x  1)^{2} = 9
Taking square roots on both the sides
x^{2}  1 = ± 3
x1=3 or x1=3
x=1+3 or x=3+1
x=4 or x=2
Example 3
Solve by completing the square
3x^{2}5x= 2
Taking the square roots on both sides
Try these questions:
Solve by completing the square

(2x3) (3x+4) = 0

3x^{2}+7x = 6

4x^{2}2x = 3

x^{2} = 0.6x  0.05

5(x+2) = 4(2x1)(x+1)16

2x^{2}7x = 4

12x^{2}cx20c2 = 0
Answers to Practice Problems
Solutions: To solve by completing the square.

(2x3) (3x+4) = 0
On expanding we get
2x(3x+4) 3 (3x+4) = 0
6x^{2} +8x 9x 12 = 0
6x^{2}x12 = 0
⇒ 6x^{2}x = 12
Dividing throughout by 6 we get
x^{2}  x / 6= 12/6
x^{2}  x / 6 = 2 (1)
Comparing with x^{2} b / a x = c / a we get
a=6 b= 1, c= 2
To solve 3x^{2} + 7x = 6 by completing the square
Dividing throughout by 3 we get
∴ x = 2/3, 3 are the roots of the equation.

To solve 4x^{2}2x = 3 by completing the square
Dividing throughout by a = 4 we get
x^{2} 2 / 4 x = 3/4
Comparing with x^{2}+ b/a x= c/a
a=4 b=2, c=3
Taking square roots on both the sides we get

To solve by completing the square
x^{2} = 0.6x  0.05
Transposing
x^{2}0.6x = 0.05
Comparing with x^{2}+b / a x = c / a
a = 1, b =  0.6, c =  0.05
Adding (b / 2a)^{2} = (0.6/2)^{2} = (0.3)^{2} = .09 to both sides we get
x^{2}0.6x+0.09 = 0.05+0.09
x^{2}0.6x+(0.3)^{2} = 0.04
(x0.3)^{2} = (0.2)^{2}
Taking square roots on both the sides
x  0.3 = ± 0.2
⇒ x0.3 = 0.2 or x  0.3 =  0.2
⇒ x = 0.2 + 0.3 or x =  0.2 + 0.3
⇒ x = 0.5 or x = 0.1
Roots of the equations are x = 0.1, 0.5

To solve 5(x+2) = 4(2x1)(x+1) 16 by completing the square
We first obtain the form ax^{2} + bx + c = 0
5x+10 = 4[(2x 1)x+(2x  1) * 1] 16
5x+10 = 4[2x^{2}x+2x1]16
5x+10 = 4[2x^{2} + x  1] 16
5x+10 = 8x^{2} + 4x  4  16
⇒ 8x^{2} + 4x  20  5x  10 = 0
⇒ 8x^{2}  x  30 = 0
⇒ 8x^{2}  x = 30
Dividing throughout by 8

To solve 2x^{2}7x = 4 by completing the square
Dividing throughout by a = 2

To solve 12x^{2}cx20c2 = 0 by completing the square Transposing and dividing by a = 12 we get
x^{2}c / 12 x = 20 / 12 c2
Comparing with x + (b/a) x = c/a
a = 12, b = c, c =  20c2
to both sides
Taking square roots on both sides
Roots of the equations are