#### Consider the following identities

x2+2ax+a2=(x+a)2

x2-2ax+a2=(x-a)2

These trinomials are perfect squares. The highest power x2 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 x2 and x are given, the square may be completed by adding the square of half the coefficient of x.

If ax2+bx+c=0

Transposing ax2+bx=-c

Dividing by a to make the coefficient of x2 = 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

x2+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 x2-2x=8 by completing the square,

comparing x2 - 2x = 8 with x2+ b / a x = c / -a

a = 1, b = -2, c = -8 x2 - 2x + (1)2 = 8 + (1)2

(x - 1)2 = 8 + 1

(x - 1)2 = 9

Taking square roots on both the sides

x2 - 1 = ± 3

x-1=3 or x-1=-3

x=1+3 or x=-3+1

x=4 or x=-2

Example 3

Solve by completing the square

3x2-5x= 2 Taking the square roots on both sides #### Solve by completing the square

1. (2x-3) (3x+4) = 0

2. 3x2+7x = 6

3. 4x2-2x = 3

4. x2 = 0.6x - 0.05

5. 5(x+2) = 4(2x-1)(x+1)-16

6. 2x2-7x = 4

7. 12x2-cx-20c2 = 0

Solutions: To solve by completing the square.

1. (2x-3) (3x+4) = 0

On expanding we get

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

6x2 +8x -9x -12 = 0

6x2-x-12 = 0

⇒        6x2-x = 12

Dividing throughout by 6 we get

x2 - x / 6= 12/6

x2 - x / 6 = 2        (1)

Comparing with x2 b / -a x = c / -a we get

a=6 b= -1, c= -2  2. To solve 3x2 + 7x = 6 by completing the square

Dividing throughout by 3 we get  ∴ x = 2/3, -3 are the roots of the equation.

3. To solve 4x2-2x = 3 by completing the square

Dividing throughout by a = 4 we get

x2 2 / -4 x = 3/4

Comparing with x2+ b/a x= c/-a

a=4 b=-2, c=-3 Taking square roots on both the sides we get 4. To solve by completing the square

x2 = 0.6x - 0.05

Transposing

x2-0.6x = -0.05

Comparing with x2+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

x2-0.6x+0.09 = -0.05+0.09

x2-0.6x+(0.3)2 = 0.04

(x-0.3)2 = (0.2)2

Taking square roots on both the sides

x - 0.3 = ± 0.2

⇒ x-0.3 = 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

5. To solve 5(x+2) = 4(2x-1)(x+1) -16 by completing the square

We first obtain the form ax2 + bx + c = 0

5x+10 = 4[(2x -1)x+(2x - 1) * 1] -16

5x+10 = 4[2x2-x+2x-1]-16

5x+10 = 4[2x2 + x - 1] -16

5x+10 = 8x2 + 4x - 4 - 16

⇒ 8x2 + 4x - 20 - 5x - 10 = 0

⇒        8x2 - x - 30 = 0

⇒        8x2 - x = 30

Dividing throughout by 8  6. To solve 2x2-7x = 4 by completing the square

Dividing throughout by a = 2  7. To solve 12x2-cx-20c2 = 0 by completing the square Transposing and dividing by a = 12 we get

x2c / -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 