Completing Square Method

Completing square method : The standard form of quadratic equation is ax ² + bx + c = 0, where a ≠0

Steps used in finding the roots by completing square method are as follows:

1) a$x^{2}$ + bx + c = 0	2x2+ 9x – 20 = 0 a = 2, b = 9 and c = -20
2) Divide by "a" on both sides $x^{2}$ + bx/ a + c/a = 0	Divide by "2" on both sides $x^{2}$ + 9x/ 2 -10 = 0
3) add - c on both sides $x^{2}$ + bx/a + c/a = 0 - c/a = –c /a ---------------------- $x^{2}$ + bx/a = -c/a	add + 10 on both sides $x^{2}$ + 9x/2 - 10 = 0 +10 = –20 --------------------- $x^{2}$ + 9x/2 = 10
4) add $b^{2}$/ 4$a^{2}$on both sides $x^{2}$ + bx/a + $b^{2}$/ 4$a^{2}$ = - c/a + $b^{2}$/ 4$a^{2}$	add $9x^{2}$/4 x $2^{2}$= 81/16 $x^{2}$ + 9x/2 + 81/16 = 10 + 81/ 16
5) $(x + b/2a)^{2}= - c/a+ b^{2}/ 4a^{2}$	$(x + 9/4)^{2}$ = (160 + 81)/16
6) $(x + b/2a)^{2} = (-4ac + b^{2})/ 4a^{2}$	$(x + 9/4)^{2}$ = 241/16
7)$ (x + b/2a)^{2}=( b^{2} - 4ac)/4a^{2}$	$(x + 9/4)^{2} $= 241/16
8) Taking square root on both sides (x + b/2a) =$\pm ( \sqrt(b^{2}$ - 4ac )/ 2a	Taking square root on both sides (x + 9/4) = $\pm$ √241 /4
9)Add - b/2a on both sides x = - b/2a ~+mn~ (√b2 - 4ac )/2a	Add - 9/4 on both sides x = - 9/4 ~+mn~ √241 /4
10) x = (- b ~+mn~ √b2 - 4ac )/2a	x = (- 9 ~+mn~ √241 )/ 4

The roots of the equations are (- 9 + √241 )/ 4 and (- 9 - √241 )/ 4


The easiest way to find the roots of the equation in completing square method :

As we know that ax ² + bx + c = 0 is the standard quadratic equation.

1) Move 'c' to other side.
2) Make the leading coefficient as 1.
3) Add (-b/2a)² on both sides of the equation. Then the left side will be a perfect square and then solve it to find the roots of the equation.

Examples :

1) Find the roots of the equation 9x ² - 15x + 6 = 0 using completing square method.

Solution :
9x ² - 15x + 6 = 0

9x ² - 15x = - 6

Here, leading coefficient = a = 9 , b = -15 and c = 6

Divide the whole equation by 9, we get

x ² - 15x/9 = - 6/9

x ² - 5x/3 = - 2/3

Now add (-b/2a) ² = -[-5/(2 x 3)] ² = (5/6) ² both sides

x ² - 5x/3 + (5/6) ² = -2/3 + (5/6) ²

( x - 5/6) ² = -2/3 + 25/36

( x - 5/6) ² = (-24 + 25)/36

( x - 5/6) ² = 1/36

x - 5/6 = ~+mn~ 1/6

∴ x = 5/6 ~+mn~ 1/6

x = 5/6 + 1/6 or x = 5/6 - 1/6

x = 6/6 = 1 or x = 4/6 = 2/3

So, roots of the equation are 1 and 2/3.

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2) Find the roots of the equation 2x ² - 5x + 3 = 0 by Completing square method.

Solution :
2x ² - 5x + 3 = 0

2x ² - 5x = - 3

Here, leading coefficient = a = 2 , b = -5 and c = 3

Divide the whole equation by 2, we get

x ² - 5x/2 = - 3/2

x ² - 5x/2 = - 3/2

Now add (-b/2a) ² = -[-5/(2 x 2)] ² = (5/4) ² both sides

x ² - 5x/2 + (5/4) ² = -3/2 + (5/4) ²

( x - 5/4) ² = -3/2 + 25/16

( x - 5/4) ² = (-24 + 25)/16

( x - 5/4) ² = 1/36

x - 5/4 = ~+mn~ 1/4

∴ x = 5/4 ~+mn~ 1/4

∴ x = 5/4 + 1/4 or x = 5/4 - 1/4

∴ x = 6/4 = 3/2 or x = 4/4 = 1

So roots of the equation is 3/2 and 1.

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Introduction of Quadratic Equations

• Splitting of middle term
• Completing square method
• Factorization using Quadratic Formula
• Solved Problems on Quadratic Equation

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