Completing Square Method
Completing square method : The standard form of quadratic equation is ax ^{2} + bx + c = 0, where a ≠0Steps used in finding the roots by completing square method are as follows:
1) a$x^{2}$ + bx + c = 0 | 2x^{2}+ 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~ (√b^{2} - 4ac )/2a |
Add - 9/4 on both sides x = - 9/4 ~+mn~ √241 /4 |
10) x = (- b ~+mn~ √b^{2} - 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.
Introduction of Quadratic Equations
• Splitting of middle term
• Completing square method
• Factorization using Quadratic Formula
• Solved Problems on Quadratic Equation