There are many methods by which one can solve quadratic equations. The first method being the factorization method. The next is the quadratic formula method and the third which we will be seeing now is completing the square method.

Solving a quadratic equation by completing the square is a method of obtaining the roots of a quadratic equation by converting the quadratic polynomial [ ax^{2} + bx + c] to a(….)^{2} + constant. Here, the constant is not ‘x’, but (x – constant).

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# Steps to solve quadratic equations by completing the square

Here are the steps to be followed to solve quadratic equations by completing the square method.

1) Let the quadratic equation be ax^{2} + bx + c = 0.

2) Take ‘a’ out from the equation and multiply and divide coefficient of x by 2.

Then equation becomes x^{2} + 2(b/2a)x + c/a = 0.

3) Add (b/2a)^{2} on both sides of equation.

x^{2} + 2(b/2a)x + c/a + (b/2a)^{2} = (b/2a)^{2}

(x + b/2a)^{2} + c/a = (b/2a)^{2}

4) Send the constant term on left hand side to the other side and take square root on both sides.

5) Then solve for x.

Let us see with examples on solving quadratic equations by completing the square so that you can understand more clearly.

## Examples on solving quadratic equations by completing the square

1) **Find the roots of the quadratic equation 2x ^{2} – 7x + 3 = 0 by the method of completing the square.**

Solution: 2x^{2} – 7x + 3 = 0 [take 2 out of the equation]

2(x^{2} – (7/2)x + 3/2) = 0

x^{2} – (7/2)x + 3/2 = 0 [multiply and divide coefficient of x by 2]

x^{2} – 2(7/4)x + 3/2 = 0 [add (7/4)^{2} to both sides of equation]

x^{2} – 2(7/4)x + 3/2 + (7/4)^{2} = (7/4)^{2}

[x – (7/4)]^{2} + 3/2 = 49/16

[x – (7/4)]^{2} = 49/16 – 3/2

[x – (7/4)]^{2} = (49 – 24)/16

[x – (7/4)]^{2} = 25/16

[x – (7/4)] = ± 5/4

x = 7/4 ± 5/4

x= 12/4 or 2/4

** x = 3 or ½**

2) **Find the roots of the quadratic equation x ^{2} -3x -1 = 0 by the method of completing the square.**

Solution: x^{2} -3x -1 = 0 [multiply and divide coefficient of x by 2]

x^{2} – 2(3/2)x -1 = 0 [add (3/2)^{2} to both sides of equation]

x^{2} – 2(3/2)x -1 + (3/2)^{2 }= (3/2)^{2 }

[x – (3/2)]^{2} = 1 + 9/4 = 13/4

x – (3/2) = ± √13/2

x = 3/2 ± √13/2

** x = (3 – ****√13)/2, (3 + ****√13)/2**

3) **The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.**

Solution: Let the larger number be x and smaller number be y.

Difference of squares of two numbers is 180

x^{2} – y^{2} = 180 ……(1)

The square of the smaller number is 8 times the larger number

y^{2} = 8x

Substitute y^{2} = 8x in equation (1)

x^{2} – 8x = 180

x^{2} – 2(4)x = 180 [add 16 to both sides of equation]

x^{2} – 2(4)x + 16 = 180 + 16

(x – 4)^{2} = 196

x – 4 = ± 14

x = 4 ± 14

x = 4 + 14 = 18.

If x = 4 – 14 = -10, then y^{2} = 8(-10) = -80, y don’t have real roots. So, x cannot be

equal to -10

y^{2} = 8x

y^{2} = 8(18) = 144

y = -12, 12

**Therefore the two numbers are (18,-12) or (18, 12).**

the explanation is easy to understand.

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