To solve a quadratic equation by the method of completing square is illustrated through the following examples.

### Example

Solve the equation *x ^{2} − 3x − 4 = 0* by completing square.

### Solution:

*x ^{2} − 3x − 4 = 0 (i)*

Shifting constant term −4 to the right, we have

*x ^{2} − 3x = 4 (ii)*

Adding the square of 1⁄2 × coefficient of x, that is,

(−3⁄2)^{2} on both sides of equation (ii), we get

*x ^{2} − 3x + (−3/2)^{2} = 4 + (−3/2)^{2} (ii)*

As x^{2} **− 3x** + (−3/2)^{2} = x^{2} **− 1⁄2 × x × 3⁄2** + (−3⁄2)^{2} = (x−3⁄2)^{2}, so we have

*(x−3⁄2) ^{2} = 4 + 9⁄4 = 4 + (16 + 9)/4*

*(x−3⁄2) ^{2} = 25⁄4*

Taking square root of both sides of the above equation,

*√(x−3⁄2) ^{2} = ±√(25⁄4)*

⇒ *x−3⁄2 = ± 5⁄2 or x = 3⁄2 ± 5⁄2*

Either *x = 3⁄2 + 5⁄2 = (3 + 5)/2 = 8⁄2 = 4 or x = 3⁄2 − 5⁄2 = (3 – 5)/2 = −2⁄2 = −1
*

∴ 4, −1 are the roots of the given equation.

Thus, the solution set is {−1, 4}.

### Example

Solve the equation *2x ^{2} − 5x − 3 = 0* by completing square.

### Solution:

*2x ^{2} − 5x − 3 = 0*

Dividing each term by 2, to make coefficient of *x ^{2}* equal to 1.

*x ^{2} − (5/2)x – 3/2 = 0*

**Remember that **for our convenience, we make the co-efficient of *x ^{2} equal to 1 in the method of completing square.*

Shifting constant term −3/2 to the right, we have

*x ^{2} − 5⁄2 x = 3/2 (i)*

Adding the square of 1/2 × coefficient of x , that is, *(1/2 × −5/2) ^{2} = (−5/4)^{2}* on both sides of equation (i), we get

*x ^{2} − 5⁄2 x + (−5/4)^{2} = 3/2 + (−5/4)^{2}*

*, that is,*

*(x − 5/4) ^{2} = 3/2 + 25/16 = (24 + 25)/16*

*(x − 5/4) ^{2} = 49/16*

Taking square root of both sides of the above equation,

*√(x− 5/4) ^{2} = ±√(49/16)*

⇒ *x− 5/4 = ± 7/4 or x = 5/4 ± 7/4
*

Either *x = 5/4 + 7/4 = (5 + 7)/4 = 12/4 = 3 *

*or x = 5/4 − 7/4 = (5 − 7)/4 = − 2/4 = − 1/2
*

∴ 3, −1/2 are the roots of the given equation

Thus, the solution set is {−1/2, 3}.

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