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Solution of Quadratic Equation by Completing Square

Geometry of Solution of Quadratic Equation by Completing Square

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

Example

Solve the equation x2 − 3x − 4 = 0 by completing square.

Solution:

x2 − 3x − 4 = 0          (i)

Shifting constant term −4 to the right, we have

x2 − 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

x2 − 3x + (−3/2)2 = 4 + (−3/2)2          (ii)

As    x2 − 3x + (−3/2)2 = x2 − 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 2x2 − 5x − 3 = 0 by completing square.

Solution:

2x2 − 5x − 3 = 0

Dividing each term by 2, to make coefficient of x2 equal to 1.

x2 − (5/2)x – 3/2 = 0

Remember that for our convenience, we make the co-efficient of x2 equal to 1 in the method of completing square.

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

x2 − 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

x2 − 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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