Completing the Square Method

In completing the square method for solving a quadratic equation of the form ax² + bx + c = 0, we transpose the constant term c to the right-hand side of the equation, and then the left-hand side is expressed as a perfect square expression.

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We use the following steps for solving quadratic equation of the form ax² + bx + c = 0 by completing the square method:

Steps:

1.  Rewrite the equation so that the terms containing x² and x (i.e. variable) are on the left side and the constant term is on the right side.

2.  If the coefficient of x² is other than 1, then reduce the coefficient of x² as 1 by dividing to both sides by the coefficient of x².

3.  Add the square of half of the coefficient of x on both sides of the equation to make the left-hand side of the equation a complete square.

4.  Express the left hand side of the equation as a perfect square of a binomial.

5.  Use the square root property on both sides and calculate the roots.

This process of Completing the Square Method will be clear by the following worked-out examples.

Workout Examples

Example 1: Solve the following quadratic equations by completing the square method.

            a.    x² – 2x – 8 =0

            b.    2x² + 5x – 12 = 0

            c.     3x² – 5x + 2 = 0

Solution:  a) Here,

                     x² – 2x – 8 = 0

            or,     x² – 2x = 8

            or,     x² – 2x + 1 = 8 + 1  [adding both sides by 1]

            or,     (x – 1)² = 9

            or,     x – 1 = ±√9

            or,     x – 1 = ±3

            or,     x = 1 ± 3

            Taking +ve sign,

                      x = 1 + 3 = 4

            Taking –ve sign,

                      x = 1 – 3 = –2

            ∴  x = 4, –2

Solution:  b) Here,

                     2x² + 5x – 12 = 0

            or,     2x² + 5x = 12

            or,     (2x² + 5x)/2 = 12/2  [dividing both sides by 2]

            or,     x² +5x/2 = 6

            or,     x² + 2.x.5/4 + (5/4)² = 6 + (5/4)²  [adding both sides by (5/4)²]

            or,     (x + 5/4)² = 6 + 25/16

            or,     (x + 5/4)² = 121/16

            or,     x + 5/4 = ±√(121/16)

            or,     x + 5/4 = ± 11/4

            or,     x = – 5/4 ± 11/4

            Taking +ve sign,

                      x = – 5/4 + 11/4 = (–5 + 11)/4 = 6/4 = 3/2

            Taking –ve sign,

                      x = – 5/4 – 11/4 = (–5 – 11)/4 = –16/4 = –4

            ∴  x = 3/2, –4

Solution:  c) Here,

                     3x² – 5x + 2 = 0

            or,     3x² – 5x = –2

            or,     (3x² – 5x)/3 = –2/3  [dividing both sides by 3]

            or,     x² –5x/3 = –2/3

            or,     x² – 2.x.5/6 + (5/6)² = –2/3 + (5/6)²  [adding both sides by (5/6)²]

            or,     (x – 5/6)² = –2/3 + 25/36

            or,     (x – 5/6)² = (–24 + 25)/36

            or,     x – 5/6 = ±√(1/36)

            or,     x – 5/6 = ± 1/6

            or,     x = 5/6 ± 1/6

            Taking +ve sign,

                      x = 5/6 + 1/6 = (5 + 1)/6 = 6/6 = 1

            Taking –ve sign,

                      x = 5/6 – 1/6 = (5 –1)/6 = 4/6 = 2/3

            ∴  x = 1, 2/3

If you have any questions regarding completing the square method, you can comment here in the comment section below.

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