# Completing a Square

Here we will learn how to completing a square.

a$$^{2}$$x$$^{2}$$ + bx = a$$^{2}$$x$$^{2}$$ + 2 ∙ ax ∙ $$\frac{b}{2a}$$

= {(ax)$$^{2}$$ + 2ax ∙ $$\frac{b}{2a}$$ + ($$\frac{b}{2a}$$)$$^{2}$$} - ($$\frac{b}{2a}$$)$$^{2}$$

= (ax + $$\frac{b}{2a}$$)$$^{2}$$ - ($$\frac{b}{2a}$$)$$^{2}$$

a$$^{2}$$x$$^{2}$$ - bx = a{x$$^{2}$$ - $$\frac{b}{a}$$x}

= a{x$$^{2}$$ - 2x ∙ $$\frac{b}{2a}$$ + ($$\frac{b}{2a}$$)$$^{2}$$} - a ∙ ($$\frac{b}{2a}$$)$$^{2}$$

= a(x - $$\frac{b}{2a}$$)$$^{2}$$ - $$\mathrm{\frac{b^{2}}{4a}}$$

Solved Examples on Completing a Square:

1. What should be added to the polynomial 4m$$^{2}$$ + 8m so that it becomes perfect square?

Solution:

4m$$^{2}$$ + 8m

= (2m)$$^{2}$$ + 2 ∙ (2m) ∙ 2

= (2m)$$^{2}$$ + 2 ∙ (2m) ∙ 2 + 2$$^{2}$$ – 2$$^{2}$$

= (2m + 2)$$^{2}$$ – 4.

Therefore, (4m$$^{2}$$ + 8m) + 4 = (2m + 2)$$^{2}$$ – 4 + 4 = (2m + 2)$$^{2}$$.

So, 4 is to be added to 4m$$^{2}$$ + 8m to make it a perfect square.

2. What should be added to the polynomial 9k$$^{2}$$ – 4k so that it becomes perfect square?

Solution:

9k$$^{2}$$ – 4k

= (3k)$$^{2}$$ - 2 ∙ (3k) ∙ $$\frac{2}{3}$$

= (3k)$$^{2}$$ - 2 ∙ (3k) ∙ $$\frac{2}{3}$$ + $$\mathrm{(\frac{2}{3})^{2}}$$ - $$\mathrm{(\frac{2}{3})^{2}}$$

= (3x - $$\frac{2}{3}$$)$$^{2}$$ – $$\frac{4}{9}$$

Therefore, (9k$$^{2}$$ – 4k) + $$\frac{4}{9}$$ = (3x - $$\frac{2}{3}$$)$$^{2}$$ – $$\frac{4}{9}$$ + $$\frac{4}{9}$$  = (3x - $$\frac{2}{3}$$)$$^{2}$$

So, $$\frac{4}{9}$$ is to be added to 9k$$^{2}$$ - 4k to make it a perfect square.

3. What should be added to 16m$$^{4}$$ + 9 to make it a whole square of a polynomial of the second degree?

Solution:

16m$$^{4}$$ + 9

= (4m$$^{2}$$)$$^{2}$$ + 3$$^{2}$$

= (4m$$^{2}$$)$$^{2}$$ ± 2 ∙ (4m$$^{2}$$) ∙ 3 + 3$$^{2}$$ ∓ 2 ∙ (4m$$^{2}$$) ∙ 3

= (4m$$^{2}$$ ± 3)$$^{2}$$ ∓ 24m$$^{2}$$

Therefore, (16m$$^{4}$$ + 9) ± 24m$$^{2}$$ = (4m$$^{2}$$ ± 3)$$^{2}$$ ∓ 24m$$^{2}$$ ± 24m$$^{2}$$

= (4m$$^{2}$$ ± 3)$$^{2}$$.

So, ± 24m$$^{2}$$ is to be added to 16m$$^{4}$$ + 9 to make it s whole square of a polynomial of the second degree.

4. Find k so that p$$^{2}$$ – 5p + k can be a perfect square of a linear polynomial.

Solution:

p$$^{2}$$ – 5p + k

= p$$^{2}$$ – 2p ∙ $$\frac{5}{2}$$ + k

= p$$^{2}$$ – 2p ∙ $$\frac{5}{2}$$ + ($$\frac{5}{2}$$)$$^{2}$$ + k - ($$\frac{5}{2}$$)$$^{2}$$

= (p - $$\frac{5}{2}$$)$$^{2}$$ + (k - $$\frac{25}{4}$$)

So, p$$^{2}$$ – 5p + k can be perfect square if k - $$\frac{25}{4}$$ = 0, i.e., k = $$\frac{25}{4}$$.