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 = $\frac{25}{4}$.

9th Grade Math

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