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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