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 = \(\frac{25}{4}\).
From Completing a Square to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
May 24, 24 06:42 PM
May 24, 24 06:23 PM
May 24, 24 06:22 PM
May 24, 24 05:37 PM
May 24, 24 05:09 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.