A binomial is an algebraic expression which has exactly two terms, for example, a ± b. Its power is indicated by a superscript. For example, (a ± b)^{2} is a power of the binomial a ± b, the index being 2.
A trinomial is an algebraic expression which has exactly three terms, for example, a ± b ± c. Its power is also indicated by a superscript. For example, (a ± b ± c)^{3} is a power of the trinomial a ± b ± c, whose index is 3.
Expansion of (a ± b)^{2}
(a +b)\(^{2}\)
= (a + b)(a + b)
= a(a + b) + b(a+ b)
= a\(^{2}\) + ab + ab + b\(^{2}\)
= a\(^{2}\) + 2ab + b\(^{2}\).
(a - b)\(^{2}\)
= (a - b)(a - b)
= a(a - b) - b(a - b)
= a\(^{2}\) - ab - ab + b\(^{2}\)
= a\(^{2}\) - 2ab + b\(^{2}\).
Therefore, (a + b)\(^{2}\) + (a - b)\(^{2}\)
= a\(^{2}\) + 2ab + b\(^{2}\) + a\(^{2}\) - 2ab + b\(^{2}\)
= 2(a\(^{2}\) + b\(^{2}\)), and
(a + b)\(^{2}\) - (a - b)\(^{2}\)
= a\(^{2}\) + 2ab + b\(^{2}\) - {a\(^{2}\) - 2ab + b\(^{2}\)}
= a\(^{2}\) + 2ab + b\(^{2}\) - a\(^{2}\) + 2ab - b\(^{2}\)
= 4ab.
Corollaries:
(i) (a + b)\(^{2}\) - 2ab = a\(^{2}\) + b\(^{2}\)
(ii) (a - b)\(^{2}\) + 2ab = a\(^{2}\) + b\(^{2}\)
(iii) (a + b)\(^{2}\) - (a\(^{2}\) + b\(^{2}\)) = 2ab
(iv) a\(^{2}\) + b\(^{2}\) - (a - b)\(^{2}\) = 2ab
(v) (a - b)\(^{2}\) = (a + b)\(^{2}\) - 4ab
(vi) (a + b)\(^{2}\) = (a - b)\(^{2}\) + 4ab
(vii) (a + \(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) + 2a ∙ \(\frac{1}{a}\) + (\(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) + \(\frac{1}{a^{2}}\) + 2
(viii) (a - \(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) - 2a ∙ \(\frac{1}{a}\) + (\(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) + \(\frac{1}{a^{2}}\) - 2
Thus, we have
1. (a +b)\(^{2}\) = a\(^{2}\) + 2ab + b\(^{2}\).
2. (a - b)\(^{2}\) = a\(^{2}\) - 2ab + b\(^{2}\).
3. (a + b)\(^{2}\) + (a - b)\(^{2}\) = 2(a\(^{2}\) + b\(^{2}\))
4. (a + b)\(^{2}\) - (a - b)\(^{2}\) = 4ab.
5. (a + \(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) + \(\frac{1}{a^{2}}\) + 2
6. (a - \(\frac{1}{a}\))\(^{2}\) = a\(^{2}\) + \(\frac{1}{a^{2}}\) - 2
Solved Example on Expansion of (a ± b)^{2}
1. Expand (2a + 5b)\(^{2}\).
Solution:
(2a + 5b)\(^{2}\)
= (2a)\(^{2}\) + 2 ∙ 2a ∙ 5b + (5b)\(^{2}\)
= 4a\(^{2}\) + 20ab + 25b\(^{2}\)
2. Expand (3m - n)\(^{2}\)
Solution:
(3m - n)\(^{2}\)
= (3m)\(^{2}\) - 2 ∙ 3m ∙ n + n\(^{2}\)
= 9m\(^{2}\) - 6mn + n\(^{2}\)
3. Expand (2p + \(\frac{1}{2p}\))\(^{2}\)
Solution:
(2p + \(\frac{1}{2p}\))\(^{2}\)
= (2p)\(^{2}\) + 2 ∙ 2p ∙ \(\frac{1}{2p}\) + (\(\frac{1}{2p}\))\(^{2}\)
= 4p\(^{2}\) + 2 + \(\frac{1}{4p^{2}}\)
4. Expand (a - \(\frac{1}{3a}\))\(^{2}\)
Solution:
(a - \(\frac{1}{3a}\))\(^{2}\)
= a\(^{2}\) - 2 ∙ a ∙ \(\frac{1}{3a}\) + (\(\frac{1}{3a}\))\(^{2}\)
= a\(^{2}\) - \(\frac{2}{3}\) + \(\frac{1}{9a^{2}}\).
5. If a + \(\frac{1}{a}\) = 3, find (i) a\(^{2}\) + \(\frac{1}{a^{2}}\) and (ii) a\(^{4}\) + \(\frac{1}{a^{4}}\)
Solution:
We know, x\(^{2}\) + y\(^{2}\) = (x + y)\(^{2}\) – 2xy.
Therefore, a\(^{2}\) + \(\frac{1}{a^{2}}\)
= (a + \(\frac{1}{a}\))\(^{2}\) – 2 ∙ a ∙ \(\frac{1}{a}\)
= 3\(^{2}\) – 2
= 9 – 2
= 7.
Again, Therefore, a\(^{4}\) + \(\frac{1}{a^{4}}\)
= (a\(^{2}\) + \(\frac{1}{a^{2}}\))\(^{2}\) – 2 ∙ a\(^{2}\) ∙ \(\frac{1}{a^{2}}\)
= 7\(^{2}\) – 2
= 49 – 2
= 47.
6. If a - \(\frac{1}{a}\) = 2, find a\(^{2}\) + \(\frac{1}{a^{2}}\)
Solution:
We know, x\(^{2}\) + y\(^{2}\) = (x - y)\(^{2}\) + 2xy.
Therefore, a\(^{2}\) + \(\frac{1}{a^{2}}\)
= (a - \(\frac{1}{a}\))\(^{2}\) + 2 ∙ a ∙ \(\frac{1}{a}\)
= 2\(^{2}\) + 2
= 4 + 2
= 6.
7. Find ab if a + b = 6 and a – b = 4.
Solution:
We know, 4ab = (a + b)\(^{2}\) – (a – b)\(^{2}\)
= 6\(^{2}\) – 4\(^{2}\)
= 36 – 16
= 20
Therefore, 4ab = 20
So, ab = \(\frac{20}{4}\) = 5.
8. Simplify: (7m + 4n)\(^{2}\) + (7m - 4n)\(^{2}\)
Solution:
(7m + 4n)\(^{2}\) + (7m - 4n)\(^{2}\)
= 2{(7m)\(^{2}\) + (4n)\(^{2}\)}, [Since (a + b)\(^{2}\) + (a – b)\(^{2}\) = 2(a\(^{2}\) + b\(^{2}\))]
= 2(49m\(^{2}\)+ 16n\(^{2}\))
= 98m\(^{2}\) + 32n\(^{2}\).
9. Simplify: (3u + 5v)\(^{2}\) - (3u - 5v)\(^{2}\)
Solution:
(3u + 5v)\(^{2}\) - (3u - 5v)\(^{2}\)
= 4(3u)(5v), [Since (a + b)\(^{2}\) - (a – b)\(^{2}\) = 4ab]
= 60uv.
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