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Expansion of (a ± b)2
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)
= a2 + ab + ab + b2
= a2 + 2ab + b2.
(a - b)2
= (a - b)(a - b)
= a(a - b) - b(a - b)
= a2 - ab - ab + b2
= a2 - 2ab + b2.
Therefore, (a + b)2 + (a - b)2
= a2 + 2ab + b2 + a2 - 2ab + b2
= 2(a2 + b2), and
(a + b)2 - (a - b)2
= a2 + 2ab + b2 - {a2 - 2ab + b2}
= a2 + 2ab + b2 - a2 + 2ab - b2
= 4ab.
Corollaries:
(i) (a + b)2 - 2ab = a2 + b2
(ii) (a - b)2 + 2ab = a2 + b2
(iii) (a + b)2 - (a2 + b2) = 2ab
(iv) a2 + b2 - (a - b)2 = 2ab
(v) (a - b)2 = (a + b)2 - 4ab
(vi) (a + b)2 = (a - b)2 + 4ab
(vii) (a + 1a)2 = a2 + 2a ∙ 1a + (1a)2 = a2 + 1a2 + 2
(viii) (a - 1a)2 = a2 - 2a ∙ 1a + (1a)2 = a2 + 1a2 - 2
Thus, we have
1. (a +b)2 = a2 + 2ab + b2.
2. (a - b)2 = a2 - 2ab + b2.
3. (a + b)2 + (a - b)2 = 2(a2 + b2)
4. (a + b)2 - (a - b)2 = 4ab.
5. (a + 1a)2 = a2 + 1a2 + 2
6. (a - 1a)2 = a2 + 1a2 - 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
= 4a2 + 20ab + 25b2
2. Expand (3m - n)2
Solution:
(3m - n)2
= (3m)2 - 2 ∙ 3m ∙ n + n2
= 9m2 - 6mn + n2
3. Expand (2p + 12p)2
Solution:
(2p + 12p)2
= (2p)2 + 2 ∙ 2p ∙ 12p + (12p)2
= 4p2 + 2 + 14p2
4. Expand (a - 13a)2
Solution:
(a - 13a)2
= a2 - 2 ∙ a ∙ 13a + (13a)2
= a2 - 23 + 19a2.
5. If a + 1a = 3, find (i) a2 + 1a2 and (ii) a4 + 1a4
Solution:
We know, x2 + y2 = (x + y)2 – 2xy.
Therefore, a2 + 1a2
= (a + 1a)2 – 2 ∙ a ∙ 1a
= 32 – 2
= 9 – 2
= 7.
Again, Therefore, a4 + 1a4
= (a2 + 1a2)2 – 2 ∙ a2 ∙ 1a2
= 72 – 2
= 49 – 2
= 47.
6. If a - 1a = 2, find a2 + 1a2
Solution:
We know, x2 + y2 = (x - y)2 + 2xy.
Therefore, a2 + 1a2
= (a - 1a)2 + 2 ∙ a ∙ 1a
= 22 + 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
= 62 – 42
= 36 – 16
= 20
Therefore, 4ab = 20
So, ab = 204 = 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(a2 + b2)]
= 2(49m2+ 16n2)
= 98m2 + 32n2.
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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