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)

= 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.









9th Grade Math

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