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

From Expansion of (a ± b)^2 to HOME PAGE