Expand - i 2a - b2 ii a - 2b2 iii a - 1-2a2 iv 2a - 1-a2 v a - b - c2 vi a - b - c2 vii 3x - 1-3x2 viii 2x - 1-2x2

(i) (2a + b)^2 = (2a)^2 + (b)^2 + 2 × 2a × b [(a+b)^2 = a^2 + b^2 + 2ab] =4a^2 + b^2 + 4ab (ii) (a 2b)^2 = (a)^2 + (2b)^2 2 × a × 2b [(a b)^2 = a^2 + b^2 ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x+y+z)^2

Expand : i 2a...

Question

Expand :

(i) $(2 a + b)^{2}$

(ii) $(a - 2 b)^{2}$

(iii) ${(a + \frac{1}{2 a})}^{2}$

(iv) ${(2 a - \frac{1}{a})}^{2}$

(v) $(a + b - c)^{2}$

(vi) $(a - b + c)^{2}$

(vii) ${(3 x + \frac{1}{3 x})}^{2}$

(viii) ${(2 x - \frac{1}{2 x})}^{2}$

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Solution

(i) $(2 a + b)^{2} = (2 a)^{2} + (b)^{2} + 2 \times 2 a \times b [(a + b)^{2} = a^{2} + b^{2} + 2 a b]$

$= 4 a^{2} + b^{2} + 4 a b$

(ii) $(a - 2 b)^{2} = (a)^{2} + (2 b)^{2} - 2 \times a \times 2 b [(a - b)^{2} = a^{2} + b^{2} - 2 a b]$

$= a^{2} + 4 b^{2} - 4 a b$ .

(iii) ${(a + \frac{1}{2 a})}^{2} = (a)^{2} + {(\frac{1}{2 a})}^{2} + 2 \times a \times \frac{1}{2 a}$

$= a^{2} + \frac{1}{4 a^{2}} + \frac{2 a}{2 a}$

$= a^{2} + \frac{1}{4 a^{2}} + 1$

(iv) ${(2 a - \frac{1}{a})}^{2} = (2 a)^{2} + {(\frac{1}{a})}^{2} - 2 \times 2 a \times \frac{1}{a}$

$= 4 a^{2} + \frac{1}{a^{2}} - 4$ .

(v) $(a + b - c)^{2} = (a)^{2} + (b)^{2} + (- c)^{2} + 2 \times a \times b + 2 \times b \times (- c) + 2 \times (- c) \times (a) = a^{2} + b^{2} + c^{2} + 2 a b - 2 b c - 2 c a$

Note : \((a+ b + c)^2 = a^2 +b^2 + c^2 + \2ab - 2bc - 2ca)

(vi) $(a - b + c)^{2} = (a)^{2} + (- b)^{2} + (c)^{2} + 2 \times a \times - b + 2 (- b) (c) + 2 \times c \times a$

$= a^{2} + b^{2} + c^{2} - 2 a b - 2 b c + 2 c a$ .

(vii) ${(3 x + \frac{1}{3 x})}^{2} = (3 x)^{2} + {(\frac{1}{3 x})}^{2} + 2 \times 3 x \times \frac{1}{3 x}$

$= 9 x^{2} + \frac{1}{9 x^{2}} + 2$ .

(viii) ${(2 x - \frac{1}{2 x})}^{2} = (2 x)^{2} + {(\frac{1}{2 x})}^{2} - 2 \times 2 x \times \frac{1}{2 x}$

$= 4 x^{2} + \frac{1}{4 x^{2}} - 2$ .

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Similar questions

Q. Solve for x.

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}

Solve for $x : \frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$

Q. If

f (x) =

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then

Solve by factorisation (quadratic equations)

1/2a+b+2x = 1/2a+ 1/b +1/2x

Q. Solve the following for x :

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} . \frac{+ 1}{b} + \frac{1}{2 x}

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Expand : (i) (2a+b)2 (ii) (a−2b)2 (iii) (a+12a)2 (iv) (2a−1a)2 (v) (a+b−c)2 (vi) (a−b+c)2 (vii) (3x+13x)2 (viii) (2x−12x)2

Expand :

(i) $(2 a + b)^{2}$

(ii) $(a - 2 b)^{2}$

(iii) ${(a + \frac{1}{2 a})}^{2}$

(iv) ${(2 a - \frac{1}{a})}^{2}$

(v) $(a + b - c)^{2}$

(vi) $(a - b + c)^{2}$

(vii) ${(3 x + \frac{1}{3 x})}^{2}$

(viii) ${(2 x - \frac{1}{2 x})}^{2}$