Find the value of 1-2 a-3-4 b2A- 1-4 a2-9-16 b2-4-3 abB- 1-4 a2-y-16 b2-3-4 a bC- 1-16 a2-y-16 b2-b-3 a bD- 1-4 a2-3-16 b2-3-4 a b

The correct option is B 1/4a^2 + 9/16b^2 + 3/4ab Using the identity (a+b)^2 = a^2 + b^2 + 2ab (1/2a+3/4b)^2 = 1/4a^2 + 9/16b^2 + 2×1/2a×3/4b = 1/4a^2 + 9/16b^2 ...

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Byju's Answer

Standard IX

Mathematics

Algebraic Identity 1

Find the valu...

Question

Find the value of ${(\frac{1}{2 a} + \frac{3}{4 b})}^{2}$

$\frac{1}{4 a^{2}} + \frac{9}{16 b^{2}} + \frac{3}{4 a b}$

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$\frac{1}{16 a^{2}} + \frac{9}{16 b^{2}} + \frac{3}{4 a b}$

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$\frac{1}{4 a^{2}} + \frac{3}{16 b^{2}} + \frac{3}{4 a b}$

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$\frac{1}{4 a^{2}} + \frac{9}{16 b^{2}} + \frac{4}{3 a b}$

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Solution

The correct option is A
$\frac{1}{4 a^{2}} + \frac{9}{16 b^{2}} + \frac{3}{4 a b}$

Using the identity $(a + b)^{2} = a^{2} + b^{2} + 2 a b$

${(\frac{1}{2 a} + \frac{3}{4 b})}^{2} = \frac{1}{4 a^{2}} + \frac{9}{16 b^{2}} + 2 \times \frac{1}{2 a} \times \frac{3}{4 b}$

$= \frac{1}{4 a^{2}} + \frac{9}{16 b^{2}} + \frac{3}{4 a b}$

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Similar questions

Q. Find

a^{2}

and

b^{2}

16 b^{2} + 9 a^{2} = a^{2} b^{2}

ii)

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

Q. Find the value of

(25 a^{2} + 16 b^{2} + 9 + 40 a b - 24 b - 30 a)

a = - 1

and

b = 2

Q. Factorise:

27 a^{3} + \frac{1}{64 b^{3}} + \frac{27 a^{2}}{4 b} + \frac{9 a}{16 b^{2}}

Factorise Completely : $a^{2} - 16 b^{2} - 2 a - 8 b$

Q. If

2 a + \frac{1}{2 a} = 8

, then find

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

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Find the value of (12a+34b)2

The correct option is A 14a2+916b2+34ab Using the identity (a+b)2=a2+b2+2ab (12a+34b)2=14a2+916b2+2×12a×34b =14a2+916b2+34ab

Find the value of ${(\frac{1}{2 a} + \frac{3}{4 b})}^{2}$