Find the square of: $3 a - 4 b$ .

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

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9 a^{2} - 24 a b + 16 b^{2}

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

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9 a^{2} + 24 a b + b^{2}

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Solution

The correct option is B $9 a^{2} - 24 a b + 16 b^{2}$
Given, $(3 a - 4 b)$ .

On squaring, we get, $(3 a - 4 b)^{2}$ .

We know, $(x + y)^{2} = x^{2} + y^{2} + 2 x y$ .

Then,
$(3 a - 4 b)^{2}$
$= (3 a)^{2} - 2 (3 a) (4 b) + (4 b)^{2}$
$= 9 a^{2} - 24 a b + 16 b^{2}$ .

Therefore, option $B$ is correct.

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