Find the sum of all values of $x$ so that $16^{(x^{2} + 3 x - 1)} = 8^{(x^{2} + 3 x + 2)}$

0

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3

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- 5

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- 3

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Solution

The correct option is D $- 3$
$16^{x^{2} + 3 x - 1} = 8^{x^{2} + 3 x + 2}$
$\Rightarrow 8^{2 (x^{2} + 3 x - 1)} = 8^{x^{2} + 3 x + 2}$
$\Rightarrow 2 x^{2} + 6 x - 2 = x^{2} + 3 x + 2$
$\Rightarrow x^{2} + 3 x - 4 = 0$
$\Rightarrow (x + 4) (x - 1) = 0$
$\Rightarrow x = - 4, 1$
sum of values of $x = - 4 + 1 = - 3$
Option D is correct.

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