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

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Solution

The correct option is C $- 3$
Given, $16^{(x^{2} + 3 x - 1)} = 8^{(x^{2} + 3 x + 2)}$

As, 16 = $2^{4} a n d 8 = 2^{3}$

$2^{4 (x^{2} + 3 x - 1)} = 2^{3 (x^{2} + 3 x + 2)}$

So, we can write

$4 (x^{2} + 3 x - 1) = 3 (x^{2} + 3 x + 2)$

$4 x^{2} + 12 x - 4 = 3 x^{2} + 9 x + 6$

$4 x^{2} + 12 x - 4 - (3 x^{2} + 9 x + 6) = 0$
$4 x^{2} + 12 x - 4 - 3 x^{2} - 9 x - 6 = 0$

$x^{2} + 3 x - 10 = 0$

$x^{2} + 5 x - 2 x - 10 = 0$

$x (x + 5) - 2 (x + 5) = 0$

$(x - 2) (x + 5) = 0$

So, $x = 2, - 5$

Sum of values of $x = 2 + (- 5) = - 3$

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