If $\sqrt{16 x^{2} - 48 x + 36} = 2 x + 12$ , then the sum of integral values of $x$ satisfying the equation is

9

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- 9

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- 8

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8

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Solution

The correct option is D $8$
$\sqrt{16 x^{2} - 48 x + 36} = 2 x + 12$
$\Rightarrow \sqrt{4 (4 x^{2} - 12 x + 9)} = 2 (x + 6)$
$\Rightarrow 2 \sqrt{(2 x - 3)^{2}} = 2 (x + 6)$
$\Rightarrow | 2 x - 3 | = x + 6 (∵ \sqrt{a^{2}} = | a |)$

Case $1 :$ If $x \geq \frac{3}{2}$
then $2 x - 3 = x + 6$
$\Rightarrow x = 9$

Case $2 :$ If $x < \frac{3}{2}$
then $3 - 2 x = x + 6$
$\Rightarrow x = - 1$

Sum $= 9 + (- 1) = 8$

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