Question #e1762

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Question #e1762

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Answer 1 · 2017-10-21T08:12:53

$x = \pm 10$ $x=+-10$

Explanation:

For this question, $\sqrt{x^{2} - 36} = 8$ $sqrt(x^2-36)=8$ have the form of square root which is not a good form to solve the question. So, first, we try to take square of both sides to eliminate the square root.

${(\sqrt{x^{2} - 36})}^{2} = {(8)}^{2}$ $(sqrt(x^2-36))^2=(8)^2$
$x^{2} - 36 = 64$ $x^2-36=64$
$x^{2} - 100 = 0$ $x^2-100=0$
$(x + 10) (x - 10) = 0$ $(x+10)(x-10)=0$
$x = 10 or x = - 10$ $x=10 or x=-10$

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Question #e1762

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