How do you solve $sqrt(x+7) = x-5$ and find any extraneous solutions?

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Answer 1 · 2017-01-22T06:18:46

$x=9$

$sqrt(x+7) = x-5$

$=> (sqrt(x+7))^2 = (x-5)^2$

$=> x+7 = x^2 - 10x + 25$

$=> x^2 - 11x + 18 = 0$

$=> (x-2)(x-9) = 0$

$=> x-2 = 0 or x-9 = 0$

$=> x = 2 or x = 9$

Test for extraneous solutions:

If $x=2$ , we have

$sqrt(2+7) = 3$
$2-5 = -3$

so $x=2$ is not a solution.

If $x-9$ , we have

$sqrt(9+7) = 4$
$9-5 = 4$

so $x=9$ is a solution.

Thus the only solution is $x=9$ .

How do you solve sqrt(x+7) = x-5sqrt(x+7) = x-5 and find any extraneous solutions?