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

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Answer 1 · 2017-01-22T06:18:46

$x = 9$ $x=9$

$\sqrt{x + 7} = x - 5$ $sqrt(x+7) = x-5$

$\Rightarrow {(\sqrt{x + 7})}^{2} = {(x - 5)}^{2}$ $=> (sqrt(x+7))^2 = (x-5)^2$

$\Rightarrow x + 7 = x^{2} - 10 x + 25$ $=> x+7 = x^2 - 10x + 25$

$\Rightarrow x^{2} - 11 x + 18 = 0$ $=> x^2 - 11x + 18 = 0$

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

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

$\Rightarrow x = 2 or x = 9$ $=> x = 2 or x = 9$

Test for extraneous solutions:

If $x = 2$ $x=2$ , we have

$\sqrt{2 + 7} = 3$ $sqrt(2+7) = 3$
$2 - 5 = - 3$ $2-5 = -3$

so $x = 2$ $x=2$ is not a solution.

If $x - 9$ $x-9$ , we have

$\sqrt{9 + 7} = 4$ $sqrt(9+7) = 4$
$9 - 5 = 4$ $9-5 = 4$

so $x = 9$ $x=9$ is a solution.

Thus the only solution is $x = 9$ $x=9$ .

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