How do you solve $sqrt(x+2) = x$ and are there any extraneous solutions?

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Answer 1 · 2018-03-13T14:17:22

$x = 2$

$sqrt (x + 2) = x$

square both sides:

$x + 2 = x^2$

$x^2 - x - 2 = 0$

$(x - 2)(x + 1) = 0$

$x = -1, x = 2$

Check:

$sqrt (-1 + 2) = -1$
$sqrt 1 = -1 : " extraneous solution"$

$sqrt (2 + 2) = 2$
$sqrt 4 = 2$ : this works in the original equation.

How do you solve sqrt(x+2) = xsqrt(x+2) = x and are there any extraneous solutions?