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

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Answer 1 · 2016-05-30T08:22:17

$x=-1$ or $x=-2$

First square both sides (noting that this is where any extraneous solutions may be introduced) to give:

$x+2 = (x+2)^2 = x^2+4x+4$

Subtract $x+2$ from both ends to get:

$0 = x^2+3x+2 = (x+1)(x+2)$

So $x=-1$ or $x=-2$

Checking each of these:

$sqrt((-1)+2) = sqrt(1) = 1 = (-1)+2$

$sqrt((-2)+2) = sqrt(0) = 0 = (-2)+2$

So both of the solutions of our derived quadratic equation are also solutions of the original equation.

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