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

May 30, 2016

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

#### Explanation:

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

$x + 2 = {\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4$

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

$0 = {x}^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right)$

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

Checking each of these:

$\sqrt{\left(- 1\right) + 2} = \sqrt{1} = 1 = \left(- 1\right) + 2$

$\sqrt{\left(- 2\right) + 2} = \sqrt{0} = 0 = \left(- 2\right) + 2$

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