# How do you solve x=sqrt(x+6) and find any extraneous solutions?

Apr 30, 2018

$x = 3 , x = - 2 \text{ is extraneous}$

#### Explanation:

$\textcolor{b l u e}{\text{square both sides}}$

$\text{note that } \sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a$

${x}^{2} = {\left(\sqrt{x + 6}\right)}^{2}$

$\Rightarrow {x}^{2} = x + 6$

"express in "color(blue)"standard form ";ax^2+bx+c=0

$\Rightarrow {x}^{2} - x - 6 = 0$

$\text{the factors of - 6 which sum to - 1 are - 3 and + 2}$

$\Rightarrow \left(x - 3\right) \left(x + 2\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x + 2 = 0 \Rightarrow x = - 2$

$x - 3 = 0 \Rightarrow x = 3$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the equation and if both sides equate then they are the solutions.

$x = - 2 \to \text{ left "=-2" right } = \sqrt{4} = 2$

$- 2 \ne 2 \Rightarrow x = - 2 \text{ is extraneous}$

$x = 3 \to \text{left "=3" right } = \sqrt{9} = 3$

$\Rightarrow x = 3 \text{ is the solution}$