# How do you solve sqrt(x + 2)+4 = x?

Aug 5, 2015

$x = 7$

#### Explanation:

Start by isolating the radical on one side of the equation by adding $- 4$ to both sides

$\sqrt{x + 2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} = x - 4$

$\sqrt{x + 2} = x - 4$

This means that, in order to be considered a valid solution, the value(s) of $x$ must satisfy the condition

$x - 4 \ge 0 \iff x \ge 4$

Next, square both sides of the equation to get rid of the square root

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

$x + 2 = {x}^{2} - 8 x + 16$

Rearrange this equation to get all the terms on one side

${x}^{2} - 9 x + 14 = 0$

You can find the two solutions to this quadratic by using the quadratic formula

${x}_{1 , 2} = \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \cdot 1 \cdot \left(- 14\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{9 + \sqrt{25}}{2} = \frac{9 \pm 5}{2} = \left\{\begin{matrix}{x}_{1} = \frac{9 + 5}{2} = \textcolor{g r e e n}{7} \\ {x}_{2} = \frac{9 - 5}{2} = \cancel{\textcolor{red}{2}}\end{matrix}\right.$

The only valid solution will be $x = 7$; $x = 2$ will be an extraneous solution, since it does not satisfy the aforementioned condition $x \ge 4$.