# How do you solve sqrt(x+6)=x?

Oct 12, 2015

$x = 3$

#### Explanation:

Start by squaring both sides of the equation to get rid of the square root

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

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

Next, move all the tems on one side of the equation

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

Use the quadratic formula to find the two roots of this quadratic equation

${x}_{1 , 2} = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \cdot \left(- 6\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2} = \left\{\begin{matrix}{x}_{1} = \frac{1 + 5}{2} = 3 \\ {x}_{2} = \frac{1 - 5}{2} = - 2\end{matrix}\right.$

Notice that $x = - 2$ does not satisfy the initial equation, since you would get

$\sqrt{- 2 + 6} = - 2$

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

This means that $x = - 2$ will eb an extraneous solution. The only valid solution will thus be $x = 3$, for which

$\sqrt{3 + 6} = 3$

$\sqrt{9} = 3 \text{ } \textcolor{g r e e n}{\sqrt{}}$