When you want to solve an equation with containing a radical expression you have to isolate the radical on one side from all other terms and then square both sides of the equation.

Example

$3\sqrt{x}=9$

$\sqrt{x}=\frac{9}{3}=3$

$\begin{pmatrix} \sqrt{x} \end{pmatrix}^{2}=\begin{pmatrix} 3 \end{pmatrix}^{2}$

$x=9$

When you square a radical equation you sometimes get a solution to the squared equation that is not a solution to the original equation. Such an equation is called an extraneous solution. Remember to always check your solutions in the original equation to discard the extraneous solutions.

Example

$\sqrt{2-x}=x$

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

$2-x=x^{2}$

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

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

$x=\frac{-1\pm \sqrt{1+8}}{2}$

$x_{1}=\frac{-1+\sqrt{9}}{2}=\frac{-1+3}{2}=\frac{2}{2}=1$

$x_{2}=\frac{-1-\sqrt{9}}{2}=\frac{-1-3}{2}=\frac{-4}{2}=-2$

Here we've got two solutions x = 1 or x = (-2). We check both solutions in the original equation to test whether they are true solutions or extraneous solutions.

$\begin{matrix} \sqrt{2-1}\overset{?}{=}1& \: or\: & \sqrt{2-\left ( -2 \right )}\overset{?}{=}-2\\ 1=1& &2=-2 \\ & & {\color{red} {Wrong!}} \end{matrix}$

As we could see when we checked our numbers in the original equation x =1 is the only true solution for this equation and that x = -2 is an extraneous solution.

## Video lesson

$\sqrt{10-x}=x+2$