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:

$\\ \sqrt{2-x}=x \\\\\left ( \sqrt{2-x} \right )^{2}=x^{2} \\\\2-x=x^{2} \\x^{2}+x-2=0 \\\\x=\frac{-1\pm \sqrt{1^{2}-4\cdot \left ( -2 \right )}}{2} \\\\x=\frac{-1\pm \sqrt{1+8}}{2} \\\\\left\{\begin{matrix} x_{1}=\frac{-1+\sqrt{9}}{2}=\frac{-1+3}{2}=\frac{2}{2}=1\: \: \: \: \\\\ x_{2}=\frac{-1-\sqrt{9}}{2}=\frac{-1-3}{2}=\frac{-4}{2}=-2 \end{matrix}\right. \\$

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: Solve the radical equation

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