A radical equation is one where the variable appears under a square root sign. (For the time being we only worry about square roots, not cube roots or other interesting things.) The general technique for solving a radical equation is: isolate the square root sign (and whatever's under it) on one side of the equation. Then square both sides. You should end up with an equation which can be solved by normal methods.

Example 1:

Solve.

$\sqrt{x+2}=3$

Square both sides: ${\left(\sqrt{x+2}\right)}^{2}={3}^{2}$

$x+2=9$

Solve the equation: $x+2=9$

$\begin{array}{l}\sqrt{7+2}\stackrel{?}{=}3\\ \sqrt{9}\stackrel{?}{=}3\\ 3=3\end{array}$

Therefore, $7$ is the solution to the equation.

Example 2:

Solve.

$\sqrt{12x+4}+5=x$

Get the square root alone on one side, by subtracting $5$ from both sides.

$\sqrt{12x+4}=x-5$

Square both sides.

${\left(\sqrt{12x+4}\right)}^{2}={\left(x-5\right)}^{2}$

$12x+4={x}^{2}-10x+25$

Simplify.

${x}^{2}-22x+21=0$

In this case, the result is a quadratic equation , which can be solved by factoring .

$\left(x-21\right)\left(x-1\right)=0$

So, by the zero product property ,

$x=21\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1$ .

Important Note: The step when we square both sides can introduce some " extraneous solutions ". You must check for validity of solutions.

Substituting $x=21$ :

$\begin{array}{l}\sqrt{12\left(21\right)+4}+5\stackrel{?}{=}21\\ \sqrt{256}+5\stackrel{?}{=}21\\ 21=21\end{array}$

So the first solution checks out.

Substituting $x=1$ :

$\begin{array}{l}\sqrt{12\left(1\right)+4}+5\stackrel{?}{=}1\\ \sqrt{16}+5\stackrel{?}{=}1\\ 9\ne 1\end{array}$

So the second solution doesn't check out. Therefore $x=1$ is the extraneous solution. The only valid solution is $x=21$ .