Solving Radical Equations

Example 1:

Solve.

$\sqrt{x+2}=3$

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

$x+2=9$

Solve the equation: $x+2=9$

Always check your solution:

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

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 .

$(x-21)(x-1)=0$

So, by the zero product property ,

$x=21\text{or}x=1$ .

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 $x=21$ .