 Radical equations are equations where there is an $x$ in the radicand of a square root. With this type of equation, we might end up with extraneous solutions, therefore testing answers to the equation is a part of the solution process.

Here is how you do it:

Rule

1.
Isolate the square root on one side.
2.
Square both sides.
3.
Calculate and solve the equation normally.
4.
Test the answer to the equation.

Testing the answer to the equation is a part of the solving method for these equations. It often happens that you get extraneous solutions. It’s in “Testing the answer” where you detect and remove these extraneous solutions.

Example 1

Solve the equation $x-\sqrt{{x}^{2}-1}=1$

$\begin{array}{llll}\hfill x-\sqrt{{x}^{2}-1}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -\sqrt{{x}^{2}-1}& =1-x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-1& =\sqrt{{x}^{2}-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(x-1\right)}^{2}& ={\sqrt{{x}^{2}-1}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}-2x+1& ={x}^{2}-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Now you test the answer:

Since Left side $=$ Right side, the answer is $x=1$.

Example 2

Solve the equation $\sqrt{x+1}=x-3$

$\begin{array}{llll}\hfill \sqrt{x+1}& =x-3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\sqrt{x+1}}^{2}& =\phantom{\rule{-0.17em}{0ex}}{\left(x-3\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+1& ={x}^{2}-6x+9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}-7x+8& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Now you solve the quadratic expression with the quadratic formula, or with inspection. We’ll use the quadratic formula: $\begin{array}{llll}\hfill x& =\frac{7±\sqrt{\phantom{\rule{-0.17em}{0ex}}{\left(7\right)}^{2}-4\cdot 1\cdot \phantom{\rule{-0.17em}{0ex}}\left(8\right)}}{2\cdot 1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7±\sqrt{49-32}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7±\sqrt{17}}{2},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Possible solutions are then: $\begin{array}{l}\hfill \end{array}$

x = 7 + 17 2 1.438  and  x = 7 17 2 5.562
x = 7 + 17 2 1.438 and x = 7 17 2 5.562

Now you have to test the answer by inserting

 $x=\frac{7+\sqrt{17}}{2}\approx 1.438$

(If you get a small deviation on the decimals when you test the solutions, it is due to the rounding):

Since Left side $\ne$ Right side is $x=\frac{7+\sqrt{17}}{2}\approx 1.438$ an extraneous solution.

Now you insert

 $x=\frac{7-\sqrt{17}}{2}\approx 5.562$

(If you get a small deviation on the decimals when you test the solutions, it is the rounding).

Since Left side $=$ Right side, then $x=\frac{7-\sqrt{17}}{2}\approx 5.562$ is a solution.

The solution to the equation is then

 $x=\frac{7-\sqrt{17}}{2}\approx 5.562.$

Note! Testing the answer is a part of the solving method, because squaring can sometimes lead to extraneous solutions.