Radical equations are equations involving a radical expression.  They are usually not too difficult to solve as long as you follow three steps:

• Isolate the radical expression (isolate them one at a time if there is more than one)
• Square (or cube, etc.) both sides
• Check for extraneous solutions.  Remember that when we multiply both sides of an equation by an expression involving a variable, an extraneous solution may arise (in the case where the variable causes a division by 0 or some other undefined situation…  you can do personal research to learn more about why extraneous solutions appear in radical equations).

Let’s jump straight to an example.  Solve the equation

\(x = \sqrt {4x} \)

We follow the steps from above. The radical is already isolated, so the first step is complete. Next, since the radical in question is a square root, we will square both sides of the equation to undo it. Then we will solve for \(x\). We have

\(x = \sqrt {4x} \)

\({x^2} = 4x\)

\({x^2} - 4x = 0\)

\(x\left( {x - 4} \right) = 0\)

\(x = 0,4\)

Finally, we must check our solutions to make sure that none of them are extraneous.  First we check \(x = 0\). We have

\(\begin{gathered}
  0 = \sqrt {4 \cdot 0}   \\
  0 = \sqrt 0   \\
  0 = 0  \\
\end{gathered} \)

The solution checks.  Next we check the solution  \(x = 4\). We have

\(\begin{gathered}
  4 = \sqrt {4 \cdot 4}   \\
  4 = \sqrt {16}   \\
  4 = 4  \\
\end{gathered} \)

The solution checks.  We have no extraneous solutions.  Let’s take a look at another example.  We want to solve the equation

\(\sqrt {30 - m}  = m\)

Again, we follow the same steps as before.  Here the radical is isolated, so the first step is complete.  Next, we want to square both sides.  We have

\(\sqrt {30 - m}  = m\)

\(30 - m = {m^2}\)

\({m^2} + m - 30 = 0\)

\(\left( {m + 6} \right)\left( {m - 5} \right) = 0\)

\(m =  - 6,5\)

Finally, we check for extraneous solutions.  Check \(m =  - 6\). We have

\(\sqrt {30 - \left( { - 6} \right)}  =  - 6\)

\(\sqrt {36}  =  - 6\)

\(6 =  - 6\)

The solution does not check!  So \( - 6\) is an extraneous, or false, solution.  We better check the solution  \(m = 5\). We have

\(\sqrt {30 - 5}  = 5\)

\(\sqrt {25}  = 5\)

\(5 = 5\)

The solution checks.  So the only solution to the equation \(\sqrt {30 - m}  = m\) is \(m = 5\).

Below you can download some free math worksheets and practice.