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.

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**