### Equations

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 $$m = 5$$. 2452 x

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: 2643 x

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: 2508 x

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:

### Geometry

Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles

### Algebra and Pre-Algebra

Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
Percents
Polynomials