# 7.7: Solving Simple Radical Equations

**At Grade**Created by: CK-12

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**Practice**Radical Equations

The legs of a right triangle measure 3 and \begin{align*}2\sqrt {x}\end{align*} . The hypotenuse measures 5. What is the length of the leg with the unknown value?

### Guidance

Solving radical equations are very similar to solving other types of equations. The objective is to get @$\begin{align*}x\end{align*}@$ by itself. However, now there are radicals within the equations. Recall that the opposite of the square root of something is to square it.

#### Example A

Is @$\begin{align*}x = 5\end{align*}@$ the solution to @$\begin{align*}\sqrt{2x+15}=8\end{align*}@$ ?

**
Solution:
**
Plug in 5 for
@$\begin{align*}x\end{align*}@$
to see if the equation holds true. If it does, then 5 is the solution.

@$$\begin{align*}\sqrt{2 \left(5\right)+15}&=8 \\ \sqrt{10+15}&=9 \\ \sqrt{25} &\neq 8\end{align*}@$$

We know that @$\begin{align*}\sqrt{25}=5\end{align*}@$ , so @$\begin{align*}x = 5\end{align*}@$ is not the solution.

#### Example B

Solve @$\begin{align*}\sqrt{2x-5}+7=16\end{align*}@$ .

**
Solution:
**
To solve for
@$\begin{align*}x\end{align*}@$
, we need to isolate the radical. Subtract 7 from both sides.

@$$\begin{align*}\sqrt{2x-5}+7&=16 \\ \sqrt{2x-5}&=9\end{align*}@$$

Now, we can square both sides to eliminate the radical. Only square both sides when the radical is alone on one side of the equals sign.

@$$\begin{align*}\sqrt{2x-5}^2&=9^2 \\ 2x-5&=81 \\ 2x&=86 \\ x&=43\end{align*}@$$

Check: @$\begin{align*}\sqrt{2 \left(43\right)-5}+7=\sqrt{86-5}+7=\sqrt{81}+7=9+7=16\end{align*}@$

ALWAYS check your answers when solving radical equations. Sometimes, you will solve an equation, get a solution, and then plug it back in and it will not work. These types of solutions are called
**
extraneous solutions
**
and are not actually considered solutions to the equation.

#### Example C

Solve @$\begin{align*}3\sqrt[3]{x-8}-2=-14\end{align*}@$ .

**
Solution:
**
Again, isolate the radical first. Add 2 to both sides and divide by 3.

@$$\begin{align*}3\sqrt[3]{x-8}-2&=-14\\ 3\sqrt[3]{x-8}&=-12\\ \sqrt[3]{x-8}&=-4\end{align*}@$$

Now, cube both sides to eliminate the radical.

@$$\begin{align*}\sqrt[3]{x-8}^3&=(-4)^3\\ x-8&=-64\\ x&=-56\end{align*}@$$

Check: @$\begin{align*}3\sqrt[3]{-56-8}-2=3 \sqrt[3]{-64}-2=3 \cdot -4-2=-12-2=-14\end{align*}@$

**
Intro Problem Revisit
**
Use the Pythagorean Theorem and solve for
*
x
*
then substitute that value in to solve for the leg with the unknown.

@$$\begin{align*}3^2 +(2\sqrt {x})^2) = 5^2\\ 9 + 4x = 25\\ 4x = 16\\ x = 4\end{align*}@$$

Now substitute this value into the leg with the unknown.

@$\begin{align*}2 \sqrt{4} = 2 \cdot 2 = 4\end{align*}@$

Therefore the leg with the unknown has a length of 4.

### Guided Practice

Solve the equations and check your answers.

1. @$\begin{align*}\sqrt{x+5}=6\end{align*}@$

2. @$\begin{align*}5\sqrt{2x-1}+1=26\end{align*}@$

3. @$\begin{align*}\sqrt[4]{3x+11}-2=3\end{align*}@$

#### Answers

1. The radical is already isolated here. Square both sides and solve for @$\begin{align*}x\end{align*}@$ .

@$$\begin{align*}\sqrt{x+5}^2&=6^2 \\ x+5&=36 \\ x&=31\end{align*}@$$

Check: @$\begin{align*}\sqrt{31+5}=\sqrt{36}=6 \end{align*}@$

2. Isolate the radical by subtracting 1 and then dividing by 5.

@$$\begin{align*}5\sqrt{2x-1}+1&=26 \\ 5\sqrt{2x-1}&=25 \\ \sqrt{2x-1}&=5\end{align*}@$$

Square both sides and continue to solve for @$\begin{align*}x\end{align*}@$ .

@$$\begin{align*}\sqrt{2x-1}^2&=5^2 \\ 2x-1&=25 \\ 2x&=26 \\ x&=13\end{align*}@$$

Check: @$\begin{align*}5\sqrt{2 \left(13\right)-1}+1=5\sqrt{26-1}=5\sqrt{25}+1=5 \cdot 5+1=25+1=26\end{align*}@$

3. In this problem, we have a fourth root. That means, once we isolate the radical, we must raise both sides to the fourth power to eliminate it.

@$$\begin{align*}\sqrt[4]{3x+11}-2&=3\\ \sqrt[4]{3x-11}^4&=5^4\\ 3x-11&=625\\ 3x&=636\\ x&=212\end{align*}@$$

Check: @$\begin{align*}\sqrt[4]{3 \left(212\right)+11}-2=\sqrt[4]{636-11}-2=\sqrt[4]{625}-2=5-2=3\end{align*}@$

### Explore More

Determine if the given values of
*
x
*
are solutions to the radical equations below.

- @$\begin{align*}\sqrt{x-3}=7; x = 32\end{align*}@$
- @$\begin{align*}\sqrt[3]{6+x}=3; x = 21\end{align*}@$
- @$\begin{align*}\sqrt[4]{2x+3}-11=-9; x = 6\end{align*}@$

Solve the equations and check your answers.

- @$\begin{align*}\sqrt{x+5}=6\end{align*}@$
- @$\begin{align*}2- \sqrt{x+1}=0\end{align*}@$
- @$\begin{align*}4 \sqrt{5-x}=12\end{align*}@$
- @$\begin{align*}\sqrt{x+9}+7=11\end{align*}@$
- @$\begin{align*}\frac{1}{2}\sqrt[3]{x-2}=1\end{align*}@$
- @$\begin{align*}\sqrt[3]{x+3}+5=9\end{align*}@$
- @$\begin{align*}5\sqrt{15-x}+2=17\end{align*}@$
- @$\begin{align*}-5=\sqrt[5]{x-5}-7\end{align*}@$
- @$\begin{align*}\sqrt[4]{x-6}+10=13\end{align*}@$
- @$\begin{align*}\frac{8}{5}\sqrt[3]{x+5}=8\end{align*}@$
- @$\begin{align*}3 \sqrt{x+7}-2=25\end{align*}@$
- @$\begin{align*}\sqrt[4]{235+x}+9=14\end{align*}@$

Extraneous Solution

An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.Quadratic Equation

A quadratic equation is an equation that can be written in the form , where , , and are real constants and .Quadratic Formula

The quadratic formula states that for any quadratic equation in the form , .Radical Expression

A radical expression is an expression with numbers, operations and radicals in it.### Image Attributions

## Description

## Learning Objectives

Here you'll learn how to solve basic radical equations.

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## Date Created:

Mar 12, 2013## Last Modified:

Jun 04, 2015## Vocabulary

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