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# 7.7: Solving Simple Radical Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Radical Equations

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

### Solving Radical Equations

Solving radical equations are very similar to solving other types of equations. The objective is to get x\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.

Let's determine if x=5\begin{align*}x = 5\end{align*} is the solution to 2x+15=8\begin{align*}\sqrt{2x+15}=8\end{align*}.

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

2(5)+1510+1525=8=98\begin{align*}\sqrt{2 \left(5\right)+15}&=8 \\ \sqrt{10+15}&=9 \\ \sqrt{25} &\neq 8\end{align*}

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

Now, let's solve the following equations for x.

1. 2x5+7=16\begin{align*}\sqrt{2x-5}+7=16\end{align*}

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

2x5+72x5=16=9\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.

2x522x52xx=92=81=86=43\begin{align*}\sqrt{2x-5}^2&=9^2 \\ 2x-5&=81 \\ 2x&=86 \\ x&=43\end{align*}

Check: 2(43)5+7=865+7=81+7=9+7=16\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.

1. 3x832=14\begin{align*}3\sqrt[3]{x-8}-2=-14\end{align*}

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

3x8323x83x83=14=12=4\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.

x833x8x=(4)3=64=56\begin{align*}\sqrt[3]{x-8}^3&=(-4)^3\\ x-8&=-64\\ x&=-56\end{align*}

Check: 356832=36432=342=122=14\begin{align*}3\sqrt[3]{-56-8}-2=3 \sqrt[3]{-64}-2=3 \cdot -4-2=-12-2=-14\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the length of the leg with the unknown value.

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

32+(2x)2)=529+4x=254x=16x=4\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.

24=22=4\begin{align*}2 \sqrt{4} = 2 \cdot 2 = 4\end{align*}

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

#### Example 2

Solve for x: x+5=6\begin{align*}\sqrt{x+5}=6\end{align*}. Check your answer.

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

x+52x+5x=62=36=31\begin{align*}\sqrt{x+5}^2&=6^2 \\ x+5&=36 \\ x&=31\end{align*}

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

#### Example 3

Solve for x: 52x1+1=26\begin{align*}5\sqrt{2x-1}+1=26\end{align*}. Check your answer.

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

52x1+152x12x1=26=25=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 x\begin{align*}x\end{align*}.

2x122x12xx=52=25=26=13\begin{align*}\sqrt{2x-1}^2&=5^2 \\ 2x-1&=25 \\ 2x&=26 \\ x&=13\end{align*}

Check: 52(13)1+1=5261=525+1=55+1=25+1=26\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*}

#### Example 4

Solve for x: 3x+1142=3\begin{align*}\sqrt[4]{3x+11}-2=3\end{align*}. Check your answer.

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.

3x+11423x11443x113xx=3=54=625=636=212\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: 3(212)+1142=6361142=62542=52=3\begin{align*}\sqrt[4]{3 \left(212\right)+11}-2=\sqrt[4]{636-11}-2=\sqrt[4]{625}-2=5-2=3\end{align*}

### Review

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

1. x3=7;x=32\begin{align*}\sqrt{x-3}=7; x = 32\end{align*}
2. 6+x3=3;x=21\begin{align*}\sqrt[3]{6+x}=3; x = 21\end{align*}
3. 2x+3411=9;x=6\begin{align*}\sqrt[4]{2x+3}-11=-9; x = 6\end{align*}

Solve the equations and check your answers.

1. x+5=6\begin{align*}\sqrt{x+5}=6\end{align*}
2. 2x+1=0\begin{align*}2- \sqrt{x+1}=0\end{align*}
3. 45x=12\begin{align*}4 \sqrt{5-x}=12\end{align*}
4. x+9+7=11\begin{align*}\sqrt{x+9}+7=11\end{align*}
5. \begin{align*}\frac{1}{2}\sqrt[3]{x-2}=1\end{align*}
6. \begin{align*}\sqrt[3]{x+3}+5=9\end{align*}
7. \begin{align*}5\sqrt{15-x}+2=17\end{align*}
8. \begin{align*}-5=\sqrt[5]{x-5}-7\end{align*}
9. \begin{align*}\sqrt[4]{x-6}+10=13\end{align*}
10. \begin{align*}\frac{8}{5}\sqrt[3]{x+5}=8\end{align*}
11. \begin{align*}3 \sqrt{x+7}-2=25\end{align*}
12. \begin{align*}\sqrt[4]{235+x}+9=14\end{align*}

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.7.

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### Vocabulary Language: English

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 $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.

Quadratic Formula

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Radical Expression

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

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Mar 12, 2013
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Sep 07, 2016
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