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# 7.8: Solving Radical Equations with Variables on Both Sides

Difficulty Level: At Grade Created by: CK-12
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Practice Radical Equations with Variables on Both Sides
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The legs of a right triangle measure 12 and x+1\begin{align*}\sqrt {x + 1}\end{align*}. The hypotenuse measures 7x+1\begin{align*}\sqrt{7x + 1}\end{align*}. What are the lengths of the sides with the unknown values?

### Guidance

In this concept, we will continue solving radical equations. Here, we will address variables and radicals on both sides of the equation.

#### Example A

Solve 4x+1x=1\begin{align*}\sqrt{4x+1}-x=-1\end{align*}

Solution: Now we have an x\begin{align*}x\end{align*} that is not under the radical. We will still isolate the radical.

4x+1x4x1=1=x1

Now, we can square both sides. Be careful when squaring x1\begin{align*}x-1\end{align*}, the answer is not x21\begin{align*}x^2-1\end{align*}.

4x+124x+1=(x1)2=x22x+1

This problem is now a quadratic. To solve quadratic equations, we must either factor, when possible, or use the Quadratic Formula. Combine like terms and set one side equal to zero.

4x+100x=x22x+1=x26x=x(x6)=0 or 6

Check both solutions: 4(0)+11=0+11=11=01\begin{align*}\sqrt{4 \left(0\right)+1}-1=\sqrt{0+1}-1=1-1=0 \neq -1\end{align*}. 0 is an extraneous solution.

4(6)+16=24+16=56=1\begin{align*}\sqrt{4 \left(6 \right)+1}-6=\sqrt{24+1}-6=5-6=-1\end{align*}

Therefore, 6 is the only solution.

#### Example B

Solve 8x113x+19=0\begin{align*}\sqrt{8x-11}-\sqrt{3x+19}=0\end{align*}.

Solution: In this example, you need to isolate both radicals. To do this, subtract the second radical from both sides. Then, square both sides to eliminate the variable.

8x113x+198x1128x115xx=0=3x+192=3x+19=30=6

Check: 8(6)113(6)+19=481118+19=3737=0\begin{align*}\sqrt{8 \left(6 \right)-11}-\sqrt{3 \left(6 \right)+19}=\sqrt{48-11}-\sqrt{18+19}=\sqrt{37}-\sqrt{37}=0\end{align*}

#### Example C

Solve 4x+14=x\begin{align*}\sqrt[4]{4x+1}=x\end{align*}

Solution: The radical is isolated. To eliminate it, we must raise both sides to the fourth power.

2x21442x21000x=x4=x4=x42x2+1=(x21)(x21)=(x1)(x+1)(x1)(x+1)=1 or 1

Check: 2(1)214=214=14=1\begin{align*}\sqrt[4]{2(1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1 \end{align*}

2(1)214=214=14=1\begin{align*}\sqrt[4]{2(-1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1\end{align*}

Intro Problem Revisit Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the sides with unknowns.

122+(x+1)2)=(7x+1)2144+x+1=7x+1144=6xx=24

Now substitute this value into the sides with the unknowns.

x+1=24+1=5\begin{align*} \sqrt{x + 1} = \sqrt {24 + 1} = 5\end{align*} and

7x+1=[7(24)]+1=169=13\begin{align*} \sqrt{7x + 1} = \sqrt {[7(24)] + 1} = \sqrt {169} = 13\end{align*}

Therefore the leg with the unknown measures 5 and the hypotenuse measures 13.

### Guided Practice

Solve the following radical equations. Check for extraneous solutions.

1. 4x3243=x\begin{align*}\sqrt[3]{4x^3-24}=x\end{align*}

2. 5x3=3x+19\begin{align*}\sqrt{5x-3}=\sqrt{3x+19}\end{align*}

3. 6x5x=10\begin{align*}\sqrt{6x-5}-x=-10\end{align*}

1. The radical is isolated. Cube both sides to eliminate the cubed root.

4x324334x3242482=x3=x3=3x3=x3=x

Check: 4(2)3243=32243=83=2\begin{align*}\sqrt[3]{4 \left(2 \right)^3-24}=\sqrt[3]{32-24}=\sqrt[3]{8}=2\end{align*}

2. Square both sides to solve for x\begin{align*}x\end{align*}.

5x325x32xx=3x+192=3x+19=22=11

Check:

5(11)355352=3(11)+19=33+19=52

3. Add x\begin{align*}x\end{align*} to both sides and square to eliminate the radical.

6x526x500x=(x10)2=x220x+100=x226x+105=(x21)(x5)=21 or 5

Check both solutions:

xx=21:6(21)521=126521=12121=1121=10=5:6(5)521=30521=2521=52110

5 is an extraneous solution.

### Explore More

Solve the following radical equations. Be sure to check for extraneous solutions.

1. x3=x5\begin{align*}\sqrt{x-3}=x-5\end{align*}
2. x+3+15=x12\begin{align*}\sqrt{x+3}+15=x-12\end{align*}
3. 3x2+544=x\begin{align*}\sqrt[4]{3x^2+54}=x\end{align*}
4. x2+60=4x\begin{align*}\sqrt{x^2+60}=4\sqrt{x}\end{align*}
5. x4+5x3=22x+10\begin{align*}\sqrt{x^4+5x^3}=2\sqrt{2x+10}\end{align*}
6. x=5x6\begin{align*}x=\sqrt{5x-6}\end{align*}
7. 3x+4=x2\begin{align*}\sqrt{3x+4}=x-2\end{align*}
8. x3+8x9x260=0\begin{align*}\sqrt{x^3+8x}-\sqrt{9x^2-60}=0\end{align*}
9. x=4x+4x23\begin{align*}x=\sqrt[3]{4x+4-x^2}\end{align*}
10. x3+34=2x+34\begin{align*}\sqrt[4]{x^3+3}=2\sqrt[4]{x+3}\end{align*}
11. x242x2+343=0\begin{align*}x^2-\sqrt{42x^2+343}=0\end{align*}
12. xx221=2x325x+25\begin{align*}x\sqrt{x^2-21}=2\sqrt{x^3-25x+25}\end{align*}

For questions 13-15, you will need to use the method illustrated in the example below.

x15(x15)2x1524(4)216=x3=(x3)2=x6x+9=6x=(x)2=x

1. Square both sides
2. Combine like terms to isolate the remaining radical
3. Square both sides again to solve

161511=163=43=1

1. x+112=x21\begin{align*}\sqrt{x+11}-2=\sqrt{x-21}\end{align*}
2. x6=7x22\begin{align*}\sqrt{x-6}=\sqrt{7x}-22\end{align*}
3. 2+x+5=4x7\begin{align*}2+\sqrt{x+5}=\sqrt{4x-7}\end{align*}

## Date Created:

Mar 12, 2013

Jun 04, 2015
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