# 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 \begin{align*}\sqrt {x + 1}\end{align*}. The hypotenuse measures \begin{align*}\sqrt{7x + 1}\end{align*}. What are the lengths of the sides with the unknown values?

### Solving Radical Equations with Variables

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

Let's solve the following radical equations for x.

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

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

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

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

\begin{align*}\sqrt{4x+1}^2&=(x-1)^2\\ 4x+1&=x^2-2x+1\end{align*}

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.

\begin{align*}4x+1&=x^2-2x+1\\ 0&=x^2-6x\\ 0&=x(x-6)\\ x&=0 \ or \ 6\end{align*}

Check both solutions: \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.

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

Therefore, 6 is the only solution.

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

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

\begin{align*}\sqrt{8x-11}-\sqrt{3x+19}&=0 \\ \sqrt{8x-11}^2&=\sqrt{3x+19}^2 \\ 8x-11&=3x+19 \\ 5x&=30 \\ x&=6\end{align*}

Check: \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*}

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

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

\begin{align*}\sqrt[4]{2x^2-1}^4&=x^4 \\ 2x^2-1&=x^4 \\ 0&=x^4-2x^2+1 \\ 0&=(x^2-1)(x^2-1)\\ 0&=(x-1)(x+1)(x-1)(x+1)\\ x&=1 \ or \ -1\end{align*}

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

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

### Examples

#### Example 1

Earlier, you were asked to find the lengths of the sides with the unknown values.

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

\begin{align*}12^2 +(\sqrt {x + 1})^2) = (\sqrt {7x + 1})^2 \\ 144 + x + 1 = 7x + 1\\ 144 = 6x\\ x = 24\end{align*}

Now substitute this value into the sides with the unknowns.

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

\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.

Solve the following radical equations. Check for extraneous solutions.

#### Example 2

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

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

\begin{align*}\sqrt[3]{4x^3-24}^3&=x^3\\ 4x^3-24&=x^3\\ -24&=-3x^3\\ 8&=x^3\\ 2&=x\end{align*}

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

#### Example 3

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

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

\begin{align*}\sqrt{5x-3}^2&=\sqrt{3x+19}^2\\ 5x-3&=3x+19\\ 2x&=22\\ x&=11\end{align*}

Check: \begin{align*}\sqrt{5 \left(11 \right)-3}&=\sqrt{3 \left(11 \right)+19} \\ \sqrt{55-3}&=\sqrt{33+19} \quad \\ \sqrt{52}&=\sqrt{52}\end{align*}

#### Example 4

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

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

\begin{align*}\sqrt{6x-5}^2&=(x-10)^2\\ 6x-5&=x^2-20x+100\\ 0&=x^2-26x+105\\ 0&=(x-21)(x-5)\\ x&=21 \ or \ 5\end{align*}

Check both solutions: \begin{align*}x&= 21: \sqrt{6 \left(21 \right)-5}-21=\sqrt{126-5}-21=\sqrt{121}-21=11-21=-10 \\ x&= 5: \sqrt{6 \left(5 \right)-5}-21=\sqrt{30-5}-21=\sqrt{25}-21=5-21 \neq -10\end{align*}

5 is an extraneous solution.

### Review

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

1. \begin{align*}\sqrt{x-3}=x-5\end{align*}
2. \begin{align*}\sqrt{x+3}+15=x-12\end{align*}
3. \begin{align*}\sqrt[4]{3x^2+54}=x\end{align*}
4. \begin{align*}\sqrt{x^2+60}=4\sqrt{x}\end{align*}
5. \begin{align*}\sqrt{x^4+5x^3}=2\sqrt{2x+10}\end{align*}
6. \begin{align*}x=\sqrt{5x-6}\end{align*}
7. \begin{align*}\sqrt{3x+4}=x-2\end{align*}
8. \begin{align*}\sqrt{x^3+8x}-\sqrt{9x^2-60}=0\end{align*}
9. \begin{align*}x=\sqrt[3]{4x+4-x^2}\end{align*}
10. \begin{align*}\sqrt[4]{x^3+3}=2\sqrt[4]{x+3}\end{align*}
11. \begin{align*}x^2-\sqrt{42x^2+343}=0\end{align*}
12. \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.

\begin{align*}\sqrt{x-15}&=\sqrt{x}-3\\ \left(\sqrt{x-15}\right)^2&=\left(\sqrt{x}-3 \right)^2\\ x-15&=x-6 \sqrt{x}+9\\ -24&=-6 \sqrt{x}\\ (4)^2&=\left(\sqrt{x}\right)^2\\ 16&=x\end{align*}

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

\begin{align*}\sqrt{16-15}&=\sqrt{16}-3 \\ \sqrt{1}&=4-3 \\ 1&=1\end{align*}

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

To see the Review answers, open this PDF file and look for section 7.8.

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