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# Radical Equations with Variables on Both Sides

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Practice Radical Equations with Variables on Both Sides
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Solving Radical Equations with Variables on Both Sides

The legs of a right triangle measure 12 and $\sqrt {x + 1}$ . The hypotenuse measures $\sqrt{7x + 1}$ . 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 $\sqrt{4x+1}-x=-1$

Solution: Now we have an $x$ that is not under the radical. We will still isolate the radical.

$\sqrt{4x+1}-x&=-1\\\sqrt{4x-1}&=x-1$

Now, we can square both sides. Be careful when squaring $x-1$ , the answer is not $x^2-1$ .

$\sqrt{4x+1}^2&=(x-1)^2\\4x+1&=x^2-2x+1$

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.

$4x+1&=x^2-2x+1\\0&=x^2-6x\\0&=x(x-6)\\x&=0 \ or \ 6$

Check both solutions: $\sqrt{4 \left(0\right)+1}-1=\sqrt{0+1}-1=1-1=0 \neq -1$ . 0 is an extraneous solution.

$\sqrt{4 \left(6 \right)+1}-6=\sqrt{24+1}-6=5-6=-1$

Therefore, 6 is the only solution.

#### Example B

Solve $\sqrt{8x-11}-\sqrt{3x+19}=0$ .

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.

$\sqrt{8x-11}-\sqrt{3x+19}&=0 \\\sqrt{8x-11}^2&=\sqrt{3x+19}^2 \\8x-11&=3x+19 \\5x&=30 \\x&=6$

Check: $\sqrt{8 \left(6 \right)-11}-\sqrt{3 \left(6 \right)+19}=\sqrt{48-11}-\sqrt{18+19}=\sqrt{37}-\sqrt{37}=0$

#### Example C

Solve $\sqrt[4]{4x+1}=x$

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

$\sqrt[4]{2x^2-1}^4&=x^4 \\2x^2-1&=x^4 \\0&=x^4-2x^2+1 \\0&=(x^2-1)(x^2-1)\\0&=(x-1)(x+1)(x-1)(x+1)\\x&=1 \ or \ -1$

Check: $\sqrt[4]{2(1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1$

$\sqrt[4]{2(-1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1$

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

$12^2 +(\sqrt {x + 1})^2) = (\sqrt {7x + 1})^2 \\144 + x + 1 = 7x + 1\\144 = 6x\\x = 24$

Now substitute this value into the sides with the unknowns.

$\sqrt{x + 1} = \sqrt {24 + 1} = 5$ and

$\sqrt{7x + 1} = \sqrt {[7(24)] + 1} = \sqrt {169} = 13$

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. $\sqrt[3]{4x^3-24}=x$

2. $\sqrt{5x-3}=\sqrt{3x+19}$

3. $\sqrt{6x-5}-x=-10$

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

$\sqrt[3]{4x^3-24}^3&=x^3\\4x^3-24&=x^3\\-24&=-3x^3\\8&=x^3\\2&=x$

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

2. Square both sides to solve for $x$ .

$\sqrt{5x-3}^2&=\sqrt{3x+19}^2\\5x-3&=3x+19\\2x&=22\\x&=11$

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

3. Add $x$ to both sides and square to eliminate the radical.

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

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

5 is an extraneous solution.

### Explore More

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

1. $\sqrt{x-3}=x-5$
2. $\sqrt{x+3}+15=x-12$
3. $\sqrt[4]{3x^2+54}=x$
4. $\sqrt{x^2+60}=4\sqrt{x}$
5. $\sqrt{x^4+5x^3}=2\sqrt{2x+10}$
6. $x=\sqrt{5x-6}$
7. $\sqrt{3x+4}=x-2$
8. $\sqrt{x^3+8x}-\sqrt{9x^2-60}=0$
9. $x=\sqrt[3]{4x+4-x^2}$
10. $\sqrt[4]{x^3+3}=2\sqrt[4]{x+3}$
11. $x^2-\sqrt{42x^2+343}=0$
12. $x\sqrt{x^2-21}=2\sqrt{x^3-25x+25}$

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

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

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

$\sqrt{16-15}&=\sqrt{16}-3 \\\sqrt{1}&=4-3 \\1&=1$

1. $\sqrt{x+11}-2=\sqrt{x-21}$
2. $\sqrt{x-6}=\sqrt{7x}-22$
3. $2+\sqrt{x+5}=\sqrt{4x-7}$