Solving Radical Equations

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Algebra II › Solving Radical Equations

Questions 1 - 10

1

Solve the equation: $\frac{\sqrt{-3-7x} }{-3}= -9$

$-\frac{732}{7}$

$\textup{No solution.}$

$\frac{732}{7}$

$-\frac{726}{7}$

$\frac{12}{7}$

Explanation

Multiply by negative three on both sides.

$\frac{\sqrt{-3-7x} }{-3} \cdot -3= -9 \cdot -3$

$\sqrt{-3-7x} = 27$

Square both sides.

$(\sqrt{-3-7x}) ^2= (27)^2$

Add three on both sides.

Divide by negative seven on both sides.

$\frac{-7x}{-7}=\frac{732}{-7}$

The answer is: $-\frac{732}{7}$

2

Solve for .

$\sqrt{x}=4$

$\pm2$

Explanation

$\sqrt{x}=4$ To get rid of the radical, we square both sides.

3

Evaluate: $\sqrt[3]{3-9x} = 4$

$-\frac{61}{9}$

$\textup{No solution.}$

$-\frac{67}{9}$

$-\frac{67}{3}$

$-\frac{13}{9}$

Explanation

Raise both sides by the power of three.

$(\sqrt[3]{3-9x} )^3= 4^3$

Subtract three from both sides.

Divide both sides by negative nine.

$\frac{-9x}{-9}=\frac{61}{-9}$

The answer is: $-\frac{61}{9}$

4

Solve the equation: $\sqrt{8x-3} = 30$

$\frac{903}{8}$

$\frac{897}{8}$

$\frac{903}{4}$

$\frac{57}{8}$

$\frac{63}{8}$

Explanation

$(\sqrt{8x-3}) ^2= 30^2$

Add three on both sides.

Divide by 8 on both sides.

$\frac{8x}{8}=\frac{903}{8}$

The answer is: $\frac{903}{8}$

5

Solve the equation: $\sqrt{9x-2}-7 =1$

$\frac{22}{3}$

$\frac{11}{3}$

$\frac{33}{2}$

Explanation

Add 7 on both sides.

$\sqrt{9x-2}-7+7 =1+7$

$\sqrt{9x-2} =8$

Square both sides.

$(\sqrt{9x-2}) ^2=8^2$

Simplify both sides of the equation.

Add 2 on both sides.

Divide by nine on both sides.

$\frac{9x}{9}=\frac{66}{9}$

Reduce both fractions.

$x=\frac{66}{9}=\frac{3\times 11\times 2}{3\times 3} =\frac{22}{3}$

The answer is: $\frac{22}{3}$

6

Solve the equation: $\sqrt[3]{3x}+6=8$

$\frac{8}{3}$

$\frac{3}{2}$

Explanation

Subtract six from both sides.

$\sqrt[3]{3x}+6-6=8-6$

Simplify both sides.

$\sqrt[3]{3x} = 2$

Cube both sides to eliminate the cube root.

Divide by three on both sides.

$\frac{3x}{3}=\frac{8}{3}$

The answer is: $\frac{8}{3}$

7

Solve the equation: $\sqrt[4]{-3x}+8 = 11$

$\textup{No solution.}$

Explanation

Subtract eight from both sides.

$\sqrt[4]{-3x}+8-8 = 11-8$

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

Raise both sides by the power of four.

$(\sqrt[4]{-3x} )^4=3^4$

Divide both sides by three.

$\frac{-3x }{-3}=\frac{81}{-3}$

The answer is:

8

Solve the equation: $\sqrt[3]{3x} = \frac{2}{3}$

$\frac{8}{81}$

$\frac{8}{9}$

$\frac{2}{27}$

$\frac{1}{3}$

$\frac{16}{81}$

Explanation

Cube both sides of the equation.

$(\sqrt[3]{3x}) ^3= (\frac{2}{3})^3$

This will eliminate the radical on the left side.

$3x= (\frac{2}{3})(\frac{2}{3})(\frac{2}{3})$

$3x= \frac{8}{27}$

Divide by three on both sides. This is similar to multiplying one-third on both sides.

$3x \times \frac{1}{3}= \frac{8}{27} \times \frac{1}{3}=\frac{8}{81}$

The answer is: $\frac{8}{81}$

9

Solve: $\sqrt{-2x-5} = 9$

$\textup{No solution.}$

Explanation

Square both sides in order to eliminate the radical.

$(\sqrt{-2x-5}) ^2= 9^2$

Add 5 on both sides.

Divide by negative two on both sides.

$\frac{-2x}{-2}=\frac{86}{-2}$

Reduce both fractions.

The answer is:

10

Solve for .

$\sqrt{x}=-5$

No answer

$\pm25$

$\pm5$

Explanation

To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.

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