Solving Radical Equations

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Algebra 2 › Solving Radical Equations

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1

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

$\frac{8}{81}$

CORRECT

$\frac{8}{9}$

0

$\frac{2}{27}$

0

$\frac{1}{3}$

0

$\frac{16}{81}$

0

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: $\frac{8}{81}$

2

Solve for .

$\sqrt{x}=4$

CORRECT

0

0

$\pm2$

0

0

Explanation

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

3

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

$-\frac{732}{7}$

CORRECT

$\textup{No solution.}$

0

$\frac{732}{7}$

0

$-\frac{726}{7}$

0

$\frac{12}{7}$

0

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: $-\frac{732}{7}$

4

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

CORRECT

0

0

0

$\textup{No solution.}$

0

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:

5

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

$-\frac{61}{9}$

CORRECT

$\textup{No solution.}$

0

$-\frac{67}{9}$

0

$-\frac{67}{3}$

0

$-\frac{13}{9}$

0

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: $-\frac{61}{9}$

6

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

$\frac{903}{8}$

CORRECT

$\frac{897}{8}$

0

$\frac{903}{4}$

0

$\frac{57}{8}$

0

$\frac{63}{8}$

0

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: $\frac{903}{8}$

7

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

$\frac{22}{3}$

CORRECT

0

$\frac{11}{3}$

0

$\frac{33}{2}$

0

0

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: $\frac{22}{3}$

8

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

$\frac{8}{3}$

CORRECT

0

0

$\frac{3}{2}$

0

0

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: $\frac{8}{3}$

9

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

CORRECT

0

0

0

$\textup{No solution.}$

0

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:

10

Solve and simplify:

$2\sqrt{x^3}=8$

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

CORRECT

$x=\sqrt{2}$

0

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

0

$x=2\sqrt[3]{4}$

0

$x=\sqrt[3]{16}$

0

Explanation

To solve for x, first we must isolate the radical on one side:

$\sqrt{x^3}=4$

Next, square both sides to eliminate the radical:

Now, take the cube root of each side to find x:

$x=\sqrt[3]{16}$

Finally, factor the term inside the cube root and see if any cubes can be pulled out of the radical:

$x=\sqrt[3]{16}=\sqrt[3]{8\cdot 2}=2\sqrt[3]{2}$

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