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Algebra II : Solving Radical Equations

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Example Question #91 : Solving Radical Equations

Solve: $\sqrt{3x-3}-3 = 3$

Possible Answers:

Correct answer:

Explanation:

Add three on both sides.

$\sqrt{3x-3}-3+3 = 3+3$

The equation becomes:

$\sqrt{3x-3} =6$

Square both sides to eliminate the radical.

$(\sqrt{3x-3} )^2=6^2$

3x-3 = 36

Add three on both sides.

3x-3 +3= 36+3

3x=39

Divide by three on both sides.

$\frac{3x}{3}=\frac{39}{3}$

The answer is:

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Example Question #92 : Solving Radical Equations

Solve the equation: $4\sqrt[3]{-2x-3} = 8$

Possible Answers:

$-\frac{3}{2}$

$\textup{There is no solution.}$

$-\frac{11}{2}$

$-\frac{67}{2}$

$-\frac{5}{2}$

Correct answer:

$-\frac{11}{2}$

Explanation:

Divide by four on both sides.

$\frac{4\sqrt[3]{-2x-3} }{4}= \frac{8}{4}$

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

Cube both sides to eliminate the radical.

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

-2x-3=8

Add three on both sides.

-2x-3+3=8+3

-2x=11

Divide by negative two on both sides.

$\frac{-2x}{-2}=\frac{11}{-2}$

The answer is: $-\frac{11}{2}$

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Example Question #93 : Solving Radical Equations

Solve: $-\sqrt{x+8}-16 = -20$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Add 16 on both sides.

$-\sqrt{x+8}-16+16= -20+16$

$-\sqrt{x+8} = -4$

Divide both sides by negative one.

$\frac{-\sqrt{x+8} }{-1}=\frac{ -4}{-1}$

$\sqrt{x+8}=4$

Square both sides.

$(\sqrt{x+8}) ^2=4^2$

x+8=16

Subtract eight on both sides.

x+8-8=16-8

x=8

The answer is:

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Example Question #94 : Solving Radical Equations

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

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Square both sides in order to eliminate the radical.

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

-2x-5 = 81

Add 5 on both sides.

-2x-5 +5= 81+5

-2x=86

Divide by negative two on both sides.

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

Reduce both fractions.

The answer is:

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Example Question #95 : Solving Radical Equations

Solve the equation: $\sqrt{3x}-6= 11$

Possible Answers:

$\frac{23}{2}$

$\frac{34}{3}$

$\frac{289}{3}$

Correct answer:

$\frac{289}{3}$

Explanation:

Add six on both sides.

$\sqrt{3x}-6+6= 11+6$

$\sqrt{3x} = 17$

Square both sides to eliminate the radical.

$(\sqrt{3x}) ^2= 17^2$

3x=289

Divide both sides by three.

$\frac{3x}{3}=\frac{289}{3}$

The answer is: $\frac{289}{3}$

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Example Question #96 : Solving Radical Equations

Solve the radical: $\sqrt{2-3x} = 40$

Possible Answers:

$-\frac{1598}{3}$

534

$-\frac{82}{3}$

$\textup{No solution.}$

Correct answer:

$-\frac{1598}{3}$

Explanation:

Square both sides of the equation to eliminate the radical.

$(\sqrt{2-3x})^2 = 40^2$

The equation becomes:

2-3x=1600

Subtract two from both sides.

2-3x-2=1600-2

-3x= 1598

Divide by negative three on both sides.

$\frac{-3x}{-3}=\frac{ 1598}{-3}$

The answer is: $-\frac{1598}{3}$

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Example Question #97 : Solving Radical Equations

Evaluate: $2\sqrt[3]{-2x} = \frac{1}{2}$

Possible Answers:

$-\frac{1}{4}$

$\textup{No solution.}$

$-\frac{1}{64}$

$-\frac{1}{2}$

$-\frac{1}{128}$

Correct answer:

$-\frac{1}{128}$

Explanation:

Multiply by half on both sides.

$2\sqrt[3]{-2x} \cdot \frac{1}{2}= \frac{1}{2}\cdot \frac{1}{2}$

$\sqrt[3]{-2x} =\frac{1}{4}$

Cube both sides to eliminate the radical.

$(\sqrt[3]{-2x} )^3=(\frac{1}{4})^3$

$-2x = \frac{1}{64}$

Divide both sides by negative two. This is the same as multiplying by negative half on the right side.

$\frac{-2x}{-2} = \frac{1}{64}\cdot -\frac{1}{2}$

Reduce both fractions.

The answer is: $-\frac{1}{128}$

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Example Question #98 : Solving Radical Equations

Solve the radical equation: $\frac{x}{\sqrt{x-9}} = 6$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Multiply both sides by the denominator to eliminate the fraction on the left side.

$\frac{x}{\sqrt{x-9}}\cdot \sqrt{x-9}= 6\cdot \sqrt{x-9}$

The equation becomes:

$x= 6\sqrt{x-9}$

Square both sides to eliminate the radical.

$x^2=( 6\sqrt{x-9})^2$

x^2 = 36(x-9)

Distribute the 36 into the binomial.

x^2 = 36x-324

Subtract 36x and add 324 on both sides of the equation.

x^2 -[36x]+(324)= 36x-324-[36x]+(324)

The equation becomes a parabola, which can be factored.

x^2-36x+324 = 0

(x-18)^2 = 0

The value of x=18 . Check this value to see if substitution will satisfy the domain in the original equation.

$\frac{18}{\sqrt{18-9}}$ is valid since there is no negative in the radical, and there is not a zero denominator.

The answer is:

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Example Question #99 : Solving Radical Equations

Solve the equation: $\sqrt{-9x} +5=9$

Possible Answers:

$-\frac{196}{9}$

$\textup{No solution.}$

$\frac{4}{3}$

$-\frac{16}{9}$

Correct answer:

$-\frac{16}{9}$

Explanation:

Subtract both sides by five.

$\sqrt{-9x} +5-5=9-5$

$\sqrt{-9x} =4$

Square both sides.

$(\sqrt{-9x}) ^2=4^2$

-9x = 16

Divide by negative nine on both sides.

$\frac{-9x}{-9} = \frac{16}{-9}$

The answer is: $-\frac{16}{9}$

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Example Question #100 : Solving Radical Equations

Solve the radical equation: $2-\sqrt[3]{4x} = 1$

Possible Answers:

$\frac{3}{4}$

$\frac{1}{8}$

$\frac{1}{2}$

$\textup{No solution.}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

Add the radical on both sides.

$2-\sqrt[3]{4x} +\sqrt[3]{4x}= 1+\sqrt[3]{4x}$

The equation becomes:

$2= 1+\sqrt[3]{4x}$

Subtract 1 on both sides.

$2-1= 1+\sqrt[3]{4x}-1$

$1=\sqrt[3]{4x}$

Cube both sides to eliminate the radical.

$1^3=(\sqrt[3]{4x})^3$

1=4x

Divide by four on both sides.

$\frac{1}{4}=\frac{4x}{4}$

The answer is: $\frac{1}{4}$

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Algebra II : Solving Radical Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #91 : Solving Radical Equations

Example Question #92 : Solving Radical Equations

Example Question #93 : Solving Radical Equations

Example Question #94 : Solving Radical Equations

Example Question #95 : Solving Radical Equations

Example Question #96 : Solving Radical Equations

Example Question #97 : Solving Radical Equations

Example Question #98 : Solving Radical Equations

Example Question #99 : Solving Radical Equations

Example Question #100 : Solving Radical Equations

All Algebra II Resources

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