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Algebra II : Solving Radical Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #81 : Solving And Graphing Radicals

Solve the equation: $\sqrt[3]{-6x-9} = 4$

Possible Answers:

$-\frac{73}{6}$

$-\frac{7}{3}$

$-\frac{73}{3}$

$-\frac{13}{6}$

$-\frac{65}{6}$

Correct answer:

$-\frac{73}{6}$

Explanation:

Raise both sides by the power of three in order to eliminate the radical.

$(\sqrt[3]{-6x-9} )^3= 4^3$

-6x-9 =64

Add nine on both sides.

-6x-9 +9=64+9

-6x =73

Divide by negative six on both sides.

$\frac{-6x}{-6} =\frac{73}{-6}$

The answer is: $-\frac{73}{6}$

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Example Question #82 : Solving Radical Equations

Solve: $\sqrt[3]{3x-6} = \frac{1}{2}$

Possible Answers:

$-\frac{49}{24}$

$-\frac{35}{18}$

$-\frac{47}{24}$

$\frac{49}{24}$

$\frac{37}{18}$

Correct answer:

$\frac{49}{24}$

Explanation:

Cube both sides to eliminate the radical.

$(\sqrt[3]{3x-6} )^3= (\frac{1}{2})^3$

The equation becomes: $3x-6 = \frac{1}{8}$

Multiply by eight on both sides to eliminate the fraction.

$8(3x-6)= \frac{1}{8}\cdot 8$

24x-48=1

Add 48 on both sides.

24x-48+48=1+48

24x=49

Divide by 24 on both sides.

$\frac{24x}{24}=\frac{49}{24}$

The answer is: $\frac{49}{24}$

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Example Question #83 : Solving And Graphing Radicals

Solve: $\sqrt{9x}+2 = 16$

Possible Answers:

$\frac{196}{9}$

$\frac{62}{3}$

$\frac{1}{7}$

$\frac{9}{49}$

Correct answer:

$\frac{196}{9}$

Explanation:

Subtract two from both sides.

$\sqrt{9x}+2 -2= 16-2$

$\sqrt{9x} = 14$

Square both sides in order to eliminate the radical.

$(\sqrt{9x}) ^2= 14^2$

9x=196

Divide by nine on both sides.

$\frac{9x}{9}=\frac{196}{9}$

The right side is irreducible.

The answer is: $\frac{196}{9}$

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Example Question #81 : Solving Radical Equations

Solve the radical: $\sqrt{-5x+3} = 3$

Possible Answers:

$-\frac{5}{6}$

$-\frac{45}{4}$

$-\frac{1}{5}$

$-\frac{6}{5}$

Correct answer:

$-\frac{6}{5}$

Explanation:

Square both sides in order to eliminate the radical.

$(\sqrt{-5x+3} )^2= 3^2$

The equation becomes:

-5x+3=9

Subtract three from both sides.

-5x+3-3=9-3

-5x=6

Divide by negative five on both sides.

$\frac{-5x}{-5}=\frac{6}{-5}$

The answer is: $-\frac{6}{5}$

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Example Question #85 : Solving And Graphing Radicals

Solve: $\sqrt[5]{-3x}-3 = -1$

Possible Answers:

$-\frac{32}{3}$

$-\frac{20}{3}$

$\textup{No solution.}$

$-\frac{2}{3}$

$-\frac{16}{3}$

Correct answer:

$-\frac{32}{3}$

Explanation:

Add three on both sides.

$\sqrt[5]{-3x}-3+3 = -1+3$

$\sqrt[5]{-3x} = 2$

Raise both sides by the power of five to eliminate the radical.

$(\sqrt[5]{-3x} )^5= 2^5$

-3x =32

Divide by negative three on both sides.

$\frac{-3x }{-3}=\frac{32}{-3}$

The answer is: $-\frac{32}{3}$

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Example Question #86 : Solving And Graphing Radicals

Solve the equation: $\sqrt5 \cdot \sqrt{2x} -10= -9$

Possible Answers:

$\textup{No solution.}$

$\frac{1}{20}$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

The radicals can be combined as one radical by multiplying the two inner terms together.

The equation becomes:

$\sqrt{10x} -10= -9$

Add 10 on both sides.

$\sqrt{10x} -10+10= -9+10$

Simplify both sides.

$\sqrt{10x} =1$

Square both sides.

10x=1

Divide by ten on both sides.

$\frac{10x}{10}=\frac{1}{10}$

The answer is: $\frac{1}{10}$

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Example Question #401 : Radicals

Solve the radical: $\sqrt{\sqrt{2x}} = \sqrt{8}$

Possible Answers:

128

256

Correct answer:

Explanation:

Rewrite the radicals so that they are in exponential form.

$\sqrt{x} = x^\frac{1}{2}$

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

The equation becomes:

$(2x)^{\frac{1}{4}} = 8^{\frac{1}{2}}$

Raise both sides by the power of four to eliminate the fractional exponents.

$[(2x)^{\frac{1}{4}}] ^4= [8^{\frac{1}{2}}]^4$

2x=64

Divide by two on both sides.

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

The answer is:

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Example Question #82 : Solving Radical Equations

Solve: $4=2^{10x-8}$

Possible Answers:

$-\frac{1}{2}$

$\frac{3}{4}$

Correct answer:

Explanation:

Evaluate by changing the base of the left term.

$2^2=2^{10x-8}$

Now that both bases are common, set the powers equal and solve for x.

2=10x-8

Add 8 on both sides.

2+8=10x-8+8

10=10x

Divide by ten on both sides.

$\frac{10}{10}=\frac{10x}{10}$

The answer is:

Report an Error

Example Question #401 : Radicals

Evaluate: $\sqrt[3]{2x-8} = 2$

Possible Answers:

$\textup{No solution.}$

-8,8

Correct answer:

Explanation:

Cube both sides to eliminate the radical.

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

2x-8 =8

Add eight on both sides.

2x-8+(8) =8+(8)

2x=16

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{16}{2}$

The answer is:

Report an Error

Example Question #81 : Solving Radical Equations

Solve: $2\sqrt[3]{2-5x} = 10$

Possible Answers:

$\frac{13}{5}$

$-\frac{123}{5}$

$\frac{123}{5}$

$-\frac{13}{5}$

$\frac{127}{5}$

Correct answer:

$-\frac{123}{5}$

Explanation:

Solve by first dividing both sides by two.

$\frac{2\sqrt[3]{2-5x} }{2}= \frac{10}{2}$

Simplify both sides.

$\sqrt[3]{2-5x} =5$

Cube both sides.

$(\sqrt[3]{2-5x})^3 =5^3$

2-5x =125

Subtract two from both sides.

2-5x -(2)=125-(2)

-5x=123

Divide by negative five on both sides.

$\frac{-5x}{-5}=\frac{123}{-5}$

The answer is: $-\frac{123}{5}$

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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 #81 : Solving And Graphing Radicals

Example Question #82 : Solving Radical Equations

Example Question #83 : Solving And Graphing Radicals

Example Question #81 : Solving Radical Equations

Example Question #85 : Solving And Graphing Radicals

Example Question #86 : Solving And Graphing Radicals

Example Question #401 : Radicals

Example Question #82 : Solving Radical Equations

Example Question #401 : Radicals

Example Question #81 : Solving Radical Equations

All Algebra II Resources

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