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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 #51 : Solving And Graphing Radicals

Solve the equation: $\sqrt[3]{7x-1}+5 = 8$

Possible Answers:

$\frac{26}{7}$

$\frac{2196}{7}$

$\frac{1}{4}$

Correct answer:

Explanation:

In order to solve this equation, first add five on both sides.

$\sqrt[3]{7x-1}+5 -5= 8-5$

The equation becomes: $\sqrt[3]{7x-1}=3$

Cube both sides.

$(\sqrt[3]{7x-1})^3=3^3$

Simplify both sides of the equation.

7x-1 = 27

Add one on both sides.

7x-1 +1= 27+1

7x=28

Divide both sides by seven.

$\frac{7x}{7}=\frac{28}{7}$

The answer is:

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Example Question #361 : Radicals

Solve the equation: $\sqrt[4]{2x+6}+6 = 8$

Possible Answers:

Correct answer:

Explanation:

Subtract six from both sides.

$\sqrt[4]{2x+6}+6 -6= 8-6$

Simplify the equation.

$\sqrt[4]{2x+6} =2$

Take the fourth power of both sides.

$(\sqrt[4]{2x+6}) ^4=2^4$

2x+6=16

Subtract six from both sides.

2x+6-6=16-6

2x=10

Divide by two on both sides.

$\frac{2x}{2}=\frac{10}{2}$

Reduce the fractions.

The answer is:

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Example Question #51 : Solving Radical Equations

Solve the equation: $\sqrt{2x}+2= \sqrt8$

Possible Answers:

$4-\sqrt2$

$8-4\sqrt2$

$4-3\sqrt2$

$3+8\sqrt2$

$6-4\sqrt2$

Correct answer:

$6-4\sqrt2$

Explanation:

Subtract two from both sides.

$\sqrt{2x}+2-2= \sqrt8-2$

$\sqrt{2x}= \sqrt8-2$

Square both sides. Use the FOIL method to simplify the right side.

$(\sqrt{2x})^2= (\sqrt8-2)^2$

$2x=(\sqrt8)(\sqrt8)+(\sqrt8)(-2)+(-2)(\sqrt8)+(-2)(-2)$

$2x= 8-4\sqrt8+4$

Combine like terms.

$2x= 12-4\sqrt8$

The term $\sqrt8$ can be rewritten as: $2\sqrt2$

$2x= 12-4\cdot 2\sqrt2$

$2x= 12-8\sqrt2$

Divide by two on both sides.

$\frac{2x}{2}= \frac{12-8\sqrt2}{2}$

The answer is: $6-4\sqrt2$

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Example Question #361 : Radicals

Solve the radical equation: $\sqrt{4x-8}-3 = 6$

Possible Answers:

$\frac{1}{4}$

$\frac{17}{4}$

$\frac{89}{4}$

$\frac{89}{2}$

$\frac{23}{8}$

Correct answer:

$\frac{89}{4}$

Explanation:

Add three on both sides.

$\sqrt{4x-8}-3 +3= 6+3$

$\sqrt{4x-8} = 9$

Square both sides of the equation.

$(\sqrt{4x-8})^2 = 9^2$

4x-8 = 81

Add 8 on both sides.

4x=89

Divide by four on both sides.

$\frac{4x}{4}=\frac{89}{4}$

The answer is: $\frac{89}{4}$

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Example Question #4271 : Algebra Ii

Solve the radical equation: $\sqrt[4]{2x+16}+5=8$

Possible Answers:

$\frac{65}{4}$

$\frac{128}{3}$

$\frac{256}{9}$

$\frac{65}{2}$

$\frac{256}{3}$

Correct answer:

$\frac{65}{2}$

Explanation:

Subtract five from both sides.

$\sqrt[4]{2x+16}+5-5=8-5$

The equation becomes:

$\sqrt[4]{2x+16}=3$

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

$(\sqrt[4]{2x+16})^4=3^4$

2x+16=81

Subtract 16 on both sides.

2x+16-16=81-16

2x=65

Divide by two on both sides.

$\frac{2x}{2}=\frac{65}{2}$

The answer is: $\frac{65}{2}$

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Example Question #4272 : Algebra Ii

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

Possible Answers:

$\frac{57}{8}$

$\frac{903}{8}$

$\frac{63}{8}$

$\frac{897}{8}$

$\frac{903}{4}$

Correct answer:

$\frac{903}{8}$

Explanation:

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

8x-3 = 900

Add three on both sides.

8x-3 +3= 900+3

8x=903

Divide by 8 on both sides.

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

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

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Example Question #371 : Radicals

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

Possible Answers:

$-\frac{67}{9}$

$\textup{No solution.}$

$-\frac{13}{9}$

$-\frac{67}{3}$

$-\frac{61}{9}$

Correct answer:

$-\frac{61}{9}$

Explanation:

Raise both sides by the power of three.

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

3-9x = 64

Subtract three from both sides.

3-9x-3 = 64-3

-9x=61

Divide both sides by negative nine.

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

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

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Example Question #4274 : Algebra Ii

Solve: $\sqrt[4]{3-x} +2= 5$

Possible Answers:

Correct answer:

Explanation:

Subtract two from both sides.

$\sqrt[4]{3-x} +2-2= 5-2$

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

Raise both sides by the fourth power. This will eliminate the radical.

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

3-x=81

Add on both sides.

3-x+x=81+x

3=81+x

Subtract 81 on both sides.

3-81=81+x-81

-78=x

The answer is:

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Example Question #371 : Radicals

Solve the radical equation: $\sqrt{6x+7} = 3^{-1}$

Possible Answers:

$\frac{31}{27}$

$-\frac{31}{27}$

$\frac{32}{27}$

$\textup{No solution.}$

Correct answer:

$-\frac{31}{27}$

Explanation:

Square both sides to eliminate the radical.

$\sqrt{6x+7} ^2=( 3^{-1})^2$

$6x+7 = 3^{-2}$

Convert the negative exponent into a fraction.

$x^{-a} =\frac{1}{x^a}$

$6x+7 =\frac{1}{9}$

Multiply by nine on both sides to eliminate the fraction.

$9(6x+7) =9(\frac{1}{9})$

54x+63 = 1

Subtract 63 on both sides.

54x+63 -63= 1-63

54x= -62

Divide by 54 on both sides.

$\frac{54x}{54}=\frac{ -62}{54}$

Reduce the fraction.

The answer is: $-\frac{31}{27}$

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Example Question #4276 : Algebra Ii

Solve the radical: $x\sqrt{5x} = 25$

Possible Answers:

125

$\frac{1}{125}$

$\frac{1}{25}$

625

Correct answer:

Explanation:

Square both sides to eliminate the radical.

$(x\sqrt{5x}) ^2= 25^2$

The equation becomes:

$x^2\cdot 5x = 625$

Divide by five on both sides and combine the product of x-variables.

$x^3=\frac{ 625}{5}$

x^3 = 125

Cube root both sides.

$\sqrt[3]{x^3} = \sqrt[3]{125}$

The answer is:

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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 #51 : Solving And Graphing Radicals

Example Question #361 : Radicals

Example Question #51 : Solving Radical Equations

Example Question #361 : Radicals

Example Question #4271 : Algebra Ii

Example Question #4272 : Algebra Ii

Example Question #371 : Radicals

Example Question #4274 : Algebra Ii

Example Question #371 : Radicals

Example Question #4276 : Algebra Ii

All Algebra II Resources

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