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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #4273 : Algebra Ii

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

Possible Answers:

$-\frac{61}{9}$

$-\frac{67}{9}$

$-\frac{13}{9}$

$\textup{No solution.}$

$-\frac{67}{3}$

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 #4275 : Algebra Ii

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

Possible Answers:

$\textup{No solution.}$

$\frac{32}{27}$

$\frac{31}{27}$

$-\frac{31}{27}$

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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Example Question #4277 : Algebra Ii

Solve: $\sqrt{8-x}+\sqrt{9-x} =2$

Possible Answers:

$\frac{53}{5}$

$\frac{13}{7}$

$\frac{119}{16}$

$\frac{99}{8}$

$\frac{131}{24}$

Correct answer:

$\frac{119}{16}$

Explanation:

Square both sides.

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

We will need to simplify the left side using the FOIL method.

$(\sqrt{8-x}+\sqrt{9-x} )(\sqrt{8-x}+\sqrt{9-x} )$

A radical multiplied by itself becomes what's inside the radical.

The expression becomes:

$(8-x)+(\sqrt{8-x})(\sqrt{9-x})+(\sqrt{9-x})(\sqrt{8-x})+(9-x)$

Combine like-terms and rewrite the equation.

$-2x+2\sqrt{8-x}\sqrt{9-x}+17 = 4$

Isolate the radical by adding and subtract 17 on both sides. Then divide by two on both sides.

$2\sqrt{8-x}\sqrt{9-x} =2x-13$

$\frac{2\sqrt{8-x}\sqrt{9-x} }{2}=\frac{2x-13}{2}$

$\sqrt{8-x}\sqrt{9-x} = x-\frac{13}{2}$

The radicals can be combined as one whole using the FOIL method.

$\sqrt{72-17x+x^2} = x-\frac{13}{2}$

Square both sides.

$(\sqrt{72-17x+x^2}) ^2= (x-\frac{13}{2})^2$

$72-17x+x^2 = x^2-13x+\frac{169}{4}$

The x^2 terms will cancel by subtracting the term on both side, and we can solve for .

$72-17x = -13x+\frac{169}{4}$

Add 17x on both sides.

$72 = 4x+\frac{169}{4}$

Multiply by four on both sides to eliminate the fraction.

$4[72 = 4x+\frac{169}{4}]$

288 = 16x+169

Solve for x.

288-169 = 16x+169-169

119 = 16x

The answer is: $\frac{119}{16}$

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Example Question #4278 : Algebra Ii

Solve: $4\sqrt{3x-3} = 12$

Possible Answers:

Correct answer:

Explanation:

Divide by four on both sides.

$\frac{4\sqrt{3x-3} }{4}= \frac{12}{4}$

The equation becomes:

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

Square both sides.

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

3x-3= 9

Add three on both sides.

3x-3+3= 9+3

3x =12

Divide by three on both sides.

$\frac{3x}{3} =\frac{12}{3}$

The answer is:

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Example Question #4279 : Algebra Ii

Solve: $\frac{\sqrt{3x}}{x} = 6$

Possible Answers:

$\frac{1}{12}$

$\frac{2}{5}$

$\frac{1}{3}$

$\frac{1}{24}$

$\frac{1}{42}$

Correct answer:

$\frac{1}{12}$

Explanation:

Multiply by on both sides.

$\frac{\sqrt{3x}}{x} \cdot x= 6\cdot x$

Simplify both sides.

$\sqrt{3x}=6x$

Square both sides to eliminate the radical.

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

Simplify both sides.

3x= 36x^2

Divide by 36x on both sides.

$\frac{3x}{36x}= \frac{36x^2}{36x}$

The answer is: $x=\frac{1}{12}$

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Example Question #4280 : Algebra Ii

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

Possible Answers:

$\frac{1}{81}$

$\frac{3}{28}$

$\frac{64}{9}$

$\frac{1}{18}$

$\frac{1}{9}$

Correct answer:

$\frac{1}{81}$

Explanation:

Divide by two on both sides. This is similar to multiplying by one-half.

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

Simplify both sides.

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

Cube both sides to eliminate the radical.

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

$3x=\frac{1}{27}$

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

$\frac{3x}{3}=\frac{1}{27} \cdot \frac{1}{3}$

The answer is: $\frac{1}{81}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4271 : Algebra Ii

Example Question #4272 : Algebra Ii

Example Question #4273 : Algebra Ii

Example Question #4274 : Algebra Ii

Example Question #4275 : Algebra Ii

Example Question #4276 : Algebra Ii

Example Question #4277 : Algebra Ii

Example Question #4278 : Algebra Ii

Example Question #4279 : Algebra Ii

Example Question #4280 : Algebra Ii

All Algebra II Resources

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