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Algebra II : Mathematical Relationships and Basic Graphs

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 #4291 : Algebra Ii

Solve the equation: $\sqrt[3]{8x-2} = 10$

Possible Answers:

128

125

$\frac{501}{4}$

$\frac{251}{4}$

Correct answer:

$\frac{501}{4}$

Explanation:

In order to eliminate the radical on both sides, we will need to cube both sides.

$(\sqrt[3]{8x-2})^3 = 10^3$

Simplify both sides.

8x-2 = 1000

Add two on both sides.

8x-2+2 = 1000+2

8x=1002

Divide by eight on both sides.

$\frac{8x}{8}=\frac{1002}{8}$

Reduce the fractions on both sides.

The answer is: $\frac{501}{4}$

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Example Question #78 : Solving And Graphing Radicals

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

Possible Answers:

$\textup{No solution.}$

$-\frac{13}{5}$

$-\frac{26}{5}$

Correct answer:

$-\frac{13}{5}$

Explanation:

Square both sides to eliminate the radical.

$(\sqrt{-5x+3} )^2= 4^2$

-5x+3 = 16

Subtract three on both sides.

-5x+3 -3= 16-3

The equation becomes:

-5x= 13

Divide by negative five on both sides.

$\frac{-5x}{-5}= \frac{13}{-5}$

The answer is: $-\frac{13}{5}$

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Example Question #79 : Solving And Graphing Radicals

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

Possible Answers:

$\textup{No solution.}$

Correct answer:

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.

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

-3x =81

Divide both sides by three.

$\frac{-3x }{-3}=\frac{81}{-3}$

The answer is:

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Example Question #80 : Solving And Graphing Radicals

Solve: $\sqrt{6x-9} -3 =5$

Possible Answers:

$\frac{73}{6}$

$\frac{53}{3}$

$\frac{13}{3}$

$\frac{13}{6}$

$\frac{53}{6}$

Correct answer:

$\frac{73}{6}$

Explanation:

Add three on both sides.

$\sqrt{6x-9} -3 +3=5+3$

$\sqrt{6x-9} = 8$

Square both sides.

$(\sqrt{6x-9})^2 = 8^2$

6x-9 = 64

Add nine on both sides.

6x-9+9 = 64+9

Simplify both sides.

6x=73

Divide by 6 on both sides.

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

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

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Example Question #81 : Solving Radical Equations

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

Possible Answers:

$-\frac{7}{3}$

$-\frac{73}{3}$

$-\frac{13}{6}$

$-\frac{65}{6}$

$-\frac{73}{6}$

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: $-\frac{6}{5}$

Report an Error

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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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4291 : Algebra Ii

Example Question #78 : Solving And Graphing Radicals

Example Question #79 : Solving And Graphing Radicals

Example Question #80 : Solving And Graphing Radicals

Example Question #81 : Solving Radical Equations

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

All Algebra II Resources

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