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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 #61 : Solving And Graphing Radicals

Solve: $3\sqrt[4]{9-2x} = 1$

Possible Answers:

$\frac{364}{81}$

$-\frac{364}{81}$

$\frac{99}{2}$

$\frac{1459}{162}$

Correct answer:

$\frac{364}{81}$

Explanation:

Divide by three on both sides.

$\frac{3\sqrt[4]{9-2x} }{3}= \frac{1}{3}$

Rewrite the equation.

$\sqrt[4]{9-2x} = \frac{1}{3}$

Raise both sides to the fourth power.

$(\sqrt[4]{9-2x})^4 =( \frac{1}{3})^4$

Simplify both sides.

$9-2x = \frac{1}{81}$

Multiply by 81 on both sides to eliminate the fraction.

$81(9-2x )= \frac{1}{81} \cdot 81$

729-162x=1

Subtract 729 from both sides.

729-162x-729=1-729

-162x = -728

Divide by negative 162 on both sides.

$\frac{-162x}{-162} =\frac{ -728}{-162}$

Reduce the fractions. Double negatives will negate, turning the answer positive.

$\frac{ -728}{-162} =\frac{364\times 2}{81\times 2}$

The answer is: $\frac{364}{81}$

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Example Question #382 : Radicals

Evaluate: $3\sqrt[4]{4-2x}-4=2$

Possible Answers:

$-\frac{1}2$

Correct answer:

Explanation:

Isolate the radical. Add four on both sides.

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

$3\sqrt[4]{4-2x}=6$

Divide by three on both sides. The equation becomes:

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

Raise both sides by four.

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

4-2x = 16

Subtract four on both sides.

4-2x -4= 16-4

-2x=12

Divide by negative two on both sides.

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

The answer is:

Report an Error

Example Question #383 : Radicals

Solve the equation: $\frac{\sqrt[3]{5x-2}}{6} = \frac{1}{2}$

Possible Answers:

346

$\frac{514}{5}$

$\frac{29}{5}$

$-\frac{11}{27}$

$\frac{11}{27}$

Correct answer:

$\frac{29}{5}$

Explanation:

Multiply by six on both sides of the equation.

$\frac{\sqrt[3]{5x-2}}{6} \cdot 6= \frac{1}{2} \cdot 6$

The equation becomes:

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

Cube both sides to eliminate the radical.

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

5x-2 = 27

Add two on both sides.

5x-2 +2= 27+2

5x=29

Divide by five on both sides.

$\frac{5x}{5}=\frac{29}{5}$

The answer is: $\frac{29}{5}$

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Example Question #384 : Radicals

Solve the radical equation: $\sqrt{-6x-5}-7 = 10$

Possible Answers:

$-\frac{1}{7}$

$-\frac{49}{11}$

$\textup{No solution.}$

$-\frac{7}{3}$

Correct answer:

Explanation:

Add seven on both sides.

$\sqrt{-6x-5}-7 +7= 10+7$

$\sqrt{-6x-5} = 17$

Square both sides. This will eliminate the radical.

$(\sqrt{-6x-5})^2= 17^2$

-6x-5 = 289

Add five on both sides.

-6x-5 +5= 289+5

-6x=294

Divide both sides by negative 6 and reduce the fraction.

$\frac{-6x}{-6}=\frac{294}{-6}$

The answer is:

Report an Error

Example Question #71 : Solving Radical Equations

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

Possible Answers:

$\frac{8}{81}$

$\frac{16}{81}$

$\frac{1}{3}$

$\frac{8}{9}$

$\frac{2}{27}$

Correct answer:

$\frac{8}{81}$

Explanation:

Cube both sides of the equation.

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

This will eliminate the radical on the left side.

$3x= (\frac{2}{3})(\frac{2}{3})(\frac{2}{3})$

$3x= \frac{8}{27}$

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

$3x \times \frac{1}{3}= \frac{8}{27} \times \frac{1}{3}=\frac{8}{81}$

The answer is: $\frac{8}{81}$

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Example Question #72 : Solving And Graphing Radicals

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

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Subtract both sides by three.

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

The equation becomes:

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

Divide both sides by negative one.

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

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

Cube both sides to eliminate the radical.

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

2-x = 64

Add on both sides.

2-x +x= 64+x

2= 64+x

Subtract 64 on both sides.

2-64= 64+x-64

The answer is:

Report an Error

Example Question #73 : Solving And Graphing Radicals

Solve the radical equation: $\sqrt[8]{2x-1} =2$

Possible Answers:

$\frac{255}{2}$

$\frac{63}{2}$

$\frac{17}{2}$

$\frac{65}{2}$

$\frac{257}{2}$

Correct answer:

$\frac{257}{2}$

Explanation:

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

$(\sqrt[8]{2x-1} )^8=2^8$

Simplify both sides.

2x-1=256

Add one on both sides.

2x-1+1=256+1

2x=257

Divide by two on both sides.

$\frac{2x}{2}=\frac{257}{2}$

The answer is: $\frac{257}{2}$

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Example Question #74 : Solving And Graphing Radicals

Solve: $\frac{1}{\sqrt{3x}} = 4$

Possible Answers:

$\frac{1}{24}$

$\frac{1}{72}$

$\frac{1}{48}$

$\frac{16}{3}$

Correct answer:

$\frac{1}{48}$

Explanation:

Multiply by $\sqrt{3x}$ on both sides.

$\frac{1}{\sqrt{3x}} \cdot \sqrt{3x}= 4 \cdot \sqrt{3x}$

$1=4\sqrt{3x}$

Divide by 4 on both sides.

$\frac{1}{4}=\frac{4\sqrt{3x}}{4}$

$\frac{1}{4} =\sqrt{3x}$

Square both sides to eliminate the radical.

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

$\frac{1}{16} = 3x$

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

$\frac{1}{16} \cdot \frac{1}{3}= 3x\cdot \frac{1}{3}$

The answer is: $\frac{1}{48}$

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Example Question #75 : Solving And Graphing Radicals

Solve: $\frac{2}{\sqrt{8x}} = \sqrt{x}$

Possible Answers:

$\frac{\sqrt2}{2}$

$\textup{No solution.}$

$\frac{3\sqrt2}{4}$

$\frac{\sqrt5}{8}$

$\frac{\sqrt2}{4}$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

Multiply both sides by the denominator on the left side.

$\frac{2}{\sqrt{8x}}\cdot \sqrt{8x} = \sqrt{x}\cdot \sqrt{8x}$

The equation becomes: $2 = \sqrt{8x^2}$

Square both sides to eliminate the radical.

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

4=8x^2

Divide both sides by eight.

$\frac{4}{8}=\frac{8x^2}{8}$

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

Notice that the negative radical solution will not satisfy the original equation.

Rationalize the positive term.

$\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt2}{2}$

The answer is: $\frac{\sqrt2}{2}$

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Example Question #76 : Solving And Graphing Radicals

Solve: $\sqrt[4]{32x+1} = 3$

Possible Answers:

$\frac{145}{32}$

$\frac{7}{2}$

$\frac{4}{3}$

$\frac{41}{16}$

$\frac{5}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

Raise both sides to the power of four.

$(\sqrt[4]{32x+1}) ^4= 3^4$

Simplify both sides.

32x+1 = 81

Subtract 1 from both sides.

32x+1 -1= 81-1

32x=80

Divide by 32 on both sides.

$\frac{32x}{32}=\frac{80}{32}$

Reduce this fraction.

$\frac{80}{32} = \frac{16\times 5}{16\times 2}$

The answer is: $\frac{5}{2}$

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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 #61 : Solving And Graphing Radicals

Example Question #382 : Radicals

Example Question #383 : Radicals

Example Question #384 : Radicals

Example Question #71 : Solving Radical Equations

Example Question #72 : Solving And Graphing Radicals

Example Question #73 : Solving And Graphing Radicals

Example Question #74 : Solving And Graphing Radicals

Example Question #75 : Solving And Graphing Radicals

Example Question #76 : Solving And Graphing Radicals

All Algebra II Resources

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