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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #411 : Radicals

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

Possible Answers:

$-\frac{1}{128}$

$\textup{No solution.}$

$-\frac{1}{2}$

$-\frac{1}{64}$

$-\frac{1}{4}$

Correct answer:

$-\frac{1}{128}$

Explanation:

Multiply by half on both sides.

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

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

Cube both sides to eliminate the radical.

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

$-2x = \frac{1}{64}$

Divide both sides by negative two. This is the same as multiplying by negative half on the right side.

$\frac{-2x}{-2} = \frac{1}{64}\cdot -\frac{1}{2}$

Reduce both fractions.

The answer is: $-\frac{1}{128}$

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Example Question #412 : Radicals

Solve the radical equation: $\frac{x}{\sqrt{x-9}} = 6$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Multiply both sides by the denominator to eliminate the fraction on the left side.

$\frac{x}{\sqrt{x-9}}\cdot \sqrt{x-9}= 6\cdot \sqrt{x-9}$

The equation becomes:

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

Square both sides to eliminate the radical.

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

x^2 = 36(x-9)

Distribute the 36 into the binomial.

x^2 = 36x-324

Subtract 36x and add 324 on both sides of the equation.

x^2 -[36x]+(324)= 36x-324-[36x]+(324)

The equation becomes a parabola, which can be factored.

x^2-36x+324 = 0

(x-18)^2 = 0

The value of x=18 . Check this value to see if substitution will satisfy the domain in the original equation.

$\frac{18}{\sqrt{18-9}}$ is valid since there is no negative in the radical, and there is not a zero denominator.

The answer is:

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Example Question #99 : Solving Radical Equations

Solve the equation: $\sqrt{-9x} +5=9$

Possible Answers:

$-\frac{196}{9}$

$\textup{No solution.}$

$\frac{4}{3}$

$-\frac{16}{9}$

Correct answer:

$-\frac{16}{9}$

Explanation:

Subtract both sides by five.

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

$\sqrt{-9x} =4$

Square both sides.

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

-9x = 16

Divide by negative nine on both sides.

$\frac{-9x}{-9} = \frac{16}{-9}$

The answer is: $-\frac{16}{9}$

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Example Question #1652 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$\frac{1}{8}$

$\frac{3}{4}$

$\frac{1}{4}$

$\textup{No solution.}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{4}$

Explanation:

Add the radical on both sides.

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

The equation becomes:

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

Subtract 1 on both sides.

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

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

Cube both sides to eliminate the radical.

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

1=4x

Divide by four on both sides.

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

The answer is: $\frac{1}{4}$

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Example Question #101 : Solving Radical Equations

Solve the radical equation: $\frac{3}{\sqrt[4]{4x}}=2$

Possible Answers:

$\textup{The answer is not given.}$

$\frac{9}{8}$

$\frac{9}{64}$

$\frac{81}{8}$

$\frac{81}{64}$

Correct answer:

$\frac{81}{64}$

Explanation:

Multiply the radical denominator on both sides.

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

The equation becomes:

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

Divide by two on both sides.

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

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

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

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

$\frac{81}{16}= 4x$

Divide by four on both sides. This is the same as multiplying one-fourth on both sides.

$\frac{81}{16}\cdot \frac{1}{4}= 4x\cdot \frac{1}{4}$

The answer is: $\frac{81}{64}$

Report an Error

Example Question #101 : Solving Radical Equations

Solve: $\frac{9\sqrt2}{\sqrt{2x}} = \sqrt2$

Possible Answers:

$\frac{9}{2}$

$\frac{3}{4}$

$\frac{81}{2}$

$\frac{2}{3}$

$\frac{81}{4}$

Correct answer:

$\frac{81}{2}$

Explanation:

Multiply the denominator on both sides to eliminate the denominator.

$\frac{9\sqrt2}{\sqrt{2x}} \cdot\sqrt{2x}= \sqrt2 \cdot\sqrt{2x}$

$9\sqrt2 = \sqrt{4x}$

Square both sides.

$(9\sqrt2)^2 =( \sqrt{4x})^2$

$81\cdot 2 = 4x$

Divide by four on both sides.

$\frac{81\cdot 2}{4} =\frac{ 4x}{4}$

The answer is: $\frac{81}{2}$

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Example Question #4314 : Algebra Ii

Solve the equation: $\sqrt{5x-9} = 5$

Possible Answers:

$\frac{19}{25}$

$\frac{14}{5}$

$\frac{196}{5}$

$\frac{9}{5}$

$\frac{34}{5}$

Correct answer:

$\frac{34}{5}$

Explanation:

Square both sides of the radical to eliminate the square root.

$(\sqrt{5x-9} )^2= 5^2$

The equation becomes:

5x-9 = 25

Add 9 on both sides.

5x-9 +9= 25+9

5x=34

Divide by 5 on both sides.

$\frac{5x}{5}=\frac{34}{5}$

The answer is: $\frac{34}{5}$

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Example Question #4315 : Algebra Ii

Evaluate: $\frac{5\sqrt x}{\sqrt{9x}} = x$

Possible Answers:

$-\frac{5}{3}$

$\frac{5}{3}$

$\frac{\sqrt{5}}{3}$

$\frac{3\sqrt5}{5}$

$\frac{1}{3}$

Correct answer:

$\frac{5}{3}$

Explanation:

Divide terms of the radicals on the left side.

$\frac{5\sqrt x}{\sqrt{9x}} = 5\sqrt{\frac{x}{9x}} = 5\sqrt{\frac{1}{9}}$

$x=5\sqrt{\frac{1}{9}} = 5\cdot \frac{1}{3} = \frac{5}{3}$

The answer is: $\frac{5}{3}$

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Example Question #105 : Solving Radical Equations

Solve the equation: $\frac{\sqrt{-3-7x} }{-3}= -9$

Possible Answers:

$\textup{No solution.}$

$-\frac{726}{7}$

$-\frac{732}{7}$

$\frac{732}{7}$

$\frac{12}{7}$

Correct answer:

$-\frac{732}{7}$

Explanation:

Multiply by negative three on both sides.

$\frac{\sqrt{-3-7x} }{-3} \cdot -3= -9 \cdot -3$

$\sqrt{-3-7x} = 27$

Square both sides.

$(\sqrt{-3-7x}) ^2= (27)^2$

-3-7x = 729

Add three on both sides.

-3-7x +3= 729+3

-7x=732

Divide by negative seven on both sides.

$\frac{-7x}{-7}=\frac{732}{-7}$

The answer is: $-\frac{732}{7}$

Report an Error

Example Question #106 : Solving Radical Equations

Solve the equation: $\frac{5}{\sqrt{5x}} = 10$

Possible Answers:

$\frac{1}{200}$

$\frac{\sqrt{5}}{10}$

$\frac{1}{20}$

$\frac{5}{4}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{20}$

Explanation:

Multiply the denominator on both sides to eliminate the fraction.

$\frac{5}{\sqrt{5x}} \cdot \sqrt{5x}= 10 \cdot \sqrt{5x}$

The equation becomes:

$5=10\sqrt{5x}$

Square both sides.

$5^2=(10\sqrt{5x})^2$

25= 100(5x)

25 =500x

Divide by 500 on both sides.

$\frac{25}{500} =\frac{500x}{500}$

Reduce both fractions.

The answer is: $\frac{1}{20}$

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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 #411 : Radicals

Example Question #412 : Radicals

Example Question #99 : Solving Radical Equations

Example Question #1652 : Mathematical Relationships And Basic Graphs

Example Question #101 : Solving Radical Equations

Example Question #101 : Solving Radical Equations

Example Question #4314 : Algebra Ii

Example Question #4315 : Algebra Ii

Example Question #105 : Solving Radical Equations

Example Question #106 : Solving Radical Equations

All Algebra II Resources

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