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Algebra II : Radicals and Fractions

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Example Questions

Example Question #41 : Radicals And Fractions

Simplify, if possible: $\frac{9}{\sqrt[3]9}$

Possible Answers:

$6\sqrt[3]{3}$

$4\sqrt[3]{3}$

$\frac{3\sqrt[3]{3}}{2}$

$\textup{The expression is simplified as is.}$

$3\sqrt[3]{3}$

Correct answer:

$3\sqrt[3]{3}$

Explanation:

In order to eliminate the radical in the denominator, we will need to multiply the denominator twice on both the numerator and denominator.

$\frac{9}{\sqrt[3]9}\cdot \frac{\sqrt[3]9}{\sqrt[3]9} \cdot \frac{\sqrt[3]9}{\sqrt[3]9} = \frac{9\sqrt[3]9\cdot \sqrt[3]9}{9}$

Simplify the fraction.

$\sqrt[3]9\cdot \sqrt[3]9 = \sqrt[3]{81}$

Factor the inner term using common factors of numbers the power of three.

$\sqrt[3]{81}=\sqrt[3]{27\times 3} = \sqrt[3]{27}\cdot\sqrt[3]{3} = 3\sqrt[3]{3}$

The answer is: $3\sqrt[3]{3}$

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Example Question #42 : Radicals And Fractions

Simplify: $\frac{8}{\sqrt{6}}$

Possible Answers:

$8\sqrt{6}$

$\frac{2\sqrt3}{3}$

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

$\frac{4\sqrt6}{3}$

$\frac{16\sqrt{6}}{3}$

Correct answer:

$\frac{4\sqrt6}{3}$

Explanation:

Rationalize the denominator.

Multiply the denominator with the numerator and denominator in order to eliminate the radical in the denominator.

$\frac{8}{\sqrt{6}} \times \frac{\sqrt6}{\sqrt6} = \frac{8\sqrt6}{6} = \frac{4\sqrt6}{3}$

The answer is: $\frac{4\sqrt6}{3}$

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Example Question #43 : Radicals And Fractions

Simplify: $\frac{2\sqrt2}{5\sqrt5}$

Possible Answers:

$\frac{2\sqrt{10}}{25}$

$\frac{4\sqrt{2}}{25}$

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

$\frac{2\sqrt{2}}{25}$

$\frac{2\sqrt{10}}{5}$

Correct answer:

$\frac{2\sqrt{10}}{25}$

Explanation:

In order to simplify, we will need to rationalize the denominator. Multiply square root five on both the top and bottom.

$\frac{2\sqrt2}{5\sqrt5} \cdot \frac{\sqrt5}{\sqrt5} = \frac{2\sqrt{10}}{5\cdot 5} = \frac{2\sqrt{10}}{25}$

The answer is: $\frac{2\sqrt{10}}{25}$

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Example Question #44 : Radicals And Fractions

Evaluate: $\frac{8\sqrt{5}}{\sqrt{10}}$

Possible Answers:

$3\sqrt5$

$4\sqrt2$

$2\sqrt2$

$4\sqrt6$

Correct answer:

$4\sqrt2$

Explanation:

In order to eliminate the radical denominator, we will need to multiply the top and bottom of the fraction by the denominator.

$\frac{8\sqrt{5}}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{8\sqrt{50}}{10}$

Reduce the fraction.

$\frac{8\sqrt{50}}{10} = \frac{4\sqrt{50}}{5}$

We can still simplify the square root using factors of perfect squares.

$\sqrt{50} =\sqrt{25\cdot 2}=\sqrt{25} \cdot \sqrt{2} = 5\sqrt2$

The expression becomes:

$\frac{4\sqrt{50}}{5} = \frac{4 \cdot 5\sqrt{2}}{5} = 4\sqrt2$

The answer is: $4\sqrt2$

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Example Question #45 : Radicals And Fractions

Simplify the radical: $\sqrt7 \times\frac{6}{\sqrt2}$

Possible Answers:

$3\sqrt{14}$

$21\sqrt{14}$

$3\sqrt7$

$3\sqrt{21}$

Correct answer:

$3\sqrt{14}$

Explanation:

Multiply the square root seven with the numerator.

$\sqrt7 \times\frac{6}{\sqrt2}=\frac{6\sqrt7 }{\sqrt2}$

Rationalize the denominator by multiplying the top and bottom of the fraction by the denominator.

$\frac{6\sqrt7 }{\sqrt2} = \frac{6\sqrt7 }{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2} = \frac{6\sqrt{14}}{2}$

Simplify the fraction by dividing 6 with 2. The square root 14 cannot be simplified by perfect squares.

The answer is: $3\sqrt{14}$

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Example Question #46 : Radicals And Fractions

Simplify: $\frac{\sqrt{12}}{2\sqrt{24}}$

Possible Answers:

$\frac{2\sqrt{6}}{3}$

$\frac{\sqrt2}{4}$

$\frac{4\sqrt{6}}{3}$

$\frac{5\sqrt2}{4}$

$\sqrt{3}$

Correct answer:

$\frac{\sqrt2}{4}$

Explanation:

Simplify both the top and bottom of the fraction.

$\frac{\sqrt{12}}{2\sqrt{24}}=\frac{\sqrt{4}\cdot \sqrt3}{2\sqrt{4}\cdot \sqrt{6}}$

Cancel the common terms.

$\frac{\sqrt3}{2\sqrt6}$

Rationalize the denominator by multiplying the top and bottom by root six.

$\frac{\sqrt3}{2\sqrt6}\cdot \frac{\sqrt6}{\sqrt6}$

Multiply the numerator with numerator and denominator with denominator.

$\frac{\sqrt{18}}{2 \times 6} = \frac{\sqrt9\cdot \sqrt2}{12} = \frac{3\sqrt2}{12}$

Simplify the fraction.

The answer is: $\frac{\sqrt2}{4}$

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Example Question #47 : Radicals And Fractions

Evaluate: $\frac{5}{\sqrt{3}\cdot \sqrt{5}}$

Possible Answers:

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

$\frac{ \sqrt{15}}{3}$

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

$\frac{ \sqrt{15}}{5}$

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

Correct answer:

$\frac{ \sqrt{15}}{3}$

Explanation:

Multiply the denominator together to combine as one radical.

$\frac{5}{\sqrt{3}\cdot \sqrt{5}} = \frac{5}{\sqrt{15}}$

Rationalize the denominator by multiplying the top and bottom of the fraction by the denominator.

$\frac{5}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}} =\frac{ 5\sqrt{15}}{15}$

Reduce the fraction.

The answer is: $\frac{ \sqrt{15}}{3}$

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Example Question #48 : Radicals And Fractions

Solve the radical: $\frac{\sqrt{3}\times \sqrt{5}}{\sqrt{32}}$

Possible Answers:

$\frac{\sqrt{30}}{4}$

$\frac{2\sqrt{15}}{3}$

$\frac{\sqrt{15}}{8}$

$\frac{\sqrt{30}}{8}$

$\frac{\sqrt{15}}{4}$

Correct answer:

$\frac{\sqrt{30}}{8}$

Explanation:

Simplify the top of the fraction.

$\frac{\sqrt{3}\times \sqrt{5}}{\sqrt{32}}=\frac{\sqrt{15}}{\sqrt{32}}$

We can simplify the denominator first.

$\sqrt{32} = \sqrt{16}\times \sqrt2 =4\sqrt2$

Rationalize the denominator by multiplying the top and bottom of the fraction by the radical. Simplify the radicals.

$\frac{\sqrt{15}}{4\sqrt2}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{30}}{4 \times 2}= \frac{\sqrt{30}}{8}$

The answer is: $\frac{\sqrt{30}}{8}$

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Example Question #49 : Radicals And Fractions

Simplify: $\frac{4\sqrt{8}}{5\sqrt{3}}$

Possible Answers:

$\frac{\sqrt{6}}{6}$

$\frac{3\sqrt{6}}{4}$

$\frac{2\sqrt{6}}{5}$

$\frac{4\sqrt{3}}{5}$

$\frac{8\sqrt{6}}{15}$

Correct answer:

$\frac{8\sqrt{6}}{15}$

Explanation:

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\frac{4\sqrt{8}}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{24}}{5\times3} = \frac{4\sqrt{4}\times \sqrt{6}}{15}=\frac{4(2)\times \sqrt{6}}{15}$

The answer is: $\frac{8\sqrt{6}}{15}$

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Example Question #1551 : Mathematical Relationships And Basic Graphs

Simplify: $\frac{7\sqrt7}{\sqrt{\frac{1}{7}}}$

Possible Answers:

$\sqrt{14}$

$\frac{1}{7}$

$\frac{1}{49}$

Correct answer:

Explanation:

The radical $\sqrt{\frac{1}{7}}$ can be rewritten as:

$\frac{\sqrt1}{\sqrt7} = \frac{1}{\sqrt7}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\frac{1}{\sqrt7} \cdot \frac{\sqrt7}{\sqrt7} =\frac{\sqrt7}{7}$

Rewrite the given question.

$\frac{7\sqrt7}{\sqrt{\frac{1}{7}}} = \frac{7\sqrt7}{\frac{\sqrt7}{7}}$

Simplify this complex fraction by rewriting it as a division sign.

$\frac{7\sqrt7}{\frac{\sqrt7}{7}} = 7\sqrt7 \div \frac{\sqrt7}{7}$

Change the division sign to multiplication and take the reciprocal of the second term.

$7\sqrt7 \div \frac{\sqrt7}{7} = 7\sqrt7 \times \frac{7}{\sqrt7} =49$

The answer is:

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Algebra II : Radicals and Fractions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Radicals And Fractions

Example Question #42 : Radicals And Fractions

Example Question #43 : Radicals And Fractions

Example Question #44 : Radicals And Fractions

Example Question #45 : Radicals And Fractions

Example Question #46 : Radicals And Fractions

Example Question #47 : Radicals And Fractions

Example Question #48 : Radicals And Fractions

Example Question #49 : Radicals And Fractions

Example Question #1551 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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