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Algebra II : Multiplying and Dividing Radicals

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Example Question #1461 : Mathematical Relationships And Basic Graphs

Simplify.

$\frac{\sqrt{7}}{\sqrt{5}}$

Possible Answers:

$\frac{\sqrt{35}}{5}$

$\frac{7\sqrt{5}}{5}$

$\frac{5}{7}$

$\frac{5\sqrt{7}}{7}$

$\frac{7}{5}$

Correct answer:

$\frac{\sqrt{35}}{5}$

Explanation:

When dividing radicals, we multiply top and bottom by the bottom radical to ensure our denominator is an integer.

$\frac{\sqrt{7}}{\sqrt{5}}=\frac{\sqrt{7}\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}=\frac{\sqrt{35}}{5}$ . This can't be simplified and is our answer.

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Example Question #1462 : Mathematical Relationships And Basic Graphs

Simplify.

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

Possible Answers:

$3\sqrt{3}$

$\sqrt{3}$

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

Correct answer:

$\sqrt{3}$

Explanation:

When dividing radicals, we multiply top and bottom by the bottom radical to ensure our denominator is an integer.

$\frac{3}{\sqrt{3}}=\frac{3\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}$ .

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Example Question #33 : Multiplying And Dividing Radicals

Simplify.

$\frac{2}{1+\sqrt{3}}$

Possible Answers:

$-2+2\sqrt{3}$

$\frac{2}{1-\sqrt{3}}$

$1-\sqrt{3}$

$\frac{1-\sqrt{3}}{2}$

$-1+\sqrt{3}$

Correct answer:

$-1+\sqrt{3}$

Explanation:

$\frac{2}{1+\sqrt{3}}$ To get rid of the radical in the denominator, we need to multiply top and bottom by the conjugate which is the opposite sign of the radical expression. This would be $1-\sqrt{3}$ .

$\frac{2}{1+\sqrt{3}}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}}=\frac{2-2\sqrt{3}}{1-3}=\frac{2-2\sqrt{3}}{-2}$ Remember to FOIL out when multiplying out the denominators. Now, with our answer, we can factor out a .

$\frac{2-2\sqrt{3}}{-2}=\frac{-2(-1+\sqrt{3})}{-2}=-1+\sqrt{3}$

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Example Question #34 : Multiplying And Dividing Radicals

Simplify.

$\frac{3}{1-\sqrt{2}}$

Possible Answers:

$3-3\sqrt{2}$

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

$-3-3\sqrt{2}$

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

$3+3\sqrt{2}$

Correct answer:

$-3-3\sqrt{2}$

Explanation:

$\frac{3}{1-\sqrt{2}}$ To get rid of the radical in the denominator, we need to multiply top and bottom by the conjugate which is the opposite sign of the radical expression. This would be $1+\sqrt{2}$ .

$\frac{3}{1-\sqrt{2}}\cdot\frac{1+\sqrt{2}}{1+\sqrt{2}}=\frac{3+3\sqrt{2}}{1-2}=\frac{3+3\sqrt{2}}{-1}$ Remember to use FOIL when multiplying out the denominators. Now, with out answer, we can distribute out the .

$\frac{3+3\sqrt{2}}{-1}=-3-3\sqrt{2}$

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Example Question #35 : Multiplying And Dividing Radicals

Multiply: $6\sqrt{3}*2\sqrt{8}$

Possible Answers:

$24\sqrt{6}$

None of these

$12\sqrt{6}$

Cannot combine these radicals

$12\sqrt{24}$

Correct answer:

$24\sqrt{6}$

Explanation:

$6\sqrt{3}*2\sqrt{8}$

Multiply the outer numbers first:

$12(\sqrt{3}*\sqrt{8})$

Combine radicals:

$12\sqrt{24}$

Simplify the radical:

$12\sqrt{6*4}$

$\mathbf{24\sqrt{6}}$

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Example Question #4131 : Algebra Ii

$\sqrt{3}*\sqrt{7}$

Possible Answers:

$\sqrt{21}$

$\sqrt{10}$

$\sqrt{37}$

Correct answer:

$\sqrt{21}$

Explanation:

When multiplying radicals, we can just multiply the values inside the radicand.

$\sqrt{3}*\sqrt{7}=\sqrt{3*7}=\sqrt{21}$

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Example Question #34 : Multiplying And Dividing Radicals

$\sqrt{3}*\sqrt{27}$

Possible Answers:

$4\sqrt{3}$

$9\sqrt{3}$

$\sqrt{30}$

Correct answer:

Explanation:

When multiplying radicals, we can just multiply the values inside the radicand.

$\sqrt{3}*\sqrt{27}=\sqrt{3*27}=\sqrt{81}=9$

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Example Question #35 : Multiplying And Dividing Radicals

$\sqrt{5}*\sqrt{15}$

Possible Answers:

$4\sqrt{3}$

$3\sqrt{5}$

$5\sqrt{2}$

$5\sqrt{3}$

$2\sqrt{5}$

Correct answer:

$5\sqrt{3}$

Explanation:

When multiplying radicals, we can just multiply the values inside the radicand.

$\sqrt{5}*\sqrt{15}=\sqrt{5*15}=\sqrt{75}$ We can simplify this by finding a perfect square.

$\sqrt{75}=\sqrt{25}*\sqrt{3}=5\sqrt{3}$

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Example Question #35 : Multiplying And Dividing Radicals

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

Possible Answers:

$2\sqrt{2}$

$4\sqrt{2}$

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

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

Correct answer:

$2\sqrt{2}$

Explanation:

When dealing with radicals in the denominator, we simplify it by multiplying top and bottom by the radical.

$\frac{4}{\sqrt{2}}=\frac{4}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

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Example Question #40 : Multiplying And Dividing Radicals

$\frac{\sqrt{6}}{\sqrt{12}}$

Possible Answers:

$\frac{1}{2}$

$\sqrt{2}$

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

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

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

Correct answer:

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

Explanation:

We can simplify this fraction. We can combine the radicals into a giant radical expression.

$\frac{\sqrt{6}}{\sqrt{12}}=\sqrt{\frac{6}{12}}=\sqrt{\frac{1}{2}}$

When dealing with radicals in the denominator, we simplify it by multiplying top and bottom by the radical.

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

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Algebra II : Multiplying and Dividing Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1461 : Mathematical Relationships And Basic Graphs

Example Question #1462 : Mathematical Relationships And Basic Graphs

Example Question #33 : Multiplying And Dividing Radicals

Example Question #34 : Multiplying And Dividing Radicals

Example Question #35 : Multiplying And Dividing Radicals

Example Question #4131 : Algebra Ii

Example Question #34 : Multiplying And Dividing Radicals

Example Question #35 : Multiplying And Dividing Radicals

Example Question #35 : Multiplying And Dividing Radicals

Example Question #40 : Multiplying And Dividing Radicals

All Algebra II Resources

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