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Algebra II : Multiplying and Dividing Radicals

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Example Questions

Example Question #41 : Multiplying And Dividing Radicals

$\sqrt[3]{4}*\sqrt[3]16{}$ $\displaystyle \sqrt[3]{4}*\sqrt[3]16{}$

Possible Answers:

$2\sqrt[3]{4}$ $\displaystyle 2\sqrt[3]{4}$

$\displaystyle 64$

$2\sqrt[3]{16}$ $\displaystyle 2\sqrt[3]{16}$

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

$\displaystyle 4$

Correct answer:

$\displaystyle 4$

Explanation:

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

$\sqrt[3]{4}*\sqrt[3]{16}=\sqrt[3]{64}$ $\displaystyle \sqrt[3]{4}*\sqrt[3]{16}=\sqrt[3]{64}$ This can be simplified to $\displaystyle 4$ which is the cubic root of the answer.

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Example Question #1476 : Mathematical Relationships And Basic Graphs

$\frac{10}{\sqrt{50}}$ $\displaystyle \frac{10}{\sqrt{50}}$

Possible Answers:

$\displaystyle 5$

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

$2\sqrt{2}$ $\displaystyle 2\sqrt{2}$

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

$\sqrt{2}$ $\displaystyle \sqrt{2}$

Correct answer:

$\sqrt{2}$ $\displaystyle \sqrt{2}$

Explanation:

We can simplify $\sqrt{50}$ $\displaystyle \sqrt{50}$ by finding a perfect square.

$\sqrt{50}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$ $\displaystyle \sqrt{50}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$ Next we can reduce to $\frac{2}{\sqrt{2}}$ $\displaystyle \frac{2}{\sqrt{2}}$ .

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

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

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Example Question #43 : Multiplying And Dividing Radicals

$\sqrt{16x}\cdot \sqrt{x^2w^4}$ $\displaystyle \sqrt{16x}\cdot \sqrt{x^2w^4}$

Possible Answers:

$4xw\sqrt{x}$ $\displaystyle 4xw\sqrt{x}$

$4w^2\sqrt{x}$ $\displaystyle 4w^2\sqrt{x}$

$4xw^2\sqrt{x}$ $\displaystyle 4xw^2\sqrt{x}$

4xw^2 $\displaystyle 4xw^2$

Correct answer:

$4xw^2\sqrt{x}$ $\displaystyle 4xw^2\sqrt{x}$

Explanation:

The first step I'd recommend is to multiply everything and put it all underneath one radical: $\sqrt{16x^3w^4}$ $\displaystyle \sqrt{16x^3w^4}$ . Then, recall that for every two of the same term, cross them out underneath the radical and put one of them outside. Attack each term separately: $\sqrt{16}=4$ $\displaystyle \sqrt{16}=4$ , $\sqrt{x^3}=x\sqrt{x}$ $\displaystyle \sqrt{x^3}=x\sqrt{x}$ , and $\sqrt{w^4}=w^2$ $\displaystyle \sqrt{w^4}=w^2$ . Put those all together to get: $4xw^2\sqrt{x}$ $\displaystyle 4xw^2\sqrt{x}$ .

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Example Question #142 : Simplifying Radicals

$3\sqrt{3}*\sqrt{8}$ $\displaystyle 3\sqrt{3}*\sqrt{8}$

Possible Answers:

$3\sqrt{24}$ $\displaystyle 3\sqrt{24}$

$3\sqrt{6}$ $\displaystyle 3\sqrt{6}$

$6\sqrt{6}$ $\displaystyle 6\sqrt{6}$

$3\sqrt{24}$ $\displaystyle 3\sqrt{24}$

$\displaystyle 6$

Correct answer:

$6\sqrt{6}$ $\displaystyle 6\sqrt{6}$

Explanation:

$3\sqrt{3}*\sqrt{8}$ $\displaystyle 3\sqrt{3}*\sqrt{8}$

Combine radicals:

$3\sqrt{24}$ $\displaystyle 3\sqrt{24}$

Simplify the radical leftover:

$3\sqrt{6*4}$ $\displaystyle 3\sqrt{6*4}$

$\mathbf{6\sqrt{6}}$ $\displaystyle \mathbf{6\sqrt{6}}$

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Example Question #4141 : Algebra Ii

Divide and simplify: $\frac{28\sqrt{24}}{7\sqrt{6}}$ $\displaystyle \frac{28\sqrt{24}}{7\sqrt{6}}$

Possible Answers:

None of these

$\displaystyle 8$

$4\sqrt{6}$ $\displaystyle 4\sqrt{6}$

$\displaystyle 16$

$\displaystyle 6$

Correct answer:

$\displaystyle 8$

Explanation:

$\frac{28\sqrt{24}}{7\sqrt{6}}$ $\displaystyle \frac{28\sqrt{24}}{7\sqrt{6}}$

Divide outside number:

$\frac{4\sqrt{24}}{\sqrt{6}}$ $\displaystyle \frac{4\sqrt{24}}{\sqrt{6}}$

Divide radicals:

$4\sqrt{4}$ $\displaystyle 4\sqrt{4}$

Simplify radical:

$4*2=\mathbf{8}$ $\displaystyle 4*2=\mathbf{8}$

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Example Question #42 : Multiplying And Dividing Radicals

Expand, then simplify: $(\sqrt{6}+9)^2$ $\displaystyle (\sqrt{6}+9)^2$

Possible Answers:

$87+18\sqrt{6}$ $\displaystyle 87+18\sqrt{6}$

$87+6\sqrt{18}$ $\displaystyle 87+6\sqrt{18}$

$87+\sqrt{6}$ $\displaystyle 87+\sqrt{6}$

$87+6\sqrt{18}$ $\displaystyle 87+6\sqrt{18}$

$87+18\sqrt{6}$ $\displaystyle 87+18\sqrt{6}$

$87+6\sqrt{18}$ $\displaystyle 87+6\sqrt{18}$

$87+18\sqrt{6}$ $\displaystyle 87+18\sqrt{6}$

Correct answer:

$87+18\sqrt{6}$ $\displaystyle 87+18\sqrt{6}$

Explanation:

$(\sqrt{6}+9)^2$ $\displaystyle (\sqrt{6}+9)^2$

$(\sqrt{6}+9)^2=(\sqrt{6}+9)(\sqrt{6}+9)$ $\displaystyle (\sqrt{6}+9)^2=(\sqrt{6}+9)(\sqrt{6}+9)$

Foil:

$6+9\sqrt{6}+9\sqrt{6}+81$ $\displaystyle 6+9\sqrt{6}+9\sqrt{6}+81$

$\mathbf{87+18\sqrt{6}}$ $\displaystyle \mathbf{87+18\sqrt{6}}$

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Example Question #151 : Simplifying Radicals

Simplify:

$\sqrt{6}*\sqrt{8}$ $\displaystyle \sqrt{6}*\sqrt{8}$

Possible Answers:

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

$2\sqrt{6}$ $\displaystyle 2\sqrt{6}$

$4\sqrt{2}$ $\displaystyle 4\sqrt{2}$

$4\sqrt{6}$ $\displaystyle 4\sqrt{6}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

Correct answer:

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

Explanation:

$\sqrt{6}*\sqrt{8}$ $\displaystyle \sqrt{6}*\sqrt{8}$ Multiply the numbers inside the radical.

$\sqrt{6}*\sqrt{8}=\sqrt{6*8}=\sqrt{48}$ $\displaystyle \sqrt{6}*\sqrt{8}=\sqrt{6*8}=\sqrt{48}$ Factor out a perfect square of $\displaystyle 16$ .

$\sqrt{48}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$ $\displaystyle \sqrt{48}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$

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Example Question #151 : Simplifying Radicals

Simplify:

$\sqrt{14}*\sqrt{21}$ $\displaystyle \sqrt{14}*\sqrt{21}$

Possible Answers:

$14\sqrt{2}$ $\displaystyle 14\sqrt{2}$

$7\sqrt{2}$ $\displaystyle 7\sqrt{2}$

$21\sqrt{3}$ $\displaystyle 21\sqrt{3}$

$49\sqrt{3}$ $\displaystyle 49\sqrt{3}$

$7\sqrt{6}$ $\displaystyle 7\sqrt{6}$

Correct answer:

$7\sqrt{6}$ $\displaystyle 7\sqrt{6}$

Explanation:

$\sqrt{14}*\sqrt{21}$ $\displaystyle \sqrt{14}*\sqrt{21}$ Multiply the numbers inside the radical.

$\sqrt{14}*\sqrt{21}=\sqrt{14*21}=\sqrt{294}$ $\displaystyle \sqrt{14}*\sqrt{21}=\sqrt{14*21}=\sqrt{294}$ Factor out a perfect square of $\displaystyle 49$ .

$\sqrt{294}=\sqrt{49}*\sqrt{6}=7\sqrt{6}$ $\displaystyle \sqrt{294}=\sqrt{49}*\sqrt{6}=7\sqrt{6}$

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Example Question #42 : Multiplying And Dividing Radicals

Simplify:

$\frac{\sqrt{338}}{\sqrt{169}}$ $\displaystyle \frac{\sqrt{338}}{\sqrt{169}}$

Possible Answers:

$\displaystyle 2$

$\sqrt{2}$ $\displaystyle \sqrt{2}$

$\displaystyle 3$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

Correct answer:

$\sqrt{2}$ $\displaystyle \sqrt{2}$

Explanation:

$\frac{\sqrt{338}}{\sqrt{169}}$ $\displaystyle \frac{\sqrt{338}}{\sqrt{169}}$ Divide the numbers inside the radicals.

$\frac{\sqrt{338}}{\sqrt{169}}=\sqrt{\frac{338}{169}}=\sqrt{2}$ $\displaystyle \frac{\sqrt{338}}{\sqrt{169}}=\sqrt{\frac{338}{169}}=\sqrt{2}$

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Example Question #43 : Multiplying And Dividing Radicals

Simplify:

$\sqrt{7}*\sqrt{5}$ $\displaystyle \sqrt{7}*\sqrt{5}$

Possible Answers:

$7\sqrt{5}$ $\displaystyle 7\sqrt{5}$

$\sqrt{57}$ $\displaystyle \sqrt{57}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$\sqrt{35}$ $\displaystyle \sqrt{35}$

$5\sqrt{7}$ $\displaystyle 5\sqrt{7}$

Correct answer:

$\sqrt{35}$ $\displaystyle \sqrt{35}$

Explanation:

When multiplying radicals, simply multiply the numbers inside the radical with each other. Therefore:

$\sqrt{7}*\sqrt{5}=\sqrt{35}$ $\displaystyle \sqrt{7}*\sqrt{5}=\sqrt{35}$

We cannot further simplify because both of the numbers multiplied with each other were prime numbers.

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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 #41 : Multiplying And Dividing Radicals

Example Question #1476 : Mathematical Relationships And Basic Graphs

Example Question #43 : Multiplying And Dividing Radicals

Example Question #142 : Simplifying Radicals

Example Question #4141 : Algebra Ii

Example Question #42 : Multiplying And Dividing Radicals

Example Question #151 : Simplifying Radicals

Example Question #151 : Simplifying Radicals

Example Question #42 : Multiplying And Dividing Radicals

Example Question #43 : Multiplying And Dividing Radicals

All Algebra II Resources

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