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Algebra II : Algebra II

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 #4131 : Algebra Ii

Simplify.

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

Possible Answers:

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

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

$3-3\sqrt{2}$

$3+3\sqrt{2}$

$-3-3\sqrt{2}$

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 #4132 : Algebra Ii

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

Possible Answers:

$12\sqrt{6}$

$24\sqrt{6}$

None of these

$12\sqrt{24}$

Cannot combine these radicals

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:

$\sqrt{37}$

$\sqrt{10}$

$\sqrt{21}$

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 #4132 : Algebra Ii

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

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:

$5\sqrt{3}$

$3\sqrt{5}$

$2\sqrt{5}$

$4\sqrt{3}$

$5\sqrt{2}$

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 #36 : Multiplying And Dividing Radicals

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

Possible Answers:

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

$4\sqrt{2}$

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

$2\sqrt{2}$

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 #235 : Radicals

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

Possible Answers:

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

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

$\frac{1}{2}$

$\sqrt{2}$

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

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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Example Question #1471 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$2\sqrt[3]{16}$

$4\sqrt{3}$

$2\sqrt[3]{4}$

Correct answer:

Explanation:

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

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

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Example Question #4136 : Algebra Ii

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

Possible Answers:

$2\sqrt{2}$

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

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

$\sqrt{2}$

Correct answer:

$\sqrt{2}$

Explanation:

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

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

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

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

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Example Question #43 : Multiplying And Dividing Radicals

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

Possible Answers:

$4xw\sqrt{x}$

$4w^2\sqrt{x}$

$4xw^2\sqrt{x}$

4xw^2

Correct answer:

$4xw^2\sqrt{x}$

Explanation:

The first step I'd recommend is to multiply everything and put it all underneath one radical: $\sqrt{16x^3w^4}$ . 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: $\sqrt{16}=4$ , $\sqrt{x^3}=x\sqrt{x}$ , and $\sqrt{w^4}=w^2$ . Put those all together to get: $4xw^2\sqrt{x}$ .

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4131 : Algebra Ii

Example Question #4132 : Algebra Ii

Example Question #4131 : Algebra Ii

Example Question #4132 : Algebra Ii

Example Question #35 : Multiplying And Dividing Radicals

Example Question #36 : Multiplying And Dividing Radicals

Example Question #235 : Radicals

Example Question #1471 : Mathematical Relationships And Basic Graphs

Example Question #4136 : Algebra Ii

Example Question #43 : Multiplying And Dividing Radicals

All Algebra II Resources

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