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Algebra II : Radicals and Fractions

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Example Question #11 : Radicals And Fractions

Simplify the following equation:

$\frac{\sqrt{14}}{\sqrt{6}}$

Possible Answers:

$\frac{\sqrt{21}}{3}$

$\frac{\sqrt{84}}{\sqrt{6}}$

$\frac{\sqrt{14}}{4}$

None of the above

${2\sqrt{21}}$

Correct answer:

$\frac{\sqrt{21}}{3}$

Explanation:

You are not allowed to have radicals in the denominator, so to simplify this you must multiply everything by $\sqrt{6}$

So:

$\frac{\sqrt{14}}{\sqrt{6}}*\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{84}}{6}$

However, this is not the last step because you need to simplify to numerator, looking for perfect squares.

So:

$\frac{\sqrt{4}*\sqrt{21}}{6}=\frac{{2}*\sqrt{21}}{6}$

Then you need to simplify a little further to:

$\frac{\sqrt{4}*\sqrt{21}}{6}=\frac{{}\sqrt{21}}{3}$

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Example Question #12 : Radicals And Fractions

Which of the following is equivalent to the expression $\frac{2}{\sqrt{2}}$ ?

Possible Answers:

$\frac{1}{2}$

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

$\sqrt{2}$

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

Correct answer:

$\sqrt{2}$

Explanation:

Which of the following is equivalent to the expression $\frac{2}{\sqrt{2}}$ ?

When dealing with radicals in denominators, we want to bring them up to the numerator. To do so, multiply the beginning fraction by a fraction equal to 1 which looks like the beginning radical (look at the red part of the following expression).

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

Now, instead of canceling the square roots, multiply them out and then simplify.

$\frac{2\sqrt{2}}{\sqrt{2}*\sqrt{2}}=\frac{2\sqrt2}{2}=\sqrt{2}$

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Example Question #13 : Radicals And Fractions

Simplify: $\frac{2}{4+\sqrt{8}}$

Possible Answers:

None of these.

$1-\frac{\sqrt{8}}{4}$

$\frac{1-\sqrt{8}}{8}$

$1+\frac{\sqrt{8}}{4}$

$4+\sqrt{8}$

Correct answer:

$1-\frac{\sqrt{8}}{4}$

Explanation:

$\frac{2}{4+\sqrt{8}}$

Rationalize the denominator:

$\frac{2}{4+\sqrt{8}}*\frac{4-\sqrt{8}}{4-\sqrt{8}}$

Simplify:

$\frac{2(4-\sqrt{8})}{16-8}$

$\frac{2(4-\sqrt{8})}{8}$

$\frac{(4-\sqrt{8})}{4}$

$\mathbf{1-\frac{\sqrt{8}}{4}}$

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Example Question #4171 : Algebra Ii

Rationalize the numerator: $\frac{\sqrt{6}-\sqrt{8}}{2}$

Possible Answers:

$\frac{1}{2(\sqrt{6}+\sqrt{8})}$

None of these

$\frac{-1}{2(\sqrt{6}+\sqrt{8})}$

$\frac{1}{\sqrt{6}-\sqrt{8}}$

$\frac{-1}{\sqrt{6}+\sqrt{8}}$

Correct answer:

$\frac{-1}{\sqrt{6}+\sqrt{8}}$

Explanation:

To rationalize the numerator you must multiply by the "conjugate" of the numerator. To find the conjugate simply switch the sign of the expression.

$\frac{\sqrt{6}-\sqrt{8}}{2}$

$\frac{\sqrt{6}-\sqrt{8}}{2}*\frac{\sqrt{6}+\sqrt{8}}{\sqrt{6}+\sqrt{8}}$

*foil the numerator. The middle terms will cancel each other and the radical cancel each other.

$\frac{6-8}{2(\sqrt{6}+\sqrt{8})}$

$\mathbf{\frac{-1}{(\sqrt{6}+\sqrt{8})}}$

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Example Question #15 : Radicals And Fractions

Simplify the following:

$\small \frac{\sqrt{7}}{\sqrt14}$

Possible Answers:

$\small \frac{\sqrt{98}}{14}$

$\small \frac{1}{2}$

$\small \small \frac{1\sqrt{2}}{2}$

$\small \frac{\sqrt1}{\sqrt 2}$

$\small \frac{\sqrt{98}}{\sqrt{14}}$

Correct answer:

$\small \small \frac{1\sqrt{2}}{2}$

Explanation:

To rationalize the denominator you need to multiply the top and bottom by the square root of 14. Upon doing so, you get:

$\small \frac{\sqrt{98}}{14}$

However you can further simplify this because you know that two times forty-nine is ninety-eight you can simplify this to:

$\small \frac{7\sqrt{2}}{14}=\frac{1\sqrt2}{2}$

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Example Question #16 : Radicals And Fractions

Simplify the following radical:

$\sqrt{\frac{x^5}{4x^7}}$

Possible Answers:

$\frac{1}{2x}$

$\frac{1}{2}$

$\frac{1}{4x}$

$\frac{2}{x}$

Correct answer:

$\frac{1}{2x}$

Explanation:

To simplify this radical, first we can simplify what is underneath the radical:

$\sqrt{\frac{x^5}{4x^7}}=\sqrt{\frac{x^5}{4\cdot x^5\cdot x^2}}=\sqrt{\frac{1}{4x^2}}$

Finally, we take the square root of top and bottom:

$\frac{\sqrt{1}}{\sqrt{4x^2}}=\frac{1}{2x}$

The square root of 1 is 1, and the square root of $\sqrt{4x^2}=\sqrt{4}\cdot \sqrt{x^2}=2x$ .

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Example Question #17 : Radicals And Fractions

Simplify:

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

Possible Answers:

$\sqrt{3}$

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

$\frac{1}{3}$

Correct answer:

$\sqrt{3}$

Explanation:

In order to simplify this problem we need to multiply the top and bottom terms of the fraction by the radical in the denominator. This is because we always want an integer in the denominator.

$\frac{3}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{3}$

Simplify.

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

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Example Question #18 : Radicals And Fractions

Simplify:

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

Possible Answers:

$\sqrt{5}$

$\sqrt{10}$

$\frac{5}{2}$

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

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

Correct answer:

$\sqrt{5}$

Explanation:

First, we can simplify the expression by dividing the integers on the top and bottom of the fraction.

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

In order to simplify this problem we need to multiply the top and bottom terms of the fraction by the radical in the denominator. This is because we always want an integer in the denominator.

$\frac{5}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{5}$

Simplify.

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

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Example Question #19 : Radicals And Fractions

Simplify:

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

Possible Answers:

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

$2-2\sqrt{2}$

$3+2\sqrt{2}$

$2+2\sqrt{2}$

$3-2\sqrt{2}$

Correct answer:

$3+2\sqrt{2}$

Explanation:

In order to get rid of the radical on the bottom of the fraction, we need to multiply the top and bottom of the fraction by the conjugate. The conjugate is the expression that possesses the opposite sign of the radical expression on the bottom of the fraction:

$\sqrt{2}-1\rightarrow\sqrt{2}+1$

Therefore:

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

Simplify.

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

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Example Question #20 : Radicals And Fractions

Simplify:

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

Possible Answers:

$\sqrt{2}+1$

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

$\sqrt{2}-1$

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

Correct answer:

$\sqrt{2}-1$

Explanation:

$\sqrt{2}+1\rightarrow\sqrt{2}-1$

Therefore:

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

Simplify.

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

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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 #11 : Radicals And Fractions

Example Question #12 : Radicals And Fractions

Example Question #13 : Radicals And Fractions

Example Question #4171 : Algebra Ii

Example Question #15 : Radicals And Fractions

Example Question #16 : Radicals And Fractions

Example Question #17 : Radicals And Fractions

Example Question #18 : Radicals And Fractions

Example Question #19 : Radicals And Fractions

Example Question #20 : Radicals And Fractions

All Algebra II Resources

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