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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #13 : Radicals And Fractions

Simplify:

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

Possible Answers:

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

$\sqrt{2}-1$

$\sqrt{2}+1$

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

Correct answer:

$\sqrt{2}-1$

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:

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

Simplify.

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

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Example Question #281 : Radicals

Simplify:

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

Possible Answers:

$3\sqrt{3}-1$

$\sqrt{3}-1$

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

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

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

Correct answer:

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

Explanation:

$\sqrt{3}-1\rightarrow \sqrt{3}+1$

Therefore:

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

Simplify.

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

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Example Question #22 : Radicals And Fractions

Simplify: $\frac{2-2\sqrt{2}}{\sqrt{6}}$

Possible Answers:

$\frac{2\sqrt{6}-4\sqrt{3}}{6}$

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

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

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

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

Correct answer:

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

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 when multiplied to the denominator removes the radical:

$\frac{2-2\sqrt{2}}{\sqrt{6}}*\frac{\sqrt{6}}{\sqrt{6}}=\frac{2\sqrt{6}-2\sqrt{12}}{6}$

Simplify.

$\frac{2\sqrt{6}-2\sqrt{12}}{6}=\frac{2\sqrt{6}-4\sqrt{3}}{6}$

Now, factor out a .

$\frac{2\sqrt{6}-4\sqrt{3}}{6}=\frac{\sqrt{6}-2\sqrt{3}}{3}$

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Example Question #23 : Radicals And Fractions

Simplify the expression: $\frac{1}{\sqrt2}+\frac{1}{\sqrt3}$

Possible Answers:

$6\sqrt{5}$

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

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

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

$\frac{\sqrt{30}}{6}$

Correct answer:

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

Explanation:

In order to simplify this, we will need to multiply both radicals together to obtain the least common denominator.

$\sqrt{2}\cdot \sqrt{3} =\sqrt{6}$

Convert both fractions so that they have similar common denominators.

$\frac{1}{\sqrt2}+\frac{1}{\sqrt3} = \frac{\sqrt3}{\sqrt6}+ \frac{\sqrt2}{\sqrt6}$

Combine the fractions. Do not add the terms of radicals on the numerator such that they would combine as a single radical!

$\sqrt{3}+\sqrt{2}\neq \sqrt{5}$

$\frac{\sqrt3+\sqrt2}{\sqrt6}$

Rationalize the denominator by multiplying $\frac{\sqrt{6}}{\sqrt6}$ .

$(\frac{\sqrt{6}}{\sqrt6})\frac{\sqrt3+\sqrt2}{\sqrt6} = \frac{\sqrt{18}+\sqrt{12}}{6}$

$\sqrt{18} = \sqrt{9}\cdot\sqrt{2} = 3\sqrt2$

$\sqrt{12} = \sqrt{4}\cdot \sqrt{3} = 2\sqrt{3}$

Replace the simplified radicals on the numerator.

The answer is: $\frac{3\sqrt{2}+2\sqrt{3}}{6}$

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Example Question #4181 : Algebra Ii

Rationalize the denominator: $\frac{6}{\sqrt{30}}$

Possible Answers:

$\frac{2\sqrt{10}}{3}$

$\frac{\sqrt{30}}{3}$

$\sqrt{10}$

$\frac{\sqrt{30}}{6}$

$\frac{\sqrt{30}}{5}$

Correct answer:

$\frac{\sqrt{30}}{5}$

Explanation:

In order to eliminate the radical sign, multiply the top and the bottom by the denominator.

$\frac{6}{\sqrt{30}}\cdot \frac{\sqrt{30}}{\sqrt{30}} = \frac{6\sqrt{30}}{30}$

Reduce the numerator and denominator.

$\frac{6\sqrt{30}}{6\times 5}= \frac{\sqrt{30}}{5}$

The answer is: $\frac{\sqrt{30}}{5}$

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Example Question #25 : Radicals And Fractions

Simplify:

$\sqrt{\frac{81x^2}{144y^6}}$

Possible Answers:

$\frac{3x}{12y^3}$

$\frac{9x}{12y^3}$

$\frac{9x}{11y^3}$

$\frac{9}{12y^3}$

$\frac{9x}{12y^2}$

Correct answer:

$\frac{9x}{12y^3}$

Explanation:

First, make the numerators and denominators separate radicals:

$\frac{\sqrt{81x^2}}{\sqrt{144y^6}}$ .

Then, simplify each one.

$\sqrt{81x^2}=9x$

and

$\sqrt{144y^6}=12y^3$ .

Then, put those together to get your answer of

$\frac{9x}{12y^3}$ .

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Example Question #21 : Radicals And Fractions

Simplify the fraction: $\frac{2\sqrt{3}}{\sqrt{30}}$

Possible Answers:

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

$\frac{\sqrt{30}}{5}$

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

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

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

Correct answer:

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

Explanation:

In order to simplify this fraction, we will need to rationalize the denominator.

Multiply the fraction by $\frac{\sqrt{30}}{\sqrt{30}}$ .

$\frac{2\sqrt{3}}{\sqrt{30}} \cdot \frac{\sqrt{30}}{\sqrt{30}} = \frac{2\sqrt{90}}{30} = \frac{2\sqrt{9}\cdot \sqrt{10}}{30}$

Simplify the fraction.

$\frac{2\sqrt{9}\cdot \sqrt{10}}{30} = \frac{2\cdot 3\cdot \sqrt{10}}{30} =\frac{\sqrt{10}}{5}$

The answer is: $\frac{\sqrt{10}}{5}$

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Example Question #22 : Radicals And Fractions

$\sqrt{\frac{25x^4}{225y^{10}}}$

Possible Answers:

$\frac{x^2}{3y^2}$

$\frac{x^2}{y^2}$

$\frac{2x^2}{3y^2}$

$\frac{4x^2}{3y^2}$

$\frac{x^2}{3y}$

Correct answer:

$\frac{x^2}{3y^2}$

Explanation:

To simplify, first start by rewriting as:

$\frac{\sqrt{25x^4}}{\sqrt{225y^{10}}}$ .

Then, simplify the numerator and denominator separately.

$\sqrt{25x^4}=5x^2$

and

$\sqrt{225y^{10}}=15y^5$ .

Now, you have

$\frac{5x^2}{15y^5}$ .

Simplify to get

$\frac{x^2}{3y^2}$ .

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Example Question #28 : Radicals And Fractions

Simplify, and ensure that no radicals remain in the denominator.

$\frac{1}{\sqrt{252}}$

Possible Answers:

$\frac{\sqrt{7}}{36}$

$\frac{\sqrt{7}}{42}$

None of these

$\frac{\sqrt{7}}{21}$

$\frac{\sqrt{7}}{252}$

Correct answer:

$\frac{\sqrt{7}}{42}$

Explanation:

$\frac{1}{\sqrt{252}}$

Moving radical from the denominator to the numerator:

$\frac{1}{\sqrt{252}}*\frac{\sqrt{252}}{\sqrt{252}}=\frac{\sqrt{252}}{252}$

Factoring:

$\frac{\sqrt{252}}{252}=\frac{\sqrt{9}*\sqrt{4}*\sqrt{7}}{252}$

Simplifying:

$\frac{\sqrt{9}*\sqrt{4}*\sqrt{7}}{252}=\frac{\sqrt{7}}{42}$

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Example Question #191 : Simplifying Radicals

Simplify the expression: $\frac{3+\sqrt2}{\sqrt2} +1$

Possible Answers:

$\frac{3\sqrt{6}+4}{2}$

$\frac{3\sqrt2+4}{2}$

$\frac{4\sqrt2+3}{2}$

$\frac{3\sqrt2+3}{2}$

$2\sqrt2+1$

Correct answer:

$\frac{3\sqrt2+4}{2}$

Explanation:

Rationalize the denominator for the first term.

$\frac{3+\sqrt2}{\sqrt2} \times \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2(3+\sqrt2)}{2} = \frac{3\sqrt2+2}{2}$

Adding one at the end of the expression mean that we are adding $\frac{2}{2}$ .

$\frac{3+\sqrt2}{\sqrt2} +1= \frac{3\sqrt2+2}{2}+ \frac{2}{2}$

Combine like-terms and form one fraction.

The answer is: $\frac{3\sqrt2+4}{2}$

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #13 : Radicals And Fractions

Example Question #281 : Radicals

Example Question #22 : Radicals And Fractions

Example Question #23 : Radicals And Fractions

Example Question #4181 : Algebra Ii

Example Question #25 : Radicals And Fractions

Example Question #21 : Radicals And Fractions

Example Question #22 : Radicals And Fractions

Example Question #28 : Radicals And Fractions

Example Question #191 : Simplifying Radicals

All Algebra II Resources

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