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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #4161 : Algebra Ii

Evaluate: $\frac{\sqrt{3}}{\sqrt7}\cdot \frac{\sqrt{14}}{\sqrt{18}}$

Possible Answers:

$\frac{\sqrt3}{3}$

$\sqrt6$

$\sqrt3$

$\frac{1}{3}$

Correct answer:

$\frac{\sqrt3}{3}$

Explanation:

Since all terms are in radicals, we can simplify the terms by using common factors.

$\frac{\sqrt{3}}{\sqrt7}\cdot \frac{\sqrt{14}}{\sqrt{18}} = \frac{\sqrt{3}}{\sqrt7}\cdot \frac{\sqrt{2}\cdot \sqrt{7}}{\sqrt{3}\cdot \sqrt{6}} =\frac{\sqrt{2}}{\sqrt{6}}=\frac{1}{\sqrt3}$

Rationalize the denominator.

$\frac{1}{\sqrt3}\cdot \frac{\sqrt3}{\sqrt3} =\frac{\sqrt3}{3}$

The answer is: $\frac{\sqrt3}{3}$

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Example Question #4162 : Algebra Ii

Multiply: $\sqrt{50}\cdot \sqrt{18}\cdot \sqrt8$

Possible Answers:

$30\sqrt3$

$30\sqrt2$

$60\sqrt{2}$

$45\sqrt{3}$

$15\sqrt{2}$

Correct answer:

$60\sqrt{2}$

Explanation:

We can simplify the radicals before expanding by multiplication.

$\sqrt{50}\cdot \sqrt{18}\cdot \sqrt8 = [\sqrt{25}\cdot \sqrt2]\cdot [\sqrt{9}\cdot \sqrt{2}]\cdot [\sqrt{4}\cdot \sqrt2]$

Simplify the radicals.

$5\sqrt2 \cdot 3\sqrt2 \cdot 2\sqrt2$

Multiply the integers together. When a square root of a number is multiplied by itself, the radical will be eliminated, giving only the integer.

$30\cdot 2 \cdot \sqrt{2}$

The answer is: $60\sqrt{2}$

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Example Question #163 : Simplifying Radicals

$\frac{5+3\sqrt{2}}{-3-\sqrt{5}}=?$

Possible Answers:

$\frac{15-5\sqrt{5}-9\sqrt{2}+3\sqrt{10}}{4}$

$\frac{-15-5\sqrt{5}+9\sqrt{2}-3\sqrt{10}}{4}$

$\frac{-15-42\sqrt{10}}{4}$

$\frac{-15+5\sqrt{5}-9\sqrt{2}+3\sqrt{10}}{4}$

$\frac{-15+8\sqrt{5}-6\sqrt{2}}{4}$

Correct answer:

$\frac{-15+5\sqrt{5}-9\sqrt{2}+3\sqrt{10}}{4}$

Explanation:

To solve this expression, multiply the numerator and the denominator by the complex conjugate of the denominator. Since the denominator is $-3-\sqrt{5}$ , the complex conjugate of this is $-3+\sqrt{5}$ . Therefore:

$\frac{5+3\sqrt{2}}{-3-\sqrt{5}}$

$=\frac{(5+3\sqrt{2})(-3+\sqrt{5})}{(-3-\sqrt{5})(-3+\sqrt{5})}$

$=\frac{-15+5\sqrt{5}-9\sqrt{2}+3\sqrt{10}}{9-5}$

$=\frac{-15+5\sqrt{5}-9\sqrt{2}+3\sqrt{10}}{4}$

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Example Question #1 : Radicals And Fractions

Simplify the following expression:

$\frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}$

Possible Answers:

$-\frac{1}{4}$

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

$\frac{1}{4}$

$\sqrt{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

First, let's see how we can combine these two fractions. Remember the following relationships:

$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$

and

$\sqrt{a}\sqrt{b}=\sqrt{ab}$

Now, let's look at our problem. Let's first try and turn the first term into one big radical:

$\frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}$

$\frac{\sqrt{27}}{\sqrt{4}}\cdot \sqrt{\frac{1}{108}}$

$\sqrt{\frac{27}{4}}\cdot \sqrt{\frac{1}{108}}$

Great! We've used the first relationship; now let's combine the two radicals using the second relationship.

$\sqrt{\frac{27}{4}\cdot\frac{1}{108}}=\sqrt{\frac{27}{4\cdot 108}}$

I haven't multiplied out anything yet because I want to see if there's any simplifying I can do BEFORE I multiply. In this case, I ask myself: Does the denominator contain any factors of 27 (3, 9, 27)?

I know 108 is divisible by 9 because its digits add up to a number that's divisible by 9. 27 is divisible by 9 too, so I can rewrite it this way:

$\sqrt{\frac{3\cdot \mathbf{9}}{4\cdot \mathbf{9}\cdot 12}}=\sqrt{\frac{3}{4\cdot 12}}$

We also know that 3 is a factor of 12:

$\sqrt{\frac{\mathbf{3}}{4\cdot 4\cdot \mathbf{3}}}=\sqrt{\frac{1}{4\cdot 4}}$

Now, after simplifying the fraction, we have to simplify the radical. Keep this in mind:

$\sqrt{a\cdot a}=\sqrt{a^{2}}=a$

We can finally simplify this expression completely:

$\sqrt{\frac{1}{4\cdot 4}}=\frac{\sqrt{1}}{\sqrt{4\cdot 4}}=\frac{1}{4}$

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Example Question #1 : Radicals And Fractions

Rationalize the denominator.

$\frac{x^{2}}{\sqrt{11}}$

Possible Answers:

$\frac{x^{2}}{11}$

$\frac{x^{4}\sqrt{11}}{11}$

$\frac{x^{2}\sqrt{11}}{\sqrt{11}}$

$11x^{4}$

$\frac{x^{2}\sqrt{11}}{11}$

Correct answer:

$\frac{x^{2}\sqrt{11}}{11}$

Explanation:

In order to rationalize the denominator we must eliminate the root in the denominator.

To do this, we multiply the radical by $\frac{\sqrt{11}}{\sqrt{11}}$ ,

$\frac{x^{2}}{\sqrt{11}}\cdot \frac{\sqrt{11}}{\sqrt{11}}=\frac{x^{2}\sqrt{11}}{(\sqrt{11})^{2}}=\frac{x^{2}\sqrt{11}}{11}$

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Example Question #1 : Radicals And Fractions

Simplify by rationalizing the denominator:

$\frac{14}{8 - \sqrt{15}}$

Possible Answers:

$\frac{16+2 \sqrt{15}}{7}$

None of the other responses is correct.

$\frac{16+14 \sqrt{15}}{7}$

$\frac{16-14 \sqrt{15}}{7}$

$\frac{16-2 \sqrt{15}}{7}$

Correct answer:

$\frac{16+2 \sqrt{15}}{7}$

Explanation:

Multiply the numerator and the denominator by the conjugate of the denominator, which is $8 + \sqrt{15}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{14}{8 - \sqrt{15}}$

$= \frac{14(8 + \sqrt{15})}{\left (8 - \sqrt{15} \right )(8 + \sqrt{15})}$

$= \frac{14(8 + \sqrt{15})}{8^{2} - \left ( \sqrt{15} \right )^{2} }$

$= \frac{14(8 + \sqrt{15})}{64- 15}$

$= \frac{14(8 + \sqrt{15})}{49}$

$= \frac{2(8 + \sqrt{15})}{7}$

$= \frac{16+2 \sqrt{15}}{7}$

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Example Question #1 : Radicals And Fractions

Simplify:

$\sqrt{\frac{9}{16}}$

Possible Answers:

$\frac{3}{8}$

$\frac{\sqrt{9}}{8}$

$\frac{3}{4}$

$\frac{3}{16}$

Correct answer:

$\frac{3}{4}$

Explanation:

We can take the square roots of the numerator and denominator separately. Thus, we get:

$\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$

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Example Question #5 : Radicals And Fractions

Simplify the following:

$\frac{4\sqrt{5}}{\sqrt{20}}$

Possible Answers:

$\sqrt{5}$

$\pm2$

$\frac{\sqrt{4}}{5}$

$\frac{4\sqrt{5}}{5}$

$\frac{1}{4}$

Correct answer:

$\pm2$

Explanation:

You can begin by rewriting this equation as:

$4 * \frac{\sqrt{5}}{\sqrt{20}}=4 * \sqrt{\frac{5}{20}} = 4 * \sqrt{\frac{1}{4}}=\frac{4}{\sqrt{4}}$

Now, you need to rationalize the denominator. To do this, multiply both top and bottom by $\sqrt{4}$ :

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

Then, cancel the common :

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

Since is a perfect square you can take the square root to get the simplified answer.

$\sqrt4=\pm 2$

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Example Question #1 : Radicals And Fractions

Simplify $\frac{\sqrt 32}{\sqrt2}$ .

Possible Answers:

$\frac{4}{\sqrt2}$

$4\sqrt2$

$\pm 4$

None of the other answers.

Correct answer:

$\pm 4$

Explanation:

Understanding properties of radicals will help you quickly solve this problem. When two radicals are multiplied or divided, you can simply combine the two radicals. For instance: $\sqrt{x}\cdot \sqrt{y} = \sqrt {xy}$

For this equation:

$\sqrt{\frac{32}{2}} = \sqrt{16} = \pm4$

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Example Question #1 : Radicals And Fractions

Solve and simplify.

$(3+\sqrt{5})^2$

Possible Answers:

$6+\sqrt{5}$

$9+\sqrt{5}$

$6\sqrt{5}$

$14+6\sqrt{5}$

Correct answer:

$14+6\sqrt{5}$

Explanation:

When foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables.

$(3+\sqrt{5})(3+\sqrt{5})$

$=9+3\sqrt{5}+3\sqrt{5}+5$

Combining like terms we get our final answer as follows.

$=14+6\sqrt{5}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4161 : Algebra Ii

Example Question #4162 : Algebra Ii

Example Question #163 : Simplifying Radicals

Example Question #1 : Radicals And Fractions

Example Question #1 : Radicals And Fractions

Example Question #1 : Radicals And Fractions

Example Question #1 : Radicals And Fractions

Example Question #5 : Radicals And Fractions

Example Question #1 : Radicals And Fractions

Example Question #1 : Radicals And Fractions

All Algebra II Resources

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