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Algebra II : Radicals

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #3 : Multiplying And Dividing Radicals

Simplify.

$5\sqrt{2} \times4\sqrt{2}$

Possible Answers:

$20\sqrt{2}$

$\sqrt{80}$

Correct answer:

Explanation:

Note: the product of the radicals is the same as the radical of the product:

$\sqrt{2} \times \sqrt2} = \sqrt{2\times 2$ which is $\sqrt{4} =2$

Once we understand this, we can plug it into the equation:

$5\sqrt{2} \times4\sqrt{2} = 5\times4\times \sqrt{2\times 2}$

$5\times4\times2=40$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify.

$\frac{\sqrt{32}}{3\sqrt{16}}$

Possible Answers:

$\frac{4}{12}$

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

$\frac{4\sqrt{2}}{12}$

$\sqrt{2}$

Correct answer:

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

Explanation:

We can simplify the radicals:

$\sqrt{32} =\sqrt{16\times 2} =4\sqrt{2}$ and $\sqrt{16} =4$

Plug in the simplifed radicals into the equation:

$\frac{4\sqrt{2}}{3\times 4} = \frac{4\sqrt{2}}{12}$

$\frac{4 \sqrt2}{12}=\frac{\sqrt2}{3}$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify and rationalize the denominator if needed,

$\frac{\sqrt{16}}{4\sqrt{2}}$

Possible Answers:

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

$\frac{1}{2}$

$\sqrt{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

We can only simplify the radical in the numerator: $\sqrt{16} =4$

Plugging in the simplifed radical into the equation we get:

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

Note: We simplified further because both the numerator and denominator had a "4" which canceled out.

Now we want to rationalize the denominator,

$\frac{1}{\sqrt2}\bigg(\frac{\sqrt2}{\sqrt2}\bigg)=\frac{\sqrt2}{2}$

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Example Question #201 : Radicals

Simplify $2x^{2}\cdot x^{5}\cdot 3x^{3}$

Possible Answers:

$5x^{10}$

$6x^{13}$

$6x^{10}$

$6x^{30}$

$5x^{30}$

Correct answer:

$6x^{10}$

Explanation:

To simplify, you must use the Law of Exponents.

First you must multiply the coefficients then add the exponents:

$2\cdot3\cdot x^{2+5+3}=6\cdot x^{10}=6x^{10}$ .

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Example Question #332 : College Algebra

What is the product of $\sqrt{12}$ and $3\sqrt{6}$ ?

Possible Answers:

$9\sqrt{24}$

$6\sqrt{18}$

$12\sqrt{18}$

$3\sqrt{72}$

$18\sqrt{2}$

Correct answer:

$18\sqrt{2}$

Explanation:

First, simplify $\sqrt{12}=\sqrt{4\cdot3}$ to $2\sqrt{3}$ .

Then set up the multiplication problem:

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

Multiply the terms outside of the radical, then the terms under the radical:

$2\cdot 3\sqrt{3\cdot6}$ then simplfy: $6\sqrt{18}.$

The radical is still not in its simplest form and must be reduced further:

$6\sqrt{18}=6\cdot3\sqrt{2}=18\sqrt{2}$ . This is the radical in its simplest form.

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Example Question #1 : Multiplying And Dividing Radicals

Simplify

${}\frac{\sqrt{117}}{\sqrt{13}}$

Possible Answers:

$\sqrt9$

$\sqrt{13}$

$3\sqrt9$

$3\sqrt{13}$

Correct answer:

Explanation:

To divide the radicals, simply divide the numbers under the radical and leave them under the radical:

$\sqrt{\frac{117}{13}}=\sqrt9$

Then simplify this radical:

$\sqrt9=3$ .

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Example Question #202 : Radicals

Solve and simplify.

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

Possible Answers:

$\sqrt{6}$

$\sqrt{7}$

$\sqrt{5}$

$\sqrt{23}$

Correct answer:

$\sqrt{6}$

Explanation:

When multiplying radicals, just take the values inside the radicand and perfom the operation.

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

$\sqrt{6}$ can't be reduced so this is the final answer.

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Example Question #11 : Multiplying And Dividing Radicals

Solve and simplify.

$\sqrt{9}\cdot \sqrt{3}$

Possible Answers:

$\sqrt{27}$

$2\sqrt{3}$

$3\sqrt{3}$

$\sqrt{3}$

Correct answer:

$3\sqrt{3}$

Explanation:

When multiplying radicals, just take the values inside the radicand and perfom the operation.

In this case, we have a perfect square so simplify that first.

Then, take that answer and multiply that with $\sqrt{3}$ to get the final answer.

${\sqrt{9}}\cdot \sqrt{3}=3\cdot \sqrt{3}=3\sqrt{3}$ .

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Example Question #111 : Simplifying Radicals

Solve and simplify.

$\frac{\sqrt{5}}{\sqrt{3}}$

Possible Answers:

${\sqrt{15}}$

$\frac{5}{3}$

$\frac{\sqrt{3}}{5}$

$\frac{\sqrt{15}}{3}$

Correct answer:

$\frac{\sqrt{15}}{3}$

Explanation:

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed. Since there is a radical present, we need to eliminate that radical. To do this, we multiply both top and bottom by $\sqrt{3}$ . The reason is because we want a whole number in the denominator and multiplying by itself will achieve that. By multiplying itself, it creates a square number which can be reduced to .

$\frac{\sqrt{5}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{\sqrt{9}}=\frac{\sqrt{15}}{3}$

With the denominator being , the numerator is $\sqrt{5\cdot 3}=\sqrt{15}$ . Final answer is $\frac{\sqrt{15}}{3}$ .

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Example Question #112 : Simplifying Radicals

Solve and simplify.

$\frac{\sqrt{9}}{\sqrt{4}}$

Possible Answers:

$\frac{9}{4}$

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

$\frac{3}{2}$

$\sqrt{13}$

Correct answer:

$\frac{3}{2}$

Explanation:

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed.

Both and are perfect squares so they can be simplify.

Final answer is

$\frac{\sqrt9}{\sqrt4}=\frac{\sqrt{3\cdot 3}}{\sqrt{2\cdot 2}}=\frac{3}{2}$ .

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Algebra II : Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #201 : Radicals

Example Question #332 : College Algebra

Example Question #1 : Multiplying And Dividing Radicals

Example Question #202 : Radicals

Example Question #11 : Multiplying And Dividing Radicals

Example Question #111 : Simplifying Radicals

Example Question #112 : Simplifying Radicals

All Algebra II Resources

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