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Algebra II : Multiplying and Dividing Radicals

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Example Questions

Example Question #1 : Multiplying And Dividing Radicals

Multiply and express the answer in the simplest form:

$\small \sqrt3(\sqrt4-\sqrt3)$ $\displaystyle \small \sqrt3(\sqrt4-\sqrt3)$

Possible Answers:

$\small 4\sqrt3-9$ $\displaystyle \small 4\sqrt3-9$

$\small \sqrt{12}-\sqrt9$ $\displaystyle \small \sqrt{12}-\sqrt9$

$\small 2\sqrt3-3$ $\displaystyle \small 2\sqrt3-3$

$\small \sqrt7-\sqrt6$ $\displaystyle \small \sqrt7-\sqrt6$

Correct answer:

$\small 2\sqrt3-3$ $\displaystyle \small 2\sqrt3-3$

Explanation:

$\small \sqrt3(\sqrt4-\sqrt3)$ $\displaystyle \small \sqrt3(\sqrt4-\sqrt3)$

$\small \sqrt{12}-\sqrt9$ $\displaystyle \small \sqrt{12}-\sqrt9$

$\small \sqrt{12}=\sqrt4 \times \sqrt3=2\sqrt3$ $\displaystyle \small \sqrt{12}=\sqrt4 \times \sqrt3=2\sqrt3$

$\small \sqrt9=3$ $\displaystyle \small \sqrt9=3$

$\small \sqrt{12}-\sqrt9=2\sqrt3-3$ $\displaystyle \small \sqrt{12}-\sqrt9=2\sqrt3-3$

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Example Question #2 : Multiplying And Dividing Radicals

$\left ( \sqrt{x}+\sqrt{6} \right )\left ( \sqrt{x}-\sqrt{6} \right )$ $\displaystyle \left ( \sqrt{x}+\sqrt{6} \right )\left ( \sqrt{x}-\sqrt{6} \right )$

Possible Answers:

x-6 $\displaystyle x-6$

$x^{2}-36$ $\displaystyle x^{2}-36$

$x-\sqrt{12}$ $\displaystyle x-\sqrt{12}$

$x-\sqrt{6}$ $\displaystyle x-\sqrt{6}$

$\sqrt{x}-6$ $\displaystyle \sqrt{x}-6$

Correct answer:

x-6 $\displaystyle x-6$

Explanation:

FOIL with difference of squares. The multiplying cancels the square roots on both terms.

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Example Question #3 : Multiplying And Dividing Radicals

Simplify.

$\sqrt{36} \times 3\sqrt{36}$ $\displaystyle \sqrt{36} \times 3\sqrt{36}$

Possible Answers:

$\displaystyle 18$

$4\sqrt{36}$ $\displaystyle 4\sqrt{36}$

108 $\displaystyle 108$

$3\sqrt{36}$ $\displaystyle 3\sqrt{36}$

Correct answer:

108 $\displaystyle 108$

Explanation:

We can solve this by simplifying the radicals first: $\sqrt{36} =6$ $\displaystyle \sqrt{36} =6$

Plugging this into the equation gives us:

$6 \times3(6)=108$ $\displaystyle 6 \times3(6)=108$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify.

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

Possible Answers:

$\displaystyle 40$

$\displaystyle 20$

$\sqrt{80}$ $\displaystyle \sqrt{80}$

$20\sqrt{2}$ $\displaystyle 20\sqrt{2}$

Correct answer:

$\displaystyle 40$

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 $\sqrt{4} =2$ $\displaystyle \sqrt{4} =2$

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

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

$5\times4\times2=40$ $\displaystyle 5\times4\times2=40$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify.

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

Possible Answers:

$\frac{4}{12}$ $\displaystyle \frac{4}{12}$

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

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

$\sqrt{2}$ $\displaystyle \sqrt{2}$

Correct answer:

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

Explanation:

We can simplify the radicals:

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

Plug in the simplifed radicals into the equation:

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

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

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Example Question #2 : Multiplying And Dividing Radicals

Simplify and rationalize the denominator if needed,

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

Possible Answers:

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

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\sqrt{2}$ $\displaystyle \sqrt{2}$

$\frac{\sqrt2}{2}$ $\displaystyle \frac{\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2}$ $\displaystyle \frac{\sqrt2}{2}$

Explanation:

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

Plugging in the simplifed radical into the equation we get:

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

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

Now we want to rationalize the denominator,

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

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Example Question #2 : Multiplying And Dividing Radicals

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

Possible Answers:

$6x^{30}$ $\displaystyle 6x^{30}$

$6x^{10}$ $\displaystyle 6x^{10}$

$6x^{13}$ $\displaystyle 6x^{13}$

$5x^{30}$ $\displaystyle 5x^{30}$

$5x^{10}$ $\displaystyle 5x^{10}$

Correct answer:

$6x^{10}$ $\displaystyle 6x^{10}$

Explanation:

To simplify, you must use the Law of Exponents.

First you must multiply the coefficients then add the exponents:

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

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Example Question #1 : Multiplying And Dividing Radicals

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

Possible Answers:

$18\sqrt{2}$ $\displaystyle 18\sqrt{2}$

$12\sqrt{18}$ $\displaystyle 12\sqrt{18}$

$6\sqrt{18}$ $\displaystyle 6\sqrt{18}$

$3\sqrt{72}$ $\displaystyle 3\sqrt{72}$

$9\sqrt{24}$ $\displaystyle 9\sqrt{24}$

Correct answer:

$18\sqrt{2}$ $\displaystyle 18\sqrt{2}$

Explanation:

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

Then set up the multiplication problem:

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

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

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

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

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

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Example Question #2 : Multiplying And Dividing Radicals

Simplify

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

Possible Answers:

$\sqrt{13}$ $\displaystyle \sqrt{13}$

$\displaystyle 3$

$3\sqrt{13}$ $\displaystyle 3\sqrt{13}$

$\sqrt9$ $\displaystyle \sqrt9$

$3\sqrt9$ $\displaystyle 3\sqrt9$

Correct answer:

$\displaystyle 3$

Explanation:

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

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

Then simplify this radical:

$\sqrt9=3$ $\displaystyle \sqrt9=3$ .

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Example Question #1 : Multiplying And Dividing Radicals

Solve and simplify.

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

Possible Answers:

$\sqrt{5}$ $\displaystyle \sqrt{5}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

$\sqrt{23}$ $\displaystyle \sqrt{23}$

$\displaystyle 6$

$\sqrt{7}$ $\displaystyle \sqrt{7}$

Correct answer:

$\sqrt{6}$ $\displaystyle \sqrt{6}$

Explanation:

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

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

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

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Algebra II : Multiplying and Dividing Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #3 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #1 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

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