# Multiplication and Division of Radicals

## Rationalize the denominator

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Multiplying and Dividing Square Roots

The area of a rectangle is \begin{align*}\sqrt{30}\end{align*} and the length of the rectangle is \begin{align*}\sqrt{20}\end{align*}. What is the width of the rectangle?

### Dividing Square Roots

Division of radicals can be a bit more difficult than performing other operations. The main complication is that you cannot leave any radicals in the denominator of a fraction. For this reason we have to do something called rationalizing the denominator, where you multiply the top and bottom of a fraction by the same radical that is in the denominator. This will cancel out the radicals and leave a whole number.

4. \begin{align*}\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\end{align*}

5. \begin{align*}\frac{\sqrt{a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{\sqrt{ab}}{b}\end{align*}

Let's simplify the following problems by rationalizing the denominator.

1. Simplify \begin{align*}\sqrt{\frac{1}{4}}\end{align*}.

Break apart the radical by using Rule #4.

\begin{align*}\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}\end{align*}

1. Simplify \begin{align*}\frac{2}{\sqrt{3}}\end{align*}.

This might look simplified, but radicals cannot be in the denominator of a fraction. This means we need to apply Rule #5 to get rid of the radical in the denominator, or rationalize the denominator. Multiply the top and bottom of the fraction by \begin{align*}\sqrt{3}\end{align*}.

\begin{align*}\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\end{align*}

1. Simplify \begin{align*}\sqrt{\frac{32}{40}}\end{align*}.

Reduce the fraction, and then apply the rules above.

\begin{align*}\sqrt{\frac{32}{40}}= \sqrt{\frac{4}{5}}= \frac{\sqrt{4}}{\sqrt{5}}= \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the width of the rectangle.

Recall that the area of a rectangle equals the length times the width, so to find the width, we must divide the area by the length.

\begin{align*}\sqrt{\frac{30}{20}}\end{align*} = \begin{align*}\sqrt{\frac{3}{2}}\end{align*}.

Now we need to rationalize the denominator. Multiply the top and bottom of the fraction by \begin{align*}\sqrt{2}\end{align*}.

\begin{align*}\frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2}\end{align*}

Therefore, the width of the rectangle is \begin{align*}\frac{\sqrt{6}}{2}\end{align*}.

Simplify the following expressions using the Radical Rules you have learned.

#### Example 2

\begin{align*}\sqrt{\frac{1}{2}}\end{align*}

\begin{align*}\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{2}\end{align*}

#### Example 3

\begin{align*}\sqrt{\frac{64}{50}}\end{align*}

\begin{align*}\sqrt{\frac{64}{50}}=\sqrt{\frac{32}{25}}=\frac{\sqrt{16 \cdot 2}}{5}= \frac{4\sqrt{2}}{5}\end{align*}

#### Example 4

\begin{align*}\frac{4\sqrt{3}}{\sqrt{6}}\end{align*}

The only thing we can do is rationalize the denominator by multiplying the numerator and denominator by \begin{align*}\sqrt{6}\end{align*} and then simplify the fraction.

\begin{align*}\frac{4\sqrt{3}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}= \frac{4\sqrt{18}}{6}=\frac{4\sqrt{9 \cdot 2}}{6}= \frac{12\sqrt{2}}{6}=2\sqrt{2}\end{align*}

### Review

Simplify the following fractions.

1. \begin{align*}\sqrt{\frac{4}{25}}\end{align*}
2. \begin{align*}\text{-}\sqrt{\frac{16}{49}}\end{align*}
3. \begin{align*}\sqrt{\frac{96}{121}}\end{align*}
4. \begin{align*}\frac{5\sqrt{2}}{\sqrt{10}}\end{align*}
5. \begin{align*}\frac{6}{\sqrt{15}}\end{align*}
6. \begin{align*}\sqrt{\frac{60}{35}}\end{align*}
7. \begin{align*}8\frac{\sqrt{18}}{\sqrt{30}}\end{align*}
8. \begin{align*}\frac{12}{\sqrt{6}}\end{align*}
9. \begin{align*}\sqrt{\frac{208}{143}}\end{align*}
10. \begin{align*}\frac{21\sqrt{3}}{2\sqrt{14}}\end{align*}

Challenge Use all the Radical Rules you have learned to simplify the expressions.

1. \begin{align*}\sqrt{\frac{8}{12}} \cdot \sqrt{15}\end{align*}
2. \begin{align*}\sqrt{\frac{32}{45}} \cdot \frac{6\sqrt{20}}{\sqrt{5}}\end{align*}
3. \begin{align*}\frac{\sqrt{24}}{\sqrt{2}}+\frac{8\sqrt{26}}{\sqrt{8}}\end{align*}
4. \begin{align*}\frac{\sqrt{2}}{\sqrt{3}}+\frac{4\sqrt{6}}{\sqrt{3}}\end{align*}
5. \begin{align*}\frac{5\sqrt{5}}{\sqrt{12}}-\frac{2\sqrt{15}}{\sqrt{10}}\end{align*}

To see the Review answers, open this PDF file and look for section 5.6.

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