# 11.4: Multiplication and Division of Radicals

**Basic**Created by: CK-12

**Practice**Multiplication and Division of Radicals

What if you knew that the area of a rectangular mirror is \begin{align*}12 \sqrt{6}\end{align*}

### Multiplying and Dividing Radicals

#### Multiplying Radicals

To multiply radicals, the roots must be the same.

\begin{align*}\sqrt[n]{a} \cdot \sqrt[n]{b}= \sqrt[n]{ab}\end{align*}

#### Let's multiply the following radicals:

\begin{align*}\sqrt{3} \cdot \sqrt{12}\end{align*}

\begin{align*}\sqrt{3} \cdot \sqrt{12}=\sqrt{36}=6\end{align*}

#### Dividing Radicals

Dividing radicals is more complicated. A radical in the denominator of a fraction is not considered simplified by mathematicians. In order to simplify the fraction, you must **rationalize the denominator.**

To **rationalize the denominator** means to remove any radical signs from the denominator of the fraction using multiplication.

Remember that \begin{align*}\sqrt{a} \times \sqrt{a}= \sqrt{a^2}=a\end{align*}

#### Let's simplify \begin{align*}\frac{2}{\sqrt{3}}\end{align*}23√ using the property above:

We must clear the denominator of its radical using the property above. Remember, what you do to one piece of a fraction, you must do to all pieces of the fraction.

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

#### Now, let's solve the following real world problem:

A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square-feet. Find the dimensions of the pool and the area of the pool.

First, make a sketch.

Let \begin{align*}x=\end{align*}

The equation for the area of the pool is \begin{align*}Area=length \cdot width\end{align*}

The combined length of pool and walkway is \begin{align*}2x+2\end{align*}

The combined width of pool and walkway is \begin{align*}x+2\end{align*}

Therefore, the combined area of the pool and the walkway is:

\begin{align*}\text{Area}=(2x+2)(x+2)\end{align*}

Since the combined area of the pool and walkway is \begin{align*}400 \ ft^2\end{align*}

\begin{align*}(2x+2)(x+2)=400\end{align*}

Now, solve the equation:

\begin{align*}&& & (2x+2)(x+2)=400\\
& \text{Multiply in order to eliminate the parentheses}. && 2x^2+4x+2x+4=400\\
& \text{Collect like terms}. && 2x^2+6x+4=400\\
& \text{Move all terms to one side of the equation}. && 2x^2+6x-396=0\\
& \text{Divide all terms by} \ 2. && x^2+3x-198=0\end{align*}

\begin{align*}x & = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
& = \frac{-3 \pm \sqrt{3^2-4(1)(-198)}}{2(1)}\\
& = \frac{-3\pm \sqrt{801}}{2} \approx \frac{-3\pm 28.3}{2}\end{align*}

Use the quadratic formula. \begin{align*}x \approx 12.65\end{align*}

We can disregard the negative solution since it does not make sense in this context. Thus, we can check our answer of 12.65 by substituting the result into the area formula.

\begin{align*}\text{Area} = [2(12.65)+2)](12.65+2)=27.3 \cdot 14.65 \approx 400 \ ft^2.\end{align*}

The answer checks out.

Since the length is twice as long as the width, the pool is 12.65 by 25.3 feet. The area of the pool is 320.045.

### Examples

#### Example 1

Earlier, you were asked to find the length of a mirror with an area of \begin{align*}12 \sqrt{6}\end{align*}

You can find the length by plugging in the known values into the formula for area and solving.

\begin{align*}\text{Starting formula }&& Area= (length)(width)\\
\text{Plug in numbers }&& 12\sqrt{6}=(length)(2\sqrt{2})\\
\text{Divide by } 2\sqrt{2}&& \frac{12\sqrt{6}}{2\sqrt{2}}=length\\
\text{Rationalize the Denominator }&&\frac{12\sqrt{6}}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=length\\
\text{Simplify }&&\frac{12\sqrt{12}}{2\times 2}=length\\
\text{Simplify }&&\frac{12\sqrt{12}}{4}=length\\
\text{Simplify }&&3\sqrt{12}=length\\
\text{Simplify }&&3\times 2\sqrt{3}=length\\
\text{Simplify }&& 6\sqrt{3}=length\\\end{align*}

The length of the mirror is \begin{align*}6\sqrt{3}\end{align*}

Yes, if we knew the width and the length we could solve for the area. Instead of a division problem, it would be a multiplication problem with radicals.

#### Example 2

Simplify \begin{align*}\frac{7}{\sqrt[3]{5}}\end{align*}

In this case, we need to make the number inside the cube root a perfect cube. We need to multiply the numerator and the denominator by \begin{align*}\sqrt[3]{5^2}\end{align*}.

\begin{align*}\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}=\frac{7\sqrt[3]{25}}{\sqrt[3]{5^3}}=\frac{7\sqrt[3]{25}}{5}\end{align*}

### Review

Multiply the following expressions.

- \begin{align*}\sqrt{6}\left ( \sqrt{10} + \sqrt{8} \right )\end{align*}
- \begin{align*}\left ( \sqrt{a} - \sqrt{b} \right ) \left ( \sqrt{a} + \sqrt{b} \right )\end{align*}
- \begin{align*}\left ( 2\sqrt{x}+ 5 \right ) \left ( 2\sqrt{x}+5 \right )\end{align*}

Rationalize the denominator.

- \begin{align*}\frac{7}{\sqrt{15}}\end{align*}
- \begin{align*}\frac{9}{\sqrt{10}}\end{align*}
- \begin{align*}\frac{2x}{\sqrt{5}x}\end{align*}
- \begin{align*}\frac{\sqrt{5}}{\sqrt{3}y}\end{align*}
- The volume of a spherical balloon is \begin{align*}950 cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}=\frac{4}{3} \pi R^3\end{align*})
- A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 in^2\end{align*}, what is the width of the frame?
- The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 11.4.

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### Image Attributions

Here you'll learn how to multiply and divide by radicals, as well as how to rationalize denominators.

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