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# 11.4: Multiplication and Division of Radicals

Difficulty Level: Basic Created by: CK-12
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Practice Multiplication and Division of Radicals

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What if you knew that the area of a rectangular mirror is 126\begin{align*}12 \sqrt{6}\end{align*} square feet and that the width of the mirror is 22\begin{align*}2 \sqrt{2}\end{align*} feet? Could you find the length of the mirror? What operation would you have to perform? If you knew the width and the length of the mirror, could you find its area? What operation would you perform in this case? In this Concept, you'll learn about multiplying and dividing radicals so that you can answer questions like these.

### Guidance

To multiply radicands, the roots must be the same.

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

#### Example A

Simplify 312\begin{align*}\sqrt{3} \cdot \sqrt{12}\end{align*}.

Solution:

312=36=6\begin{align*}\sqrt{3} \cdot \sqrt{12}=\sqrt{36}=6\end{align*}

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: a×a=a2=a\begin{align*}\sqrt{a} \times \sqrt{a}= \sqrt{a^2}=a\end{align*}

#### Example B

Simplify 23\begin{align*}\frac{2}{\sqrt{3}}\end{align*}.

Solution:

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.

23×33=2332=233\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*}

#### Example C

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.

Solution:

1. Make a sketch.
2. Let x=\begin{align*}x=\end{align*} the width of the pool.
3. Write an equation. Area=lengthwidth\begin{align*}Area=length \cdot width\end{align*}

Combined length of pool and walkway =2x+2\begin{align*}=2x+2\end{align*}

Combined width of pool and walkway =x+2\begin{align*}=x+2\end{align*}

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

Since the combined area of the pool and walkway is 400 ft2\begin{align*}400 \ ft^2\end{align*}, we can write the equation.

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

4. Solve the equation:

Multiply in order to eliminate the parentheses.Collect like terms.Move all terms to one side of the equation.Divide all terms by 2.(2x+2)(x+2)=4002x2+4x+2x+4=4002x2+6x+4=4002x2+6x396=0x2+3x198=0\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*}

x=b±b24ac2a=3±324(1)(198)2(1)=3±80123±28.32\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. x12.65\begin{align*}x \approx 12.65\end{align*} or –15.65 feet

5. 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.

Area=[2(12.65)+2)](12.65+2)=27.314.65400 ft2.\begin{align*}\text{Area} = [2(12.65)+2)](12.65+2)=27.3 \cdot 14.65 \approx 400 \ ft^2.\end{align*}

### Guided Practice

Simplify 753\begin{align*}\frac{7}{\sqrt[3]{5}}\end{align*}.

Solution:

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 523\begin{align*}\sqrt[3]{5^2}\end{align*}.

753523523=7253533=72535\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*}

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both

Multiply the following expressions.

1. 6(10+8)\begin{align*}\sqrt{6}\left ( \sqrt{10} + \sqrt{8} \right )\end{align*}
2. (ab)(a+b)\begin{align*}\left ( \sqrt{a} - \sqrt{b} \right ) \left ( \sqrt{a} + \sqrt{b} \right )\end{align*}
3. (2x+5)(2x+5)\begin{align*}\left ( 2\sqrt{x}+ 5 \right ) \left ( 2\sqrt{x}+5 \right )\end{align*}

Rationalize the denominator.

1. 715\begin{align*}\frac{7}{\sqrt{15}}\end{align*}
2. 910\begin{align*}\frac{9}{\sqrt{10}}\end{align*}
3. 2x5x\begin{align*}\frac{2x}{\sqrt{5}x}\end{align*}
4. 53y\begin{align*}\frac{\sqrt{5}}{\sqrt{3}y}\end{align*}
5. The volume of a spherical balloon is 950cm3\begin{align*}950 cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere =43πR3\begin{align*}=\frac{4}{3} \pi R^3\end{align*})
6. 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 180in2\begin{align*}180 in^2\end{align*}, what is the width of the frame?
7. The volume of a soda can is 355 cm3\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.

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### Vocabulary Language: English Spanish

TermDefinition
rationalize the denominator To remove any radical signs from the denominator of the fraction using multiplication.

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