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5.6: Dividing Square Roots

Difficulty Level: At Grade Created by: CK-12
Practice Multiplication and Division of Radicals
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The area of a rectangle is \sqrt{30} . The length of the rectangle is \sqrt{20} . What is the width of the rectangle?

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Watch the first part of this video, until about 3:15.

Khan Academy: How to Rationalize a Denominator


Dividing radicals can be a bit more difficult that the 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.

Radical Rules

4. \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

5. \frac{\sqrt{a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{\sqrt{ab}}{b}

Example A

Simplify \sqrt{\frac{1}{4}} .

Solution: Break apart the radical by using Rule #4.


Example B

Simplify \frac{2}{\sqrt{3}} .

Solution: 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 \sqrt{3} .

\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Example C

Simplify \sqrt{\frac{32}{40}} .

Solution: Reduce the fraction, and then apply the rules above.

\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}

Intro Problem Revisit 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.

\sqrt{\frac{30}{20}} = \sqrt{\frac{3}{2}} .

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

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

Therefore, the width of the rectangle is \frac{\sqrt{6}}{2} .

Guided Practice

Simplify the following expressions using the Radical Rules learned in this concept and the previous concept.

1. \sqrt{\frac{1}{2}}

2. \sqrt{\frac{64}{50}}

3. \frac{4\sqrt{3}}{\sqrt{6}}


1. \sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{2}

2. \sqrt{\frac{64}{50}}=\sqrt{\frac{32}{25}}=\frac{\sqrt{16 \cdot 2}}{5}= \frac{4\sqrt{2}}{5}

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

\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}


Rationalize the denominator
The process used to get a radical out of the denominator of a fraction.


Simplify the following fractions.

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

Challenge Use all the Radical Rules you have learned in the last two oncepts to simplify the expressions.

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

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Difficulty Level:

At Grade


Date Created:

Mar 12, 2013

Last Modified:

Dec 16, 2014
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