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Expressions with Radicals

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What if you were asked to find the sum of \sqrt{32} and 3\sqrt{8} ? How could you combine these two terms so that you could add them? After completing this lesson, you'll be able to simplify radical terms and expressions like these.

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CK-12 Foundation: Simplifying Radicals Review

Guidance

In algebra, you learned how to simplify radicals. Let’s review it here. Some key points to remember:

  1. One way to simplify a radical is to factor out the perfect squares (see Example A).
  2. When adding radicals, you can only combine radicals with the same number underneath it. For example, 2 \sqrt{5} + 3 \sqrt{6} cannot be combined, because 5 and 6 are not the same number (see Example B).
  3. To multiply two radicals, multiply what is under the radicals and what is in front (see Example B).
  4. To divide radicals, you need to simplify the denominator, which means multiplying the top and bottom of the fraction by the radical in the denominator (see Example C).

Example A

Simplify the radicals.

a) \sqrt{50}

b) \sqrt{27}

c) \sqrt{272}

Answers: For each radical, find the square number(s) that are factors.

a) \sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2}

b) \sqrt{27} = \sqrt{9 \cdot 3} = 3 \sqrt{3}

c) \sqrt{272} = \sqrt{16 \cdot 17} = 4 \sqrt{17}

Example B

Simplify the radicals.

a) 2 \sqrt{10} + \sqrt{160}

b) 5 \sqrt{6} \cdot 4 \sqrt{18}

c) \sqrt{8} \cdot 12 \sqrt{2}

d) \left( 5 \sqrt{2} \right)^2

Answers:

a) Simplify \sqrt{160} before adding: 2 \sqrt{10} + \sqrt{160} = 2 \sqrt{10} + \sqrt{16 \cdot 10} = 2 \sqrt{10} + 4 \sqrt{10} = 6 \sqrt{10}

b) 5 \sqrt{6} \cdot 4 \sqrt{18} = 5 \cdot 4 \sqrt{6 \cdot 18} = 20 \sqrt{108} = 20 \sqrt{36 \cdot 3} = 20 \cdot 6 \sqrt{3} = 120 \sqrt{3}

c) \sqrt{8} \cdot 12 \sqrt{2} = 12 \sqrt{8 \cdot 2} = 12 \sqrt{16} = 12 \cdot 4=48

d) \left( 5 \sqrt{2} \right )^2 = 5^2 \left( \sqrt{2} \right )^2 = 25 \cdot 2 = 50 \rightarrow the \sqrt{} and the ^2 cancel each other out

Example C

Divide and simplify the radicals.

a) 4 \sqrt{6} \div \sqrt{3}

b) \frac{\sqrt{30}}{\sqrt{8}}

c) \frac{8 \sqrt{2}}{6 \sqrt{7}}

Answers: Rewrite all division problems like a fraction.

a)

b) \frac{\sqrt{30}}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{\sqrt{240}}{\sqrt{64}} = \frac{\sqrt{16 \cdot 15}}{8} = \frac{4 \sqrt{15}}{8} = \frac{\sqrt{15}}{2}

c) \frac{8 \sqrt{2}}{6 \sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{8 \sqrt{14}}{6 \cdot 7} = \frac{4 \sqrt{14}}{3 \cdot 7} = \frac{4 \sqrt{14}}{21}

Notice, we do not really “divide” radicals, but get them out of the denominator of a fraction.

CK-12 Foundation: Simplifying Radicals Review

Guided Practice

Simplify the radicals.

1. \sqrt{75}

2. 2\sqrt{5} + 3\sqrt{80}

3. \frac{\sqrt{45}}{\sqrt{2}}

Answers:

1. \sqrt{75}=\sqrt{25\cdot 3}=5\sqrt{3}

2. 2\sqrt{5}+3\sqrt{80}=2\sqrt{5}+3(\sqrt{16\cdot 5})=2\sqrt{5}+(3\cdot 4)\sqrt{5}=14\sqrt{5}

3. \frac{\sqrt{45}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{90}}{\sqrt{4}} = \frac{\sqrt{9 \cdot 10}}{2} = \frac{3 \sqrt{10}}{2}

Practice

Simplify the radicals.

  1. \sqrt{48}
  2. 2 \sqrt{5} + \sqrt{20}
  3. \sqrt{24}
  4. \left( 6 \sqrt{3} \right)^2
  5. 8 \sqrt{8} \cdot \sqrt{10}
  6. \left( 2 \sqrt{30} \right )^2
  7. \sqrt{320}
  8. \frac{4 \sqrt{5}}{\sqrt{6}}
  9. \frac{12}{\sqrt{10}}
  10. \frac{21 \sqrt{5}}{9 \sqrt{15}}

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