Simplifying Radicals by Rationalizing worksheet with exercises for rationalizing denominators in algebra.

Let's solve the problems in this worksheet step by step. The goal is to rationalize the denominator — that means eliminating radicals (square roots) from the denominator.

We'll go through Section A and Section B, explaining each step.

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🔷 Section A: Rationalize the denominator

We use the rule:
- Multiply numerator and denominator by the conjugate (if it's a binomial) or by the radical itself (if it's a single radical).

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#### 1) $ \frac{3}{\sqrt{2}} $

Multiply numerator and denominator by $ \sqrt{2} $:

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

✔ Answer: $ \boxed{\frac{3\sqrt{2}}{2}} $

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#### 2) $ \frac{11}{\sqrt{5}} $

Multiply by $ \frac{\sqrt{5}}{\sqrt{5}} $:

$$
\frac{11}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{11\sqrt{5}}{5}
$$

✔ Answer: $ \boxed{\frac{11\sqrt{5}}{5}} $

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#### 3) $ \frac{9}{\sqrt{7}} $

Multiply by $ \frac{\sqrt{7}}{\sqrt{7}} $:

$$
\frac{9}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{9\sqrt{7}}{7}
$$

✔ Answer: $ \boxed{\frac{9\sqrt{7}}{7}} $

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#### 4) $ \frac{4}{\sqrt{11}} $

Multiply by $ \frac{\sqrt{11}}{\sqrt{11}} $:

$$
\frac{4}{\sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}} = \frac{4\sqrt{11}}{11}
$$

✔ Answer: $ \boxed{\frac{4\sqrt{11}}{11}} $

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#### 5) $ \frac{26}{\sqrt{13}} $

Multiply by $ \frac{\sqrt{13}}{\sqrt{13}} $:

$$
\frac{26}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = \frac{26\sqrt{13}}{13} = 2\sqrt{13}
$$

✔ Answer: $ \boxed{2\sqrt{13}} $

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#### 6) $ \frac{15}{\sqrt{3}} $

Multiply by $ \frac{\sqrt{3}}{\sqrt{3}} $:

$$
\frac{15}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3}
$$

✔ Answer: $ \boxed{5\sqrt{3}} $

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#### 7) $ \frac{14}{\sqrt{2}} $

Multiply by $ \frac{\sqrt{2}}{\sqrt{2}} $:

$$
\frac{14}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{14\sqrt{2}}{2} = 7\sqrt{2}
$$

✔ Answer: $ \boxed{7\sqrt{2}} $

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#### 8) $ \frac{20}{\sqrt{6}} $

Multiply by $ \frac{\sqrt{6}}{\sqrt{6}} $:

$$
\frac{20}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{20\sqrt{6}}{6} = \frac{10\sqrt{6}}{3}
$$

✔ Answer: $ \boxed{\frac{10\sqrt{6}}{3}} $

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#### 9) $ \frac{8}{\sqrt{6}} $

Same as above:

$$
\frac{8}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{8\sqrt{6}}{6} = \frac{4\sqrt{6}}{3}
$$

✔ Answer: $ \boxed{\frac{4\sqrt{6}}{3}} $

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#### 10) $ \frac{8}{\sqrt{6}} $ → Already done: $ \boxed{\frac{4\sqrt{6}}{3}} $

Wait! Problem 9 and 10 are both $ \frac{8}{\sqrt{6}} $? Let's check...

Actually, problem 10 is $ \frac{8}{\sqrt{6}} $, same as #9. Probably a typo? But we'll assume it's correct.

So #10: $ \boxed{\frac{4\sqrt{6}}{3}} $

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#### 11) $ \frac{28}{3\sqrt{7}} $

Multiply numerator and denominator by $ \sqrt{7} $:

$$
\frac{28}{3\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{28\sqrt{7}}{3 \cdot 7} = \frac{28\sqrt{7}}{21} = \frac{4\sqrt{7}}{3}
$$

✔ Answer: $ \boxed{\frac{4\sqrt{7}}{3}} $

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#### 12) $ \frac{18}{5\sqrt{3}} $

Multiply by $ \frac{\sqrt{3}}{\sqrt{3}} $:

$$
\frac{18}{5\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{18\sqrt{3}}{5 \cdot 3} = \frac{18\sqrt{3}}{15} = \frac{6\sqrt{3}}{5}
$$

✔ Answer: $ \boxed{\frac{6\sqrt{3}}{5}} $

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#### 13) $ \frac{2}{7\sqrt{4}} $

Note: $ \sqrt{4} = 2 $, so:

$$
\frac{2}{7 \cdot 2} = \frac{2}{14} = \frac{1}{7}
$$

No radical needed here.

✔ Answer: $ \boxed{\frac{1}{7}} $

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#### 14) $ \frac{25}{3\sqrt{5}} $

Multiply by $ \frac{\sqrt{5}}{\sqrt{5}} $:

$$
\frac{25}{3\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{25\sqrt{5}}{3 \cdot 5} = \frac{25\sqrt{5}}{15} = \frac{5\sqrt{5}}{3}
$$

✔ Answer: $ \boxed{\frac{5\sqrt{5}}{3}} $

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#### 15) $ \frac{54}{9\sqrt{2}} $

Simplify first: $ \frac{54}{9} = 6 $, so $ \frac{6}{\sqrt{2}} $

Now rationalize:

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

✔ Answer: $ \boxed{3\sqrt{2}} $

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#### 16) $ \frac{\sqrt{27}}{\sqrt{3}} $

Simplify: $ \frac{\sqrt{27}}{\sqrt{3}} = \sqrt{\frac{27}{3}} = \sqrt{9} = 3 $

Or:

$$
\frac{\sqrt{27}}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3
$$

✔ Answer: $ \boxed{3} $

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#### 17) $ \frac{\sqrt{75}}{\sqrt{3}} $

Simplify: $ \frac{\sqrt{75}}{\sqrt{3}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5 $

Or:

$ \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3} $, so $ \frac{5\sqrt{3}}{\sqrt{3}} = 5 $

✔ Answer: $ \boxed{5} $

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#### 18) $ \frac{\sqrt{64}}{\sqrt{4}} $

$ \sqrt{64} = 8 $, $ \sqrt{4} = 2 $, so $ \frac{8}{2} = 4 $

✔ Answer: $ \boxed{4} $

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#### 19) $ \frac{\sqrt{98}}{\sqrt{2}} $

$ \sqrt{98} = \sqrt{49 \cdot 2} = 7\sqrt{2} $, so:

$$
\frac{7\sqrt{2}}{\sqrt{2}} = 7
$$

✔ Answer: $ \boxed{7} $

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#### 20) $ \frac{\sqrt{24}}{\sqrt{6}} $

$ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} $, so:

$$
\frac{2\sqrt{6}}{\sqrt{6}} = 2
$$

✔ Answer: $ \boxed{2} $

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🔷 Section B: Rationalize the denominators of the following

These involve binomial denominators. Use conjugates.

> Conjugate of $ a + b $ is $ a - b $. Multiply numerator and denominator by it.

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#### 1) $ \frac{11}{2 + \sqrt{3}} $

Multiply numerator and denominator by conjugate $ 2 - \sqrt{3} $:

$$
\frac{11}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{11(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2} = \frac{11(2 - \sqrt{3})}{4 - 3} = \frac{11(2 - \sqrt{3})}{1} = 22 - 11\sqrt{3}
$$

✔ Answer: $ \boxed{22 - 11\sqrt{3}} $

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#### 2) $ \frac{1}{2 - \sqrt{5}} $

Multiply by conjugate $ 2 + \sqrt{5} $:

$$
\frac{1}{2 - \sqrt{5}} \cdot \frac{2 + \sqrt{5}}{2 + \sqrt{5}} = \frac{2 + \sqrt{5}}{(2)^2 - (\sqrt{5})^2} = \frac{2 + \sqrt{5}}{4 - 5} = \frac{2 + \sqrt{5}}{-1} = -2 - \sqrt{5}
$$

✔ Answer: $ \boxed{-2 - \sqrt{5}} $

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#### 3) $ \frac{12}{3 - \sqrt{3}} $

Multiply by $ 3 + \sqrt{3} $:

$$
\frac{12}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{12(3 + \sqrt{3})}{9 - 3} = \frac{12(3 + \sqrt{3})}{6} = 2(3 + \sqrt{3}) = 6 + 2\sqrt{3}
$$

✔ Answer: $ \boxed{6 + 2\sqrt{3}} $

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#### 4) $ \frac{7}{1 + \sqrt{2}} $

Multiply by $ 1 - \sqrt{2} $:

$$
\frac{7}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{7(1 - \sqrt{2})}{1 - 2} = \frac{7(1 - \sqrt{2})}{-1} = -7 + 7\sqrt{2}
$$

✔ Answer: $ \boxed{-7 + 7\sqrt{2}} $

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#### 5) $ \frac{4}{3 + \sqrt{11}} $

Multiply by $ 3 - \sqrt{11} $:

$$
\frac{4}{3 + \sqrt{11}} \cdot \frac{3 - \sqrt{11}}{3 - \sqrt{11}} = \frac{4(3 - \sqrt{11})}{9 - 11} = \frac{4(3 - \sqrt{11})}{-2} = -2(3 - \sqrt{11}) = -6 + 2\sqrt{11}
$$

✔ Answer: $ \boxed{-6 + 2\sqrt{11}} $

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#### 6) $ \frac{4}{\sqrt{3} - 1} $

Multiply by conjugate $ \sqrt{3} + 1 $:

$$
\frac{4}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{4(\sqrt{3} + 1)}{3 - 1} = \frac{4(\sqrt{3} + 1)}{2} = 2(\sqrt{3} + 1) = 2\sqrt{3} + 2
$$

✔ Answer: $ \boxed{2\sqrt{3} + 2} $

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✔ Final Answers Summary:

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Section A:

1) $ \frac{3\sqrt{2}}{2} $
2) $ \frac{11\sqrt{5}}{5} $
3) $ \frac{9\sqrt{7}}{7} $
4) $ \frac{4\sqrt{11}}{11} $
5) $ 2\sqrt{13} $
6) $ 5\sqrt{3} $
7) $ 7\sqrt{2} $
8) $ \frac{10\sqrt{6}}{3} $
9) $ \frac{4\sqrt{6}}{3} $
10) $ \frac{4\sqrt{6}}{3} $
11) $ \frac{4\sqrt{7}}{3} $
12) $ \frac{6\sqrt{3}}{5} $
13) $ \frac{1}{7} $
14) $ \frac{5\sqrt{5}}{3} $
15) $ 3\sqrt{2} $
16) $ 3 $
17) $ 5 $
18) $ 4 $
19) $ 7 $
20) $ 2 $

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Section B:

1) $ 22 - 11\sqrt{3} $
2) $ -2 - \sqrt{5} $
3) $ 6 + 2\sqrt{3} $
4) $ -7 + 7\sqrt{2} $
5) $ -6 + 2\sqrt{11} $
6) $ 2\sqrt{3} + 2 $

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Let me know if you'd like these written in a printable format or explained further!