Simplifying Radicals by Rationalizing Worksheet | PDF Printable ... - Free Printable

Let's solve the problems in Section A and Section B step by step. The goal is to rationalize the denominator, which means removing any radicals (square roots) from the denominator.

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🔷 Section A: Rationalizing Simple Radicals

We'll rationalize each expression by multiplying numerator and denominator by the radical in the denominator.

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

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

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

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

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

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

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

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

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

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

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

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#### 5) $\frac{20}{\sqrt{13}}$

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

$$
\frac{20\sqrt{13}}{13}
$$

✔ Answer: $\boxed{\frac{20\sqrt{13}}{13}}$

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

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

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

✔ Answer: $\boxed{5\sqrt{3}}$

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

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

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

✔ Answer: $\boxed{7\sqrt{2}}$

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#### 8) $\frac{20}{\sqrt{5}}$

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

$$
\frac{20\sqrt{5}}{5} = 4\sqrt{5}
$$

✔ Answer: $\boxed{4\sqrt{5}}$

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#### 9) $\frac{22}{\sqrt{11}}$

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

$$
\frac{22\sqrt{11}}{11} = 2\sqrt{11}
$$

✔ Answer: $\boxed{2\sqrt{11}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

Note: $\sqrt{4} = 2$, so denominator is $7 \cdot 2 = 14$

So:
$$
\frac{2}{14} = \frac{1}{7}
$$

✔ Answer: $\boxed{\frac{1}{7}}$

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

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

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

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

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

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

Now multiply by $\frac{\sqrt{2}}{\sqrt{2}}$:

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

✔ Answer: $\boxed{3\sqrt{2}}$

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

Use rule: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$

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

✔ Answer: $\boxed{3}$

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

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

✔ Answer: $\boxed{5}$

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

$$
\sqrt{\frac{64}{4}} = \sqrt{16} = 4
$$

✔ Answer: $\boxed{4}$

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

$$
\sqrt{\frac{98}{2}} = \sqrt{49} = 7
$$

✔ Answer: $\boxed{7}$

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

$$
\sqrt{\frac{24}{6}} = \sqrt{4} = 2
$$

✔ Answer: $\boxed{2}$

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## ✔ Section A Summary (Final Answers):

| Problem | Answer |
|--------|--------|
| 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) | $\frac{20\sqrt{13}}{13}$ |
| 6) | $5\sqrt{3}$ |
| 7) | $7\sqrt{2}$ |
| 8) | $4\sqrt{5}$ |
| 9) | $2\sqrt{11}$ |
| 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: Rationalizing Binomial Denominators

These involve expressions like $a + \sqrt{b}$ or $a - \sqrt{b}$. We use the conjugate to rationalize.

> Conjugate of $a - \sqrt{b}$ is $a + \sqrt{b}$, and vice versa.

We multiply numerator and denominator by the conjugate.

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

Multiply 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 $2 + \sqrt{5}$:

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

✔ Answer: $\boxed{-2 - \sqrt{5}}$

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

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

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

✔ Answer: $\boxed{6 + 2\sqrt{3}}$

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

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

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

✔ Answer: $\boxed{-7 + 7\sqrt{2}}$

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

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

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

✔ Answer: $\boxed{-6 + 2\sqrt{11}}$

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#### 12) $\frac{1}{\sqrt{3} - 1}$

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

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

✔ Answer: $\boxed{\frac{\sqrt{3} + 1}{2}}$

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## ✔ Section B Summary (Final Answers):

| Problem | Answer |
|--------|--------|
| 1) | $22 + 11\sqrt{3}$ |
| 2) | $-2 - \sqrt{5}$ |
| 6) | $6 + 2\sqrt{3}$ |
| 7) | $-7 + 7\sqrt{2}$ |
| 11) | $-6 + 2\sqrt{11}$ |
| 12) | $\frac{\sqrt{3} + 1}{2}$ |

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📌 Key Concepts Recap:

- Rationalizing a single radical: Multiply numerator and denominator by that radical.
- Rationalizing binomials with radicals: Use the conjugate $(a + b)(a - b) = a^2 - b^2$ to eliminate the radical.
- Always simplify fractions after rationalizing.

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