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Math worksheet for simplifying radicals by rationalizing denominators, featuring problems in Section A and Section B.

Worksheet titled "Simplifying Radicals by Rationalizing" with two sections (A and B) containing math problems to rationalize denominators involving square roots.

Worksheet titled "Simplifying Radicals by Rationalizing" with two sections (A and B) containing math problems to rationalize denominators involving square roots.

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Let's solve each problem step-by-step by rationalizing the denominator.

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🔷 Section A: Rationalize the denominator (simple radicals)



We rationalize by multiplying numerator and denominator by the radical in the denominator to eliminate the radical from the denominator.

---

#### 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: $\boxed{\frac{3\sqrt{2}}{2}}$

---

#### 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: $\boxed{\frac{11\sqrt{5}}{5}}$

---

#### 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: $\boxed{\frac{9\sqrt{7}}{7}}$

---

#### 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: $\boxed{\frac{4\sqrt{11}}{11}}$

---

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

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

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

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

---

#### 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: $\boxed{5\sqrt{3}}$

---

#### 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: $\boxed{7\sqrt{2}}$

---

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

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

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

Answer: $\boxed{4\sqrt{5}}$

---

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

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

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

Answer: $\boxed{2\sqrt{11}}$

---

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

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

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

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

---

#### 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: $\boxed{\frac{4\sqrt{7}}{3}}$

---

#### 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: $\boxed{\frac{6\sqrt{3}}{5}}$

---

#### 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 — already rational.

Answer: $\boxed{\frac{1}{7}}$

---

#### 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: $\boxed{\frac{5\sqrt{5}}{3}}$

---

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

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

$$
\frac{6}{\sqrt{2}} \Rightarrow \text{Multiply by } \frac{\sqrt{2}}{\sqrt{2}}: \quad \frac{6\sqrt{2}}{2} = 3\sqrt{2}
$$

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

---

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

Simplify:

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

Answer: $\boxed{3}$

---

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

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

Answer: $\boxed{5}$

---

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

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

Answer: $\boxed{4}$

---

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

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

Answer: $\boxed{7}$

---

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

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

Answer: $\boxed{2}$

---

🔷 Section B: Rationalize denominators with binomials



Use conjugate multiplication to eliminate radicals.

---

#### 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: $\boxed{22 + 11\sqrt{3}}$

---

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

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

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

Answer: $\boxed{-2 - \sqrt{5}}$

---

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

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: $\boxed{6 + 2\sqrt{3}}$

---

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

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: $\boxed{-7 + 7\sqrt{2}}$

---

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

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: $\boxed{-6 + 2\sqrt{11}}$

---

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

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

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

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

---

Final Answers Summary



#### 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. $\frac{20\sqrt{13}}{13}$
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. $2$

#### Section B:
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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