Adding and Subtracting Radicals Worksheets with Answer Key - Free Printable

Let's solve each problem step by step, focusing on adding and subtracting radical expressions. The key idea is:

> You can only combine like radicals — that is, radicals with the same index and same radicand (the number under the root).

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Problem 1: $-\sqrt{12} + \sqrt[3]{3}$

- $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$, so $-\sqrt{12} = -2\sqrt{3}$
- $\sqrt[3]{3}$ is a cube root; it cannot be combined with a square root.
- Different indices: one is square root (index 2), one is cube root (index 3)

✔ Cannot combine

Answer: $-2\sqrt{3} + \sqrt[3]{3}$

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Problem 2: $\sqrt[5]{3} - \sqrt{12}$

- $\sqrt[5]{3}$ is fifth root of 3
- $\sqrt{12} = 2\sqrt{3}$ (square root)
- Different indices and different types

✔ Cannot combine

Answer: $\sqrt[5]{3} - 2\sqrt{3}$

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Problem 3: $2\sqrt{11} + 4\sqrt{11}$

- Both terms have same radical: $\sqrt{11}$
- Combine coefficients: $2 + 4 = 6$

Answer: $6\sqrt{11}$

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Problem 4: $-\sqrt[3]{27} - \sqrt[3]{8}$

- $\sqrt[3]{27} = 3$ because $3^3 = 27$
- $\sqrt[3]{8} = 2$ because $2^3 = 8$
- So: $-3 - 2 = -5$

Answer: $-5$

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Problem 5: $\sqrt{5} + \sqrt[5]{32} - \sqrt[5]{200}$

- $\sqrt{5}$ is square root → leave as is
- $\sqrt[5]{32}$: $32 = 2^5$, so $\sqrt[5]{32} = 2$
- $\sqrt[5]{200}$: $200 = 2^3 \cdot 5^2$, no perfect fifth power → cannot simplify further

So: $\sqrt{5} + 2 - \sqrt[5]{200}$

✔ No like terms to combine

Answer: $\sqrt{5} + 2 - \sqrt[5]{200}$

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Problem 6: $-\sqrt[3]{3} - \sqrt{8} - \sqrt[3]{3}$

- $-\sqrt[3]{3} - \sqrt[3]{3} = -2\sqrt[3]{3}$
- $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$

So we have: $-2\sqrt[3]{3} - 2\sqrt{2}$

Different radicals → cannot combine

Answer: $-2\sqrt[3]{3} - 2\sqrt{2}$

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Problem 7: $-\sqrt{27} - \sqrt[3]{45} - \sqrt{20} + \sqrt{45}$

Break down each term:

- $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$ → $-\sqrt{27} = -3\sqrt{3}$
- $\sqrt[3]{45}$: $45 = 9 \cdot 5 = 3^2 \cdot 5$, not a perfect cube → stays as is
- $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$
- $\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}$

Now plug in:
$$
-3\sqrt{3} - \sqrt[3]{45} - 2\sqrt{5} + 3\sqrt{5}
$$

Combine like terms:
- $-2\sqrt{5} + 3\sqrt{5} = \sqrt{5}$

So:
$$
-3\sqrt{3} + \sqrt{5} - \sqrt[3]{45}
$$

Answer: $-3\sqrt{3} + \sqrt{5} - \sqrt[3]{45}$

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Problem 8: $\sqrt[3]{5} + \sqrt{5} + \sqrt{5}$

- $\sqrt[3]{5}$ is cube root → cannot combine with square roots
- $\sqrt{5} + \sqrt{5} = 2\sqrt{5}$

Answer: $\sqrt[3]{5} + 2\sqrt{5}$

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Problem 9: $\sqrt[3]{24} - \sqrt{54} + \sqrt{6}$

Simplify each term:

- $\sqrt[3]{24} = \sqrt[3]{8 \cdot 3} = \sqrt[3]{8} \cdot \sqrt[3]{3} = 2\sqrt[3]{3}$
- $\sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6}$
- $\sqrt{6}$ stays

Now:
$$
2\sqrt[3]{3} - 3\sqrt{6} + \sqrt{6} = 2\sqrt[3]{3} - 2\sqrt{6}
$$

Answer: $2\sqrt[3]{3} - 2\sqrt{6}$

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Problem 10: $\sqrt[3]{27} + \sqrt{20} - \sqrt{300}$

Simplify:

- $\sqrt[3]{27} = 3$ (since $3^3 = 27$)
- $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$
- $\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}$

Now:
$$
3 + 2\sqrt{5} - 10\sqrt{3}
$$

No like terms → leave as is

Answer: $3 + 2\sqrt{5} - 10\sqrt{3}$

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✔ Final Answers:

1. $-2\sqrt{3} + \sqrt[3]{3}$
2. $\sqrt[5]{3} - 2\sqrt{3}$
3. $6\sqrt{11}$
4. $-5$
5. $\sqrt{5} + 2 - \sqrt[5]{200}$
6. $-2\sqrt[3]{3} - 2\sqrt{2}$
7. $-3\sqrt{3} + \sqrt{5} - \sqrt[3]{45}$
8. $\sqrt[3]{5} + 2\sqrt{5}$
9. $2\sqrt[3]{3} - 2\sqrt{6}$
10. $3 + 2\sqrt{5} - 10\sqrt{3}$

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