Simplifying Negative Radicals worksheet with 15 problems to practice simplifying expressions with negative radicands.

Let’s solve each problem step by step. We’re simplifying negative radicals — that means square roots (or other roots) of negative numbers, which involve imaginary numbers. Remember:
√(-1) = i
So √(-a) = √(a) × i, for any positive number a.

Also, cube roots and higher odd roots of negative numbers are real and negative. For example:
∛(-8) = -2, because (-2)³ = -8.

We’ll go one by one.

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Problem 1: √-9

√-9 = √(9 × -1) = √9 × √-1 = 3i

✔ Answer: 3i

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Problem 2: ∛-8

Cube root of -8: what number cubed gives -8?
(-2)³ = -8 → so ∛-8 = -2

✔ Answer: -2

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Problem 3: 3 + √-12

First simplify √-12:
√-12 = √(12 × -1) = √12 × i
√12 = √(4×3) = 2√3
So √-12 = 2√3 i

Then add 3:
3 + 2√3 i

✔ Answer: 3 + 2√3 i

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Problem 4: 2 + √-16

√-16 = √(16 × -1) = 4i
So 2 + 4i

✔ Answer: 2 + 4i

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Problem 5: √36 + √-169

√36 = 6
√-169 = √(169 × -1) = 13i
So total: 6 + 13i

✔ Answer: 6 + 13i

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Problem 6: √-128

√-128 = √(128 × -1) = √128 × i
Simplify √128:
128 = 64 × 2 → √128 = √64 × √2 = 8√2
So √-128 = 8√2 i

✔ Answer: 8√2 i

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Problem 7: -2√-25

First, √-25 = √(25 × -1) = 5i
Then multiply by -2: -2 × 5i = -10i

✔ Answer: -10i

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Problem 8: ¹²√-90

This is the 12th root of -90. Since 12 is even, we cannot take an even root of a negative number in real numbers — but in complex numbers, it’s possible, though very advanced.

Wait — this might be a typo? Because 12th root of a negative number isn’t typically simplified at this level unless specified. But let’s think:

Actually, in standard high school math, if you see an even root of a negative number, it’s not real — but here it’s written as ¹²√-90. That would be a complex number, but simplifying it exactly is beyond basic radical simplification.

But looking back at the worksheet title: “Simplifying Negative Radicals” — and most problems are square roots or cube roots. Maybe this is meant to be √-90? Or perhaps it's a mistake?

Wait — let me check the original image again mentally... The user wrote "¹²√-90" — that’s 12th root.

But in context of the rest of the problems, this seems out of place. Perhaps it’s a typo and should be √-90? Let’s assume that for now, since otherwise it’s too advanced.

If it’s √-90:

√-90 = √(90 × -1) = √90 × i
√90 = √(9×10) = 3√10
So √-90 = 3√10 i

But if it really is 12th root, then we can’t simplify it nicely without complex exponentials.

Given the level of other problems, I think it’s likely a typo and should be √-90.

I’ll proceed with √-90 = 3√10 i

✔ Answer (assuming typo): 3√10 i

*(Note: If it truly is 12th root, then answer is more complex and not expected at this level.)*

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Problem 9: -k√-48

First, √-48 = √(48 × -1) = √48 × i
√48 = √(16×3) = 4√3
So √-48 = 4√3 i
Then multiply by -k: -k × 4√3 i = -4k√3 i

✔ Answer: -4k√3 i

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Problem 10: (4 - √-36)/4

First, √-36 = 6i
So numerator: 4 - 6i
Divide by 4: (4 - 6i)/4 = 4/4 - 6i/4 = 1 - (3/2)i

✔ Answer: 1 - (3/2)i

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Problem 11: (64 - √-4)/2

√-4 = 2i
Numerator: 64 - 2i
Divide by 2: 64/2 - 2i/2 = 32 - i

✔ Answer: 32 - i

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Problem 12: √-625

√-625 = √(625 × -1) = 25i (since √625 = 25)

✔ Answer: 25i

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Problem 13: ∛-64

Cube root of -64: what number cubed is -64?
(-4)³ = -64 → so ∛-64 = -4

✔ Answer: -4

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Problem 14: √-147

√-147 = √(147 × -1) = √147 × i
Simplify √147:
147 = 49 × 3 → √147 = 7√3
So √-147 = 7√3 i

✔ Answer: 7√3 i

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Problem 15: √-72

√-72 = √(72 × -1) = √72 × i
√72 = √(36×2) = 6√2
So √-72 = 6√2 i

✔ Answer: 6√2 i

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Now, compiling all answers:

Final Answer:
1. 3i
2. -2
3. 3 + 2√3 i
4. 2 + 4i
5. 6 + 13i
6. 8√2 i
7. -10i
8. 3√10 i *(assuming typo from 12th root to square root)*
9. -4k√3 i
10. 1 - (3/2)i
11. 32 - i
12. 25i
13. -4
14. 7√3 i
15. 6√2 i