Simplifying Radicals worksheet with 18 problems for algebra students to practice simplifying radical expressions, including square roots, cube roots, and higher-order roots with variables and exponents.

Let’s go through each problem one by one. We’ll simplify radicals step by step, using prime factorization and rules for exponents. Remember:

- For square roots (√), we look for pairs of factors.
- For cube roots (∛), we look for triples.
- For fourth roots (∜), we look for groups of four.
- If there’s a negative under an even root (like √ or ∜), it’s not real — but if it’s under an odd root (like ∛), it’s okay.
- Use absolute value signs when simplifying even roots of variables with even powers that might be negative (but in these problems, since we’re told to use them “when necessary,” we’ll only add them if the variable could be negative and the root is even).

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1) √24

Factor 24:
24 = 4 × 6 = 2² × 6
So √24 = √(2² × 6) = 2√6

✔ Final Answer: 2√6

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2) ∛1000

1000 = 10³ → so ∛1000 = 10

✔ Final Answer: 10

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3) ∛(-162)

First, ignore the sign: 162 = 2 × 81 = 2 × 3⁴
So 162 = (3³ × 3 × 2) = 3∛6
Now bring back the negative: ∛(-162) = -∛162 = -3∛6

✔ Final Answer: -3∛6

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4) √512

512 = 256 × 2 = 2⁸ × 2 = 2⁹? Wait — let’s do it right.

Actually:
512 ÷ 2 = 256
256 = 2⁸ → so 512 = 2⁹? No: 2⁹ = 512? Let’s check:
2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32, 2⁶=64, 2⁷=128, 2⁸=256, 2⁹=512 → yes!

So √512 = √(2) = √(2 × 2) = 2⁴√2 = 16√2

✔ Final Answer: 16√2

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5) ∜(128n⁴)

128 = 64 × 2 = 2⁷
n⁴ is already a perfect 4th power.

So ∜(2⁷ × n⁴) = ∜(2⁴ × 2³ × n⁴) = 2 × |n| × ∜8
Wait — why |n|? Because it’s a 4th root (even root), and n⁴ is always positive, but n could be negative. So we need absolute value for n.

But 2³ = 8, which doesn’t simplify further under 4th root.

So: 2|n|∜8

But can we write ∜8 as ∜(2³)? Yes, but it doesn’t simplify more.

Alternatively, leave as 2|n|∜8 — but maybe better to write as 2|n|·2^(3/4)? No, keep radical form.

Actually, 128 = 16 × 8 = 2⁴ × 8 → so ∜(2⁴ × 8 × n⁴) = 2|n|∜8

Yes.

✔ Final Answer: 2|n|∜8

Wait — let me double-check:
Is 128 = 2^7? Yes.
∜(2^7 * n^4) = 2^(7/4) * n^(4/4) = 2^(1 + 3/4) * |n| = 2 * 2^(3/4) * |n| = 2|n|∜8 — correct.

But sometimes they prefer to write ∜8 as is.

Alternatively, note that 8 = 2³, so no simpler form.

✔ Final Answer: 2|n|∜8

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6) √(98k)

98 = 49 × 2 = 7² × 2
So √(98k) = √(7² × 2k) = 7√(2k)

No absolute value needed because k is to the first power — we don’t know if it’s negative, but since it’s under a square root, we assume 2k ≥ 0, or just leave as is unless specified. The problem says “use absolute value signs when necessary” — here, since k is to the first power inside, and we’re taking square root, technically if k were negative, it wouldn’t be real. But typically in such problems, we assume variables represent non-negative numbers unless otherwise stated. However, the instruction says “when necessary”, so perhaps we don’t need it here.

Looking at other problems — for example, #7 has r⁷ under cube root — no abs needed. For even roots with even powers, we may need abs.

Here, k is to the first power — so probably no abs needed.

✔ Final Answer: 7√(2k)

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7) ∛(224r⁷)

Factor 224:
224 ÷ 2 = 112
112 ÷ 2 = 56
56 ÷ 2 = 28
28 ÷ 2 = 14
14 ÷ 2 = 7 → so 224 = 2⁵ × 7

r⁷ = r × r = (r²)³ × r

So ∛(2 × 7 × r⁷) = ∛(2³ × 2² × 7 × r⁶ × r) = 2 r² ∛(4 × 7 × r) = 2r² ∛(28r)

Because 2² × 7 = 4×7=28

✔ Final Answer: 2r²∛(28r)

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8) ∛(24m⁵)

24 = 8 × 3 = 2³ × 3
m⁵ = m³ × m²

So ∛(2³ × 3 × m³ × m²) = 2m ∛(3m²)

✔ Final Answer: 2m∛(3m²)

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9) √(392x²)

392: divide by 2 → 196, which is 14² → so 392 = 2 × 196 = 2 × 14²

x² is perfect square.

So √(392x²) = √(2 × 14² × x²) = 14|x|√2

Why |x|? Because x² is always positive, but x could be negative, and square root gives non-negative result, so we need absolute value for x.

✔ Final Answer: 14|x|√2

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10) √(512x²)

From earlier, 512 = 2⁹ = 2⁸ × 2 = (2⁴)² × 2 = 256 × 2

x² is perfect square.

So √(512x²) = √(256 × 2 × x²) = 16|x|√2

Same reasoning as above.

✔ Final Answer: 16|x|√2

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11) ∜(405x³y²)

Factor 405:
405 ÷ 5 = 81 → 81 = 3⁴ → so 405 = 5 × 3⁴

So ∜(3⁴ × 5 × x³ × y²)

We can take out 3 from the 4th root, since 3⁴ is perfect 4th power.

x³ and y² are less than 4, so stay inside.

So: 3 ∜(5x³y²)

No absolute values needed because the exponents on x and y are less than 4, and we’re not extracting them fully — also, since it’s a 4th root, if we had even powers extracted, we’d need abs, but here nothing is being extracted except the 3.

Wait — actually, y² is even, but we’re not taking it out — it stays inside. So no issue.

✔ Final Answer: 3∜(5x³y²)

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12) ∛(-16a³b⁶)

First, handle the negative: cube root of negative is negative.

16 = 8 × 2 = 2³ × 2
a³ is perfect cube
b⁶ = (b²)³ → perfect cube

So ∛(-1 × 2³ × 2 × a³ × b⁶) = -1 × 2 × a × b² × ∛2 = -2ab²∛2

No absolute values needed because cube root handles negatives, and all variables are raised to multiples of 3 or handled properly.

✔ Final Answer: -2ab²∛2

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13) ∜(128x⁵y⁵)

128 = 2⁷ (as before)
x⁵ = x⁴ × x
y⁵ = y × y

So ∜(2⁷ × x⁴ × x × y⁴ × y) = ∜(2 × 2³ × x × y⁴ × xy) = 2|x||y| ∜(8xy)

Why |x| and |y|? Because we’re taking 4th root (even root) of x⁴ and y⁴, which are always positive, but x and y could be negative, so we need absolute values when pulling them out.

And 2³ = 8, so ∜8 remains.

✔ Final Answer: 2|x||y|∜(8xy)

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14) ∛(16xy)

16 = 8 × 2 = 2³ × 2
So ∛(2³ × 2 × x × y) = 2 ∛(2xy)

No simplification possible for x and y.

✔ Final Answer: 2∛(2xy)

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15) ∜(448x⁵y⁷)

Factor 448:
448 ÷ 2 = 224
224 ÷ 2 = 112
112 ÷ 2 = 56
56 ÷ 2 = 28
28 ÷ 2 = 14
14 ÷ 2 = 7 → so 448 = 2⁶ × 7

x⁵ = x × x
y⁷ = y⁴ × y³

So ∜(2⁶ × 7 × x⁴ × x × y⁴ × y³) = ∜(2⁴ × 2² × 7 × x⁴ × y⁴ × x y³) = 2|x||y| ∜(4 × 7 × x y³) = 2|x||y| ∜(28x y³)

Check: 2² × 7 = 4×7=28, and x y³ left inside.

✔ Final Answer: 2|x||y|∜(28x y³)

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16) ∛(56x²y)

56 = 8 × 7 = 2³ × 7
x² and y have no cubes.

So ∛(2³ × 7 × x² × y) = 2 ∛(7x²y)

✔ Final Answer: 2∛(7x²y)

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Critical Thinking Questions:

17) What simplifies into 2mn² ∛(5mn²)?

This means: find the original expression that, when simplified, becomes 2mn² ∛(5mn²)

To reverse this, we cube the entire thing, because it’s a cube root.

So [2mn² ∛(5mn²)]³ = ?

First, (2mn²)³ = 8 m³ n⁶

Then [∛(5mn²)]³ = 5mn²

So overall: 8 m³ n⁶ × 5 m n² = 40 m⁴ n⁸

Wait — let me compute:

[ A × B ]³ = A³ × B³

A = 2mn² → A³ = 8 m³ n⁶

B = ∛(5mn²) → B³ = 5mn²

So product: 8 m³ n⁶ × 5 m n² = 40 m^{4} n^{8}

So the original expression was ∛(40 m⁴ n⁸)

But let’s verify by simplifying ∛(40 m⁴ n⁸):

40 = 8 × 5 = 2³ × 5
m⁴ = m³ × m
n⁸ = n × n² = (n²)³ × n²

So ∛(2³ × 5 × m³ × m × n⁶ × n²) = 2 m n² ∛(5 m n²) — yes! Matches.

So answer is ∛(40 m⁴ n⁸)

But the question says “What simplifies into...” — so the unsimplified form is ∛(40 m⁴ n⁸)

✔ Final Answer: ∛(40m⁴n⁸)

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18) Simplify ∛(3 · 2ⁿ · x²ⁿ yⁿ⁺³)

We need to simplify this cube root.

Look at each part:

- 3: no cube factors
- 2ⁿ: depends on n. We can write 2ⁿ = 2^{3k + r}, but since n is variable, we leave as is unless we can extract full cubes.
Similarly for x²ⁿ and yⁿ⁺³.

Note: yⁿ⁺³ = yⁿ · y³ → so y³ is a perfect cube.

So let’s rewrite:

∛(3 · 2ⁿ · x²ⁿ · yⁿ · y³) = ∛(y³) · ∛(3 · 2ⁿ · x²ⁿ · yⁿ) = y · ∛(3 · 2ⁿ · x²ⁿ · yⁿ)

Now, look at 2ⁿ · x²ⁿ · yⁿ = (2 x² y)ⁿ

So: y · ∛(3 · (2 x² y)ⁿ )

Can we simplify ∛((2x²y)ⁿ)? Only if n is multiple of 3, but since n is general, we leave it.

But notice: if we write (2x²y)ⁿ = [(2x²y)^{n/3}]³ only if n divisible by 3 — not generally.

So best to write as:

y · ∛(3) · ∛((2x²y)ⁿ) = y ∛(3 (2x²y)ⁿ )

Alternatively, combine:

= y ∛(3 · 2ⁿ · x²ⁿ · yⁿ)

But we already pulled out y from y³.

Another way: since yⁿ⁺³ = y^{n} y^3, and we took y out, leaving y^n inside.

But perhaps we can write the whole thing as:

∛(3 · 2ⁿ · x²ⁿ · yⁿ⁺³) = ∛(3) · ∛(2) · ∛(x²ⁿ) · ∛(yⁿ⁺³)

Now, ∛(yⁿ⁺³) = y^{(n+3)/3} = y^{n/3 + 1} = y · y^{n/3}

Similarly, ∛(x²ⁿ) = x^{2n/3}, ∛(2ⁿ) = 2^{n/3}

So overall: 2^{n/3} · x^{2n/3} · y^{n/3 + 1} · ∛3

Which is y · (2 x² y)^{n/3} · ∛3

But this uses fractional exponents, which might not be desired.

In radical form, perhaps leave as:

y ∛(3 · 2ⁿ · x²ⁿ · y)

Or factor the inside:

Notice that 2ⁿ · x²ⁿ · yⁿ = (2 x² y)ⁿ

So: y ∛(3 (2 x² y)ⁿ )

I think this is acceptable.

But let’s see if we can write it as:

If n is a multiple of 3, say n=3k, then it would simplify nicely, but since n is general, we can’t assume that.

Perhaps the expected answer is to pull out what we can.

We have y³ inside, so pull out y.

The rest: 3 · 2ⁿ · x²ⁿ · yⁿ — no obvious cube factors unless n is known.

So I think the simplified form is:

y ∛(3 · 2ⁿ · x²ⁿ · y)

We can write it as y ∛(3 (2x²y)^n )

Either way.

But let me check with a number. Suppose n=3:

Original: ∛(3 · 2³ · x⁶ · y) = ∛(3 · 8 · x⁶ · y⁶) = ∛(24 x⁶ y⁶) = ∛(8 · 3 · x⁶ · y⁶) = 2 x² y² ∛3

According to our formula: y ∛(3 · 2³ · x⁶ · y³) = y ∛(3 · 8 · x⁶ · y³) = y · 2 x² y ∛3 = 2 x² y² ∛3 — matches.

Another test: n=0:

Original: ∛(3 · 1 · 1 · y³) = ∛(3 y³) = y ∛3

Our formula: y ∛(3 · 1 · 1 · 1) = y ∛3 — good.

n=1:

Original: ∛(3 · 2 · x² · y⁴) = (6 x² y) = ∛(6 x² y³ y) = y ∛(6 x² y)

Our formula: y ∛(3 · 2¹ · x² · y¹) = y ∛(6 x² y) — same.

Perfect.

So the simplified form is y times the cube root of (3 times 2^n times x^{2n} times y^n)

We can write it as:

y \sqrt[3]{3 \cdot 2^{n} \cdot x^{2n} \cdot y^{n}}

Or factor: y \sqrt[3]{3 (2 x^{2} y)^{n}}

Both are fine. I’ll go with the factored version for compactness.

✔ Final Answer: y ∛[3 (2x²y)ⁿ]

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Now, compiling all final answers:

Final Answer:
1) 2√6
2) 10
3) -3∛6
4) 16√2
5) 2|n|∜8
6) 7√(2k)
7) 2r²∛(28r)
8) 2m∛(3m²)
9) 14|x|√2
10) 16|x|√2
11) 3∜(5x³y²)
12) -2ab²∛2
13) 2|x||y|∜(8xy)
14) 2∛(2xy)
15) 2|x||y|∜(28x y³)
16) 2∛(7x²y)
17) ∛(40m⁴n⁸)
18) y ∛[3 (2x²y)ⁿ]