Simplifying Radicals (Algebra II) Color by Number - Free Printable

Let’s simplify each radical expression one by one. We’ll break down the numbers and variables inside the radicals into perfect powers (squares, cubes, fourth powers, etc.) that match the root index, then pull them out.

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1. √24

24 = 4 × 6 → 4 is a perfect square
√24 = √(4×6) = √4 × √6 = 2√6

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2. √2197

Wait — let’s check: 2197 is actually 13³? But we’re taking a square root. Let me factor it.

Actually, 2197 ÷ 13 = 169 → 169 = 13² → so 2197 = 13³

But for square root: √(13³) = √(13² × 13) = 13√13

So: 13√13

*(Note: The problem says “color the expression inside the radicand” — but since we’re simplifying, we just need the simplified form.)*

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3. ⁺√(80x⁸) ← This is 4th root

80 = 16 × 5 → 16 is 2⁴ → perfect 4th power
x⁸ = (x²)⁴ → also perfect 4th power

So: ⁺√(16 × 5 × x⁸) = ⁺√16 × ⁺√5 × ⁺√x = 2 × √5 × x² = 2x²⁺√5

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4. √(98x)

98 = 49 × 2 → 49 is 7²
So: √(49 × 2 × x) = √49 × √(2x) = 7√(2x)

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5. ³√(125x³y⁴)

125 = 5³ → perfect cube
x³ → perfect cube
y⁴ = y³ × y → so y³ comes out, y stays in

So: ³√(5³ × x³ × y³ × y) = 5xy × ³√y = 5xy³√y

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6. ³√(125x⁴)

125 = 5³
x⁴ = x³ × x

So: ³√(5³ × x³ × x) = 5x × ³√x = 5x³√x

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7. ³√(8x²y⁷)

8 = 2³
x² → can’t come out (less than 3)
y⁷ = y × y = (y²)³ × y → so y² comes out, y stays

So: ³√(2³ × x² × y⁶ × y) = 2y² × ³√(x²y) = 2y²³√(x²y)

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8. ³√(-16x³y⁸)

First, negative under odd root → okay, stays negative.

-16 = -8 × 2 → -8 = (-2)³ → perfect cube
x³ → perfect cube
y⁸ = y⁶ × y² = (y²)³ × y² → y² comes out, y² stays

So: ³√[(-2)³ × 2 × x³ × (y²)³ × y²] = (-2)xy² × ³√(2y²) = -2xy²³√(2y²)

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9. ⁺√(256x⁶y³) ← 4th root

256 = 4⁴? Wait: 4⁴=256? 4²=16, 4⁴=256 → yes! Or 2⁸ → but 2⁸ = (2²)⁴ = 4⁴ → same.

Better: 256 = 16², but for 4th root: 256 = 4⁴ → yes.

x⁶ = x⁴ × x² → x⁴ comes out as x, x² stays
y³ → less than 4, stays

So: ⁺√(4⁴ × x⁴ × x² × y³) = 4x × ⁺√(x²y³) = 4x⁺√(x²y³)

*(We could write x²y³ as is — no further simplification without fractional exponents.)*

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10. ⁺√243

243 = 81 × 3 = 3⁴ × 3 → because 3⁴=81, 81×3=243

So: ⁺√(3⁴ × 3) = 3 × ⁺√3 = 3⁺√3

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11. ⁶√(448x⁷y⁷) ← 6th root

Factor 448:

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

x⁷ = x⁶ × x
y⁷ = y⁶ × y

So: ⁶√(2⁶ × 7 × x⁶ × x × y⁶ × y) = 2xy × ⁶√(7xy) = 2xy⁶√(7xy)

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12. ³√(56x⁵y)

56 = 8 × 7 = 2³ × 7
x⁵ = x³ × x²
y → stays

So: ³√(2³ × 7 × x³ × x² × y) = 2x × ³√(7x²y) = 2x³√(7x²y)

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Now, let’s list all simplified expressions clearly:

1. 2√6
2. 13√13
3. 2x²⁺√5
4. 7√(2x)
5. 5xy³√y
6. 5x³√x
7. 2y²³√(x²y)
8. -2xy²³√(2y²)
9. 4x⁺√(x²y³)
10. 3⁺√3
11. 2xy⁶√(7xy)
12. 2x³√(7x²y)

✔ All checked and verified.

Final Answer:
1. 2√6
2. 13√13
3. 2x²∜5
4. 7√(2x)
5. 5xy∛y
6. 5x∛x
7. 2y²∛(x²y)
8. -2xy²∛(2y²)
9. 4x∜(x²y³)
10. 3∜3
11. 2xy∛(7xy) — wait, correction: this is 6th root → should be 2xy∛(7xy)? No!

Wait — mistake here!

For #11: ⁶√(448x⁷y⁷) = ⁶√(2⁶ * 7 * x⁶ * x * y⁶ * y) = 2xy * ⁶√(7xy) → so it's 2xy ∛(7xy)? NO — 6th root symbol is ∛? No!

Standard notation:

- Square root: √
- Cube root: ∛
- Fourth root: ∜
- Sixth root: usually written as √ with small 6, or sometimes \sqrt[6]{}

In text, we can write:

#11: 2xy \sqrt[6]{7xy}

Similarly, #3: 2x² \sqrt[4]{5}

#9: 4x \sqrt[4]{x²y³}

#10: 3 \sqrt[4]{3}

So to be precise in final answer, use proper radical notation.

But since the original used symbols like √, ³√, ⁴√, we’ll follow that style.

In the image, they used:

- √ for square root
- ³√ for cube root
- ⁴√ for fourth root
- ⁶√ for sixth root

So in final answer, we'll use those.

Corrected Final Answers:

1. 2√6
2. 13√13
3. 2x²⁴√5
4. 7√(2x)
5. 5xy³√y
6. 5x³√x
7. 2y²³√(x²y)
8. -2xy²³√(2y²)
9. 4x⁴√(x²y³)
10. 3⁴√3
11. 2xy⁶√(7xy)
12. 2x³√(7x²y)

Yes — all correct now.

Final Answer:
1. 2√6
2. 13√13
3. 2x²⁴√5
4. 7√(2x)
5. 5xy³√y
6. 5x³√x
7. 2y²³√(x²y)
8. -2xy²³√(2y²)
9. 4x⁴√(x²y³)
10. 3⁴√3
11. 2xy⁶√(7xy)
12. 2x³√(7x²y)