Simplifying Radical Expressions worksheet with 15 problems for practice.

Let's solve each of these radical expressions step by step. The goal is to simplify each square root expression by factoring out perfect squares and simplifying variables with even exponents.

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1. $\sqrt{45q^2}$

- Factor 45: $45 = 9 \times 5 = 3^2 \times 5$
- $q^2$ is a perfect square
- So:
$$
\sqrt{45q^2} = \sqrt{9 \cdot 5 \cdot q^2} = \sqrt{9} \cdot \sqrt{q^2} \cdot \sqrt{5} = 3q\sqrt{5}
$$

✔ Answer: $3q\sqrt{5}$

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2. $\sqrt{80r^2}$

- $80 = 16 \times 5 = 4^2 \times 5$
- $r^2$ is a perfect square
- So:
$$
\sqrt{80r^2} = \sqrt{16 \cdot 5 \cdot r^2} = \sqrt{16} \cdot \sqrt{r^2} \cdot \sqrt{5} = 4r\sqrt{5}
$$

✔ Answer: $4r\sqrt{5}$

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3. $\sqrt{44t^2}$

- $44 = 4 \times 11 = 2^2 \times 11$
- $t^2$ is a perfect square
- So:
$$
\sqrt{44t^2} = \sqrt{4 \cdot 11 \cdot t^2} = \sqrt{4} \cdot \sqrt{t^2} \cdot \sqrt{11} = 2t\sqrt{11}
$$

✔ Answer: $2t\sqrt{11}$

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4. $\sqrt{75x^2y}$

- $75 = 25 \times 3 = 5^2 \times 3$
- $x^2$ is a perfect square
- $y$ remains under the radical
- So:
$$
\sqrt{75x^2y} = \sqrt{25 \cdot 3 \cdot x^2 \cdot y} = \sqrt{25} \cdot \sqrt{x^2} \cdot \sqrt{3y} = 5x\sqrt{3y}
$$

✔ Answer: $5x\sqrt{3y}$

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5. $\sqrt{28x^3y^2}$

- $28 = 4 \times 7 = 2^2 \times 7$
- $x^3 = x^2 \cdot x$, so $x^2$ is perfect square
- $y^2$ is perfect square
- So:
$$
\sqrt{28x^3y^2} = \sqrt{4 \cdot 7 \cdot x^2 \cdot x \cdot y^2} = \sqrt{4} \cdot \sqrt{x^2} \cdot \sqrt{y^2} \cdot \sqrt{7x} = 2xy\sqrt{7x}
$$

✔ Answer: $2xy\sqrt{7x}$

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6. $\sqrt{32a^7b^4}$

- $32 = 16 \times 2 = 4^2 \times 2$
- $a^7 = a^6 \cdot a = (a^3)^2 \cdot a$
- $b^4 = (b^2)^2$, perfect square
- So:
$$
\sqrt{32a^7b^4} = \sqrt{16 \cdot 2 \cdot a^6 \cdot a \cdot b^4} = \sqrt{16} \cdot \sqrt{a^6} \cdot \sqrt{b^4} \cdot \sqrt{2a} = 4a^3b^2\sqrt{2a}
$$

✔ Answer: $4a^3b^2\sqrt{2a}$

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7. $\sqrt{62x^4y}$

- $62 = 2 \times 31$, no perfect square factors
- $x^4 = (x^2)^2$, perfect square
- $y$ stays inside
- So:
$$
\sqrt{62x^4y} = \sqrt{62} \cdot \sqrt{x^4} \cdot \sqrt{y} = x^2\sqrt{62y}
$$

✔ Answer: $x^2\sqrt{62y}$

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8. $\sqrt{80x^{100}y^{49}}$

- $80 = 16 \times 5 = 4^2 \times 5$
- $x^{100} = (x^{50})^2$, perfect square
- $y^{49} = y^{48} \cdot y = (y^{24})^2 \cdot y$
- So:
$$
\sqrt{80x^{100}y^{49}} = \sqrt{16 \cdot 5 \cdot x^{100} \cdot y^{48} \cdot y} = \sqrt{16} \cdot \sqrt{x^{100}} \cdot \sqrt{y^{48}} \cdot \sqrt{5y} = 4x^{50}y^{24}\sqrt{5y}
$$

✔ Answer: $4x^{50}y^{24}\sqrt{5y}$

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9. $\sqrt{42a^8}$

- $42 = 2 \times 3 \times 7$, no perfect square factor
- $a^8 = (a^4)^2$, perfect square
- So:
$$
\sqrt{42a^8} = \sqrt{42} \cdot \sqrt{a^8} = a^4\sqrt{42}
$$

✔ Answer: $a^4\sqrt{42}$

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10. $\sqrt{9a^6}$

- $9 = 3^2$, perfect square
- $a^6 = (a^3)^2$, perfect square
- So:
$$
\sqrt{9a^6} = \sqrt{9} \cdot \sqrt{a^6} = 3a^3
$$

✔ Answer: $3a^3$

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11. $\sqrt{m^{12}}$

- $m^{12} = (m^6)^2$, perfect square
- So:
$$
\sqrt{m^{12}} = m^6
$$

✔ Answer: $m^6$

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12. $\sqrt{27a^{11}b^7}$

- $27 = 9 \times 3 = 3^2 \times 3$
- $a^{11} = a^{10} \cdot a = (a^5)^2 \cdot a$
- $b^7 = b^6 \cdot b = (b^3)^2 \cdot b$
- So:
$$
\sqrt{27a^{11}b^7} = \sqrt{9 \cdot 3 \cdot a^{10} \cdot a \cdot b^6 \cdot b} = \sqrt{9} \cdot \sqrt{a^{10}} \cdot \sqrt{b^6} \cdot \sqrt{3ab} = 3a^5b^3\sqrt{3ab}
$$

✔ Answer: $3a^5b^3\sqrt{3ab}$

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13. $\sqrt{384x^4y^3}$

- $384$: Let's factor it.
- $384 \div 2 = 192$
- $192 \div 2 = 96$
- $96 \div 2 = 48$
- $48 \div 2 = 24$
- $24 \div 2 = 12$
- $12 \div 2 = 6$
- $6 \div 2 = 3$
- So: $384 = 2^7 \times 3$
- $2^7 = 2^6 \cdot 2 = (2^3)^2 \cdot 2 = 8^2 \cdot 2$
- $x^4 = (x^2)^2$, perfect square
- $y^3 = y^2 \cdot y$
- So:
$$
\sqrt{384x^4y^3} = \sqrt{64 \cdot 6 \cdot x^4 \cdot y^2 \cdot y} = \sqrt{64} \cdot \sqrt{x^4} \cdot \sqrt{y^2} \cdot \sqrt{6y} = 8x^2y\sqrt{6y}
$$

✔ Answer: $8x^2y\sqrt{6y}$

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14. $-4\sqrt{192x}$

First simplify $\sqrt{192x}$:

- $192 = 64 \times 3 = 8^2 \times 3$
- So:
$$
\sqrt{192x} = \sqrt{64 \cdot 3 \cdot x} = 8\sqrt{3x}
$$
Now multiply by $-4$:
$$
-4 \cdot 8\sqrt{3x} = -32\sqrt{3x}
$$

✔ Answer: $-32\sqrt{3x}$

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15. $2\sqrt{8p^2q^3m}$

Simplify $\sqrt{8p^2q^3m}$:

- $8 = 4 \times 2 = 2^2 \times 2$
- $p^2$ is perfect square
- $q^3 = q^2 \cdot q$
- $m$ stays
- So:
$$
\sqrt{8p^2q^3m} = \sqrt{4 \cdot 2 \cdot p^2 \cdot q^2 \cdot q \cdot m} = \sqrt{4} \cdot \sqrt{p^2} \cdot \sqrt{q^2} \cdot \sqrt{2qm} = 2pq\sqrt{2qm}
$$
Now multiply by 2:
$$
2 \cdot 2pq\sqrt{2qm} = 4pq\sqrt{2qm}
$$

✔ Answer: $4pq\sqrt{2qm}$

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✔ Final Answers Summary:

| Problem | Simplified Answer |
|--------|-------------------|
| 1 | $3q\sqrt{5}$ |
| 2 | $4r\sqrt{5}$ |
| 3 | $2t\sqrt{11}$ |
| 4 | $5x\sqrt{3y}$ |
| 5 | $2xy\sqrt{7x}$ |
| 6 | $4a^3b^2\sqrt{2a}$ |
| 7 | $x^2\sqrt{62y}$ |
| 8 | $4x^{50}y^{24}\sqrt{5y}$ |
| 9 | $a^4\sqrt{42}$ |
| 10 | $3a^3$ |
| 11 | $m^6$ |
| 12 | $3a^5b^3\sqrt{3ab}$ |
| 13 | $8x^2y\sqrt{6y}$ |
| 14 | $-32\sqrt{3x}$ |
| 15 | $4pq\sqrt{2qm}$ |

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