Practice Problems on Simplifying Radical Expressions - Free Printable

Problem Set Analysis and Solutions

The provided image contains a set of practice problems from Algebra 2, covering topics such as radicals, solving equations, determining the validity of statements involving radicals, and simplifying expressions. Below, I will solve each problem step by step.

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Section (A): Evaluate each radical

#### 1. \( \sqrt{49a^2} \)
- The square root of a product is the product of the square roots:
\[
\sqrt{49a^2} = \sqrt{49} \cdot \sqrt{a^2}
\]
- Since \( \sqrt{49} = 7 \) and \( \sqrt{a^2} = |a| \):
\[
\sqrt{49a^2} = 7|a|
\]

#### 2. \( \sqrt[3]{27} \)
- The cube root of 27 is the number that, when cubed, gives 27:
\[
\sqrt[3]{27} = 3
\]

#### 3. \( \sqrt{-36m^2} \)
- The square root of a negative number is not real unless we use imaginary numbers. Here:
\[
\sqrt{-36m^2} = \sqrt{-1 \cdot 36 \cdot m^2} = \sqrt{-1} \cdot \sqrt{36} \cdot \sqrt{m^2}
\]
- Since \( \sqrt{-1} = i \), \( \sqrt{36} = 6 \), and \( \sqrt{m^2} = |m| \):
\[
\sqrt{-36m^2} = 6i|m|
\]

#### 4. \( -\sqrt[3]{-8} \)
- The cube root of \(-8\) is the number that, when cubed, gives \(-8\):
\[
\sqrt[3]{-8} = -2
\]
- Therefore:
\[
-\sqrt[3]{-8} = -(-2) = 2
\]

#### 5. \( \sqrt[4]{16r^6s^4} \)
- The fourth root of a product is the product of the fourth roots:
\[
\sqrt[4]{16r^6s^4} = \sqrt[4]{16} \cdot \sqrt[4]{r^6} \cdot \sqrt[4]{s^4}
\]
- Since \( \sqrt[4]{16} = 2 \), \( \sqrt[4]{r^6} = r^{6/4} = r^{3/2} \), and \( \sqrt[4]{s^4} = s \):
\[
\sqrt[4]{16r^6s^4} = 2r^{3/2}s
\]

#### 6. \( \sqrt[3]{-125u^9v^6} \)
- The cube root of a product is the product of the cube roots:
\[
\sqrt[3]{-125u^9v^6} = \sqrt[3]{-125} \cdot \sqrt[3]{u^9} \cdot \sqrt[3]{v^6}
\]
- Since \( \sqrt[3]{-125} = -5 \), \( \sqrt[3]{u^9} = u^{9/3} = u^3 \), and \( \sqrt[3]{v^6} = v^{6/3} = v^2 \):
\[
\sqrt[3]{-125u^9v^6} = -5u^3v^2
\]

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Section (B): Find the real roots of each equation

#### 7. \( 2y^3 = 128 \)
- Divide both sides by 2:
\[
y^3 = 64
\]
- Take the cube root of both sides:
\[
y = \sqrt[3]{64} = 4
\]

#### 8. \( x^2 + 16 = 0 \)
- Rearrange the equation:
\[
x^2 = -16
\]
- Since the square of a real number cannot be negative, there are no real roots.

#### 9. \( \frac{x^5}{8} - 4 = 0 \)
- Add 4 to both sides:
\[
\frac{x^5}{8} = 4
\]
- Multiply both sides by 8:
\[
x^5 = 32
\]
- Take the fifth root of both sides:
\[
x = \sqrt[5]{32} = 2
\]

#### 10. \( 8y^3 = -125 \)
- Divide both sides by 8:
\[
y^3 = -\frac{125}{8}
\]
- Take the cube root of both sides:
\[
y = \sqrt[3]{-\frac{125}{8}} = -\frac{5}{2}
\]

#### 11. \( \frac{16x^2 + 7x}{7} = \frac{4x + 7}{4} \)
- Cross-multiply to eliminate the fractions:
\[
4(16x^2 + 7x) = 7(4x + 7)
\]
- Expand both sides:
\[
64x^2 + 28x = 28x + 49
\]
- Subtract \( 28x \) from both sides:
\[
64x^2 = 49
\]
- Divide by 64:
\[
x^2 = \frac{49}{64}
\]
- Take the square root of both sides:
\[
x = \pm \sqrt{\frac{49}{64}} = \pm \frac{7}{8}
\]

#### 12. \( \frac{a^2(64a^2 - 3)}{3} = \frac{27 - 4a^2}{4} \)
- Cross-multiply to eliminate the fractions:
\[
4a^2(64a^2 - 3) = 3(27 - 4a^2)
\]
- Expand both sides:
\[
256a^4 - 12a^2 = 81 - 12a^2
\]
- Add \( 12a^2 \) to both sides:
\[
256a^4 = 81
\]
- Divide by 256:
\[
a^4 = \frac{81}{256}
\]
- Take the fourth root of both sides:
\[
a = \pm \sqrt[4]{\frac{81}{256}} = \pm \frac{3}{4}
\]

---

Section (C): For what values of the variable are the following statements true?

#### 13. \( \sqrt[3]{(x-5)^3} = x-5 \)
- The cube root of a number cubed is the number itself:
\[
\sqrt[3]{(x-5)^3} = x-5
\]
- This statement is true for all real values of \( x \).

#### 14. \( \sqrt{(3x-1)^2} = 3x-1 \)
- The square root of a squared term is the absolute value:
\[
\sqrt{(3x-1)^2} = |3x-1|
\]
- For the equation to hold, \( |3x-1| = 3x-1 \). This is true when \( 3x-1 \geq 0 \):
\[
3x-1 \geq 0 \implies x \geq \frac{1}{3}
\]

---

Section (D): Simplify

#### 15. \( \frac{12}{\sqrt{6}} \)
- Rationalize the denominator:
\[
\frac{12}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{12\sqrt{6}}{6} = 2\sqrt{6}
\]

#### 16. \( \frac{\sqrt{30} \cdot \sqrt{5}}{\sqrt{6}} \)
- Combine the square roots in the numerator:
\[
\frac{\sqrt{30} \cdot \sqrt{5}}{\sqrt{6}} = \frac{\sqrt{150}}{\sqrt{6}}
\]
- Simplify the fraction:
\[
\frac{\sqrt{150}}{\sqrt{6}} = \sqrt{\frac{150}{6}} = \sqrt{25} = 5
\]

#### 17. \( \frac{\sqrt{50s^3}}{\sqrt{108t^3}} \)
- Simplify the square roots:
\[
\frac{\sqrt{50s^3}}{\sqrt{108t^3}} = \sqrt{\frac{50s^3}{108t^3}} = \sqrt{\frac{25s^3}{54t^3}}
\]
- Factor out perfect squares:
\[
\sqrt{\frac{25s^3}{54t^3}} = \frac{\sqrt{25} \cdot \sqrt{s^2} \cdot \sqrt{s}}{\sqrt{54} \cdot \sqrt{t^2} \cdot \sqrt{t}} = \frac{5s\sqrt{s}}{3\sqrt{6}t\sqrt{t}}
\]
- Rationalize if necessary:
\[
\frac{5s\sqrt{s}}{3t\sqrt{6t}}
\]

#### 18. \( \sqrt[3]{\frac{16m^4}{3}} \)
- Separate the cube root:
\[
\sqrt[3]{\frac{16m^4}{3}} = \frac{\sqrt[3]{16m^4}}{\sqrt[3]{3}}
\]
- Simplify the numerator:
\[
\sqrt[3]{16m^4} = \sqrt[3]{8m^3 \cdot 2m} = 2m\sqrt[3]{2m}
\]
- Thus:
\[
\sqrt[3]{\frac{16m^4}{3}} = \frac{2m\sqrt[3]{2m}}{\sqrt[3]{3}}
\]

#### 19. \( \sqrt{3ab^2} \cdot \sqrt[4]{49a^4b^3} \)
- Simplify each term:
\[
\sqrt{3ab^2} = \sqrt{3} \cdot \sqrt{a} \cdot \sqrt{b^2} = \sqrt{3} \cdot \sqrt{a} \cdot b
\]
\[
\sqrt[4]{49a^4b^3} = \sqrt[4]{49} \cdot \sqrt[4]{a^4} \cdot \sqrt[4]{b^3} = \sqrt{7} \cdot a \cdot \sqrt[4]{b^3}
\]
- Multiply the results:
\[
\sqrt{3} \cdot \sqrt{a} \cdot b \cdot \sqrt{7} \cdot a \cdot \sqrt[4]{b^3} = \sqrt{21} \cdot a \cdot b \cdot \sqrt{a} \cdot \sqrt[4]{b^3}
\]

#### 20. \( \frac{\sqrt[3]{-24rs}}{\sqrt[3]{25r^4s^8}} \)
- Simplify the cube roots:
\[
\frac{\sqrt[3]{-24rs}}{\sqrt[3]{25r^4s^8}} = \sqrt[3]{\frac{-24rs}{25r^4s^8}}
\]
- Simplify the fraction inside the cube root:
\[
\sqrt[3]{\frac{-24rs}{25r^4s^8}} = \sqrt[3]{\frac{-24}{25r^3s^7}}
\]

#### 21. \( \sqrt{54} - \sqrt{6} + \sqrt{96} \)
- Simplify each square root:
\[
\sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6}
\]
\[
\sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6}
\]
- Combine like terms:
\[
3\sqrt{6} - \sqrt{6} + 4\sqrt{6} = 6\sqrt{6}
\]

#### 22. \( \sqrt{12x^4} - \sqrt{7x^4} \)
- Simplify each square root:
\[
\sqrt{12x^4} = \sqrt{4 \cdot 3 \cdot x^4} = 2x^2\sqrt{3}
\]
\[
\sqrt{7x^4} = \sqrt{7 \cdot x^4} = x^2\sqrt{7}
\]
- Combine the terms:
\[
2x^2\sqrt{3} - x^2\sqrt{7}
\]

#### 23. \( \sqrt[4]{m} \left( \sqrt[4]{m^3} + \sqrt[4]{m} \right) \)
- Simplify each term:
\[
\sqrt[4]{m^3} = m^{3/4}, \quad \sqrt[4]{m} = m^{1/4}
\]
- Distribute \( \sqrt[4]{m} \):
\[
\sqrt[4]{m} \left( m^{3/4} + m^{1/4} \right) = m^{1/4} \cdot m^{3/4} + m^{1/4} \cdot m^{1/4} = m + m^{1/2}
\]

#### 24. \( (\sqrt{3} + \sqrt{5})^2 \)
- Use the binomial expansion:
\[
(\sqrt{3} + \sqrt{5})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{5} + (\sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}
\]

#### 25. \( (5\sqrt{2} + \sqrt{3})(2\sqrt{2} - 3\sqrt{3}) \)
- Use the distributive property:
\[
(5\sqrt{2} + \sqrt{3})(2\sqrt{2} - 3\sqrt{3}) = 5\sqrt{2} \cdot 2\sqrt{2} + 5\sqrt{2} \cdot (-3\sqrt{3}) + \sqrt{3} \cdot 2\sqrt{2} + \sqrt{3} \cdot (-3\sqrt{3})
\]
- Simplify each term:
\[
5\sqrt{2} \cdot 2\sqrt{2} = 10 \cdot 2 = 20
\]
\[
5\sqrt{2} \cdot (-3\sqrt{3}) = -15\sqrt{6}
\]
\[
\sqrt{3} \cdot 2\sqrt{2} = 2\sqrt{6}
\]
\[
\sqrt{3} \cdot (-3\sqrt{3}) = -3 \cdot 3 = -9
\]
- Combine the terms:
\[
20 - 15\sqrt{6} + 2\sqrt{6} - 9 = 11 - 13\sqrt{6}
\]

#### 26. \( \frac{4}{\sqrt{3} + 2} \)
- Rationalize the denominator:
\[
\frac{4}{\sqrt{3} + 2} \cdot \frac{\sqrt{3} - 2}{\sqrt{3} - 2} = \frac{4(\sqrt{3} - 2)}{(\sqrt{3} + 2)(\sqrt{3} - 2)}
\]
- Simplify the denominator:
\[
(\sqrt{3} + 2)(\sqrt{3} - 2) = 3 - 4 = -1
\]
- Simplify the expression:
\[
\frac{4(\sqrt{3} - 2)}{-1} = -4(\sqrt{3} - 2) = -4\sqrt{3} + 8
\]

#### 27. \( \frac{2\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} \)
- Rationalize the denominator:
\[
\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} \cdot \frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} - \sqrt{5}} = \frac{(2\sqrt{3} + \sqrt{5})(\sqrt{3} - \sqrt{5})}{(\sqrt{3} + \sqrt{5})(\sqrt{3} - \sqrt{5})}
\]
- Simplify the denominator:
\[
(\sqrt{3} + \sqrt{5})(\sqrt{3} - \sqrt{5}) = 3 - 5 = -2
\]
- Simplify the numerator:
\[
(2\sqrt{3} + \sqrt{5})(\sqrt{3} - \sqrt{5}) = 2\sqrt{3} \cdot \sqrt{3} + 2\sqrt{3} \cdot (-\sqrt{5}) + \sqrt{5} \cdot \sqrt{3} + \sqrt{5} \cdot (-\sqrt{5})
\]
\[
= 2 \cdot 3 - 2\sqrt{15} + \sqrt{15} - 5 = 6 - 2\sqrt{15} + \sqrt{15} - 5 = 1 - \sqrt{15}
\]
- Combine the results:
\[
\frac{1 - \sqrt{15}}{-2} = \frac{\sqrt{15} - 1}{2}
\]

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Final Answers

\[
\boxed{
\begin{aligned}
&1. 7|a|, \quad 2. 3, \quad 3. 6i|m|, \quad 4. 2, \quad 5. 2r^{3/2}s, \quad 6. -5u^3v^2, \\
&7. 4, \quad 8. \text{No real roots}, \quad 9. 2, \quad 10. -\frac{5}{2}, \quad 11. \pm \frac{7}{8}, \quad 12. \pm \frac{3}{4}, \\
&13. \text{All } x, \quad 14. x \geq \frac{1}{3}, \\
&15. 2\sqrt{6}, \quad 16. 5, \quad 17. \frac{5s\sqrt{s}}{3t\sqrt{6t}}, \quad 18. \frac{2m\sqrt[3]{2m}}{\sqrt[3]{3}}, \\
&19. \sqrt{21} \cdot a \cdot b \cdot \sqrt{a} \cdot \sqrt[4]{b^3}, \quad 20. \sqrt[3]{\frac{-24}{25r^3s^7}}, \quad 21. 6\sqrt{6}, \\
&22. 2x^2\sqrt{3} - x^2\sqrt{7}, \quad 23. m + m^{1/2}, \quad 24. 8 + 2\sqrt{15}, \\
&25. 11 - 13\sqrt{6}, \quad 26. -4\sqrt{3} + 8, \quad 27. \frac{\sqrt{15} - 1}{2}.
\end{aligned}
}
\]