Simplifying Radical Expressions (Taking Terms Out of the Root)

 

Factors inside the radical whose exponent equals the index of the root can be taken out of the radical. For \(n \in Z^+ \),

\[ \Large\sqrt[n]{a^n} = \begin{cases} a, & \text{if } n \text{ is odd} \\ |a|, & \text{if } n \text{ is even} \end{cases} \]

 

Examples:

 

\( \bullet \sqrt[3]{ 125} = \sqrt[3]{5^3 } =5 \)

\( \bullet \sqrt[3]{ -8} = \sqrt[3]{(-2)^3 } =-2 \)

\( \bullet \sqrt[5]{\frac{1}{32}} = \sqrt[5]{\left( \frac{1}{2} \right)^5} = \frac{1}{2} \)

\( \bullet \sqrt[4]{16} = \sqrt[4]{2^4} = |2| = 2 \)

\( \bullet \sqrt{(\sqrt{3} – 2)^2} = \left| \sqrt{3} – 2 \right| \). Since \( \sqrt{3} – 2 < 0 \) here,

we get \(| \sqrt{3} – 2| =- (\sqrt{3} – 2) = 2- \sqrt{3}\)

\( \bullet \sqrt[3]{2^6} = \sqrt[3]{(2^2)^3} = 4 \) or \( \sqrt[3]{2^6} = 2^{ \frac{6}{3} } = 4 \)

\( \large\bullet \sqrt{\frac{27}{32} } = \sqrt{\frac{3 \cdot 3^2 }{2 \cdot 4^2} } = \frac{3}{4} \sqrt{\frac{3}{2} } \)

\( \bullet \sqrt[3]{-162 } = \sqrt[3]{ 6 \cdot ( -27) } = \sqrt[3]{6 \cdot (-3)^3 } = -3 \sqrt[3]{ 6} \)

\( \bullet \sqrt[4]{80} = \sqrt[4]{5 \cdot 2^4} = 2 \cdot \sqrt[4]{5 } \)

 

Question 4

 

Evaluate the expression: \[\frac{ \sqrt[4]{243 } }{\sqrt[4]{ 0.0048} }\]

 

\[ \text{A)} 1 \quad \text{B) } 5 \quad \text{C) } 10 \quad \text{D) } 15 \quad \text{E)} 20 \]

 

Solution:

 

\[ \frac{\sqrt[4]{243}}{\sqrt[4]{0.0048}} = \frac{\sqrt[4]{3 \cdot 3^4}}{\sqrt[4]{48 \cdot 10^{-4}}} = \frac{3 \cdot \sqrt[4]{3}}{\sqrt[4]{3 \cdot 2^4 \cdot (10^{-1})^4}} \]

\[ = \frac{3 \cdot \sqrt[4]{3}}{2 \cdot 10^{-1} \cdot \sqrt[4]{3}} = \frac{3 \cdot 10}{2} = 15 \]

 

\( \textbf{Answer: D}  \)