Chapter 6.5, Problem 2BGP

To determine

To Simplify:

  $\sqrt[5]{\frac{3}{4y}}$

Answer to Problem 2BGP

  $\frac{\sqrt[5]{24y^4}}{2y}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $\sqrt[5]{\frac{3}{4y}}$

Concept Used :

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$

  $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• Quotient Property of Radicals:

  $\begin{array}{l}\text{For any real numbers a and b}\ne \text{0 and any integer}\\ \text{n >1,}\\ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}},if\text{ all roots are defined}\text{.}\end{array}$

• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

Calculation:

• Given$\sqrt[5]{\frac{3}{4y}}$
• $\begin{array}{l}\sqrt[5]{\frac{3}{4y}}\\ =\frac{\sqrt[5]{3}}{\sqrt[5]{4y}}\mathrm{........................}[\text{quotient property of radicals]}\\ =\frac{\sqrt[5]{3}}{\sqrt[5]{4y}}\cdot \frac{\sqrt[5]{{\left(4y\right)}^4}}{\sqrt[5]{{\left(4y\right)}^4}}\mathrm{.............}[\text{rationalize the denominator]}\\ \text{=}\frac{\sqrt[5]{3}\cdot \sqrt[5]{{\left(4y\right)}^4}}{\sqrt[5]{4y}\cdot \sqrt[5]{{\left(4y\right)}^4}}\\ =\frac{\sqrt[5]{3\cdot {\left(4y\right)}^4}}{\sqrt[5]{4y\cdot {\left(4y\right)}^4}}\mathrm{................}[\text{product property of radicals]}\\ \text{=}\frac{\sqrt[5]{3\cdot {\left(2^2\right)}^4{y}^4}}{\sqrt[5]{{\left(4y\right)}^5}}\\ =\frac{\sqrt[5]{3\cdot 2^8\cdot y^4}}{4y}\\ =\frac{\sqrt[5]{3\cdot 2^5\cdot 2^3\cdot y^4}}{4y}\mathrm{...........}[a^{x+y}=a^x\cdot a^y]\\ =\frac{2\sqrt[5]{24y^4}}{2(2y)}\\ =\frac{\sqrt[5]{24y^4}}{2y}\end{array}$
• Since the index is 5, which is odd , so there will be only one root; hence no need to apply the absolute value symbol .