Chapter 6.2, Problem 51WE

To determine

To Evaluate:

  $\sqrt{x^{-1}{y}^{-2}}$ for x = 4 , y =3 , z =8

Answer to Problem 51WE

  $\frac{1}{6}$

Explanation of Solution

Given Information:

Given expression to evaluate:

  $\sqrt{x^{-1}{y}^{-2}}$

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 $a=\sqrt[n]{b}$
• Example : Because ${(-3)}^4=81$ , -3 is a fourth root of 81 and 3 is a principal root. That is $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

    $a^{-n}=\frac{1}{a^n}$

• The non-negative root is called the Principal Root.
• Suppose n is an integer greater than 1, and a is a real number.

a	n is even	n is odd
a >0	1 unique positive and 1 unique negative real root : $\pm \sqrt[n]{a}$ ; positive root is principal root.	1 unique positive and 0 negative real root : $\sqrt[n]{a}$
a < 0	0 real roots	0 positive and 1 negative real root : $-\sqrt[n]{a}$
a=0	1 real root : $\sqrt[n]{0}=0$	1 real root : $\sqrt[n]{0}=0$

Calculation:

• Given $\sqrt{x^{-1}{y}^{-2}}$
• $\begin{array}{l}\sqrt{x^{-1}{y}^{-2}}\\ S\text{ubstitute x = 4 , y = 3:}\\ =\sqrt{4^{-1}\cdot 3^{-2}}\\ =\frac{1}{\sqrt{4^1\cdot 3^2}}\\ =\frac{1}{\sqrt{2^2\cdot 3^2}}\\ =\frac{1}{\sqrt{6^2}}\\ =\frac{1}{6}\end{array}$
• So, $\frac{1}{6}$ is the solution.