Chapter 7.7, Problem 75HP

To determine

To Explain :

Determine whether the given statement is always , never or sometimes true for $x^{\frac{1}{n}}=a$ .

Answer to Problem 75HP

Sometimes true

Explanation of Solution

Given Statement:

 If n is even, there will be extraneous solutions.

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

  $\begin{array}{l}Given\\ x^{\frac{1}{n}}=a\\ \Rightarrow \sqrt[n]{x}=a\\ \text{n is even:}\\ \text{The radical is defined for all non-negative values of the radicand}\text{.}\\ \text{That is , the equaion will have real solutions for all non-negative values of the radicand}\text{.}\\ \text{So, if the index is even and the radicand is negative, the equation will not have any real solutions, }\\ \text{or will have extraneous solutions}\text{.}\\ \text{Hence, the given statement that if index is even , the equation will have extraneous solutions is}\\ \text{true when the radicand is negative}\text{.}\\ \text{So, the statement is sometimes true}\text{.}\end{array}$