Chapter 6.4, Problem 18PPS

To determine

To Simplify:

  $\sqrt{{(x+15)}^4}$

Answer to Problem 18PPS

  ${(x+15)}^2$ is the principle root.

Explanation of Solution

Given Information:

Given expression to simplify:

  $\sqrt{{(x+15)}^4}$

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 $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

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 : $\sqrt[n]{0}=0$

Pure Imaginary Numbers :

• Square roots of negative real numbers: for any positive real number b , $\sqrt{-b^2}=\sqrt{b^2}\sqrt{-1}$ , or $$ bi where
$\begin{array}{l}\sqrt{-1}=i\\ i=iota\end{array}$
• When you find an even root of an even power and the result is an odd power , you must use the absolute value of the result to ensure that the answer is nonnegative.
• If the index is odd, there is only one real root. Therefore, there is no principal root when index is odd , and absolute value symbol are not needed.

Calculation:

• Given $\sqrt{{(x+15)}^4}$
• $\sqrt{{(x+15)}^4}=\sqrt{{[{(x+15)}^2]}^2}={(x+15)}^2$
• Since , the result does not have odd power,no need to take absolute value to get the principal root.
• So, $\sqrt{{(x+15)}^4}={(x+15)}^2$ is the principle root.
• $-{(x+15)}^2$ is the other root.