Chapter 6.4, Problem 6CYU

To determine

To Simplify:

  $\sqrt[6]{64{(2y+1)}^{18}}$

Answer to Problem 6CYU

  $2\left|{(2y+1)}^3\right|$ is the principal root.

Explanation of Solution

Given Information:

Given expression to simplify:

  $\sqrt[6]{64{(2y+1)}^{18}}$

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 and hence is the principal root .
• 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 $\sqrt[6]{64{(2y+1)}^{18}}$
• $\sqrt[6]{64{(2y+1)}^{18}}=\sqrt[6]{{[2{(2y+1)}^3]}^6}=2{(2y+1)}^3$
• Since the index 6 is even and the exponent of the result 3 is odd . So, we must take absolute value to find the principal root.
• So, $\sqrt[6]{64{(2y+1)}^{18}}=2\left|{(2y+1)}^3\right|$ is the principal root.
• The other root is $-2\left|{(2y+1)}^3\right|$