Chapter 7.5, Problem 3BCYP

To determine

To Simplify:

  $2\sqrt[4]{8x^3{y}^2}\cdot 3\sqrt[4]{2x^5{y}^2}$

Answer to Problem 3BCYP

  $12x^2\left|y\right|\text{ is the principal root}\text{.}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $2\sqrt[4]{8x^3{y}^2}\cdot 3\sqrt[4]{2x^5{y}^2}$

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 $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• 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.
• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$

Calculation:

• Given $2\sqrt[4]{8x^3{y}^2}\cdot 3\sqrt[4]{2x^5{y}^2}$
• $\begin{array}{l}2\sqrt[4]{8x^3{y}^2}\cdot 3\sqrt[4]{2x^5{y}^2}\\ =6\sqrt[4]{8x^3{y}^2}\cdot \sqrt[4]{2x^5{y}^2}\\ =6\sqrt[4]{8x^3{y}^2\cdot 2x^5{y}^2}\mathrm{.....................}\text{[Product property of radicals]}\\ \text{=}6\sqrt[4]{16x^8{y}^4}\mathrm{...............................}[a^{x+y}=a^x\cdot a^y]\\ =6\sqrt[4]{2^4{\left(x^2\right)}^4{y}^4}\mathrm{...................}[\text{Factor into power 4]}\\ =6\cdot 2x^{2}y\mathrm{........................}[Simplify]\\ =12x^2\left|y\right|\end{array}$
• Since , the index is 4 , which is even and the result has odd exponents , so we need to use the absolute value symbol to find the principal root.
• So, $\begin{array}{l}12x^2\left|y\right|\text{ is the principal root }\text{.}\\ -12x^2\left|y\right|\text{ is the other root}\text{.}\end{array}$