Chapter 7.5, Problem 3ACYP

To determine

To Simplify:

  $6\sqrt{8c^3{d}^5}\cdot 4\sqrt{2c{d}^3}$

Answer to Problem 3ACYP

  $96c^2{d}^4\text{ is the principal root}\text{.}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $6\sqrt{8c^3{d}^5}\cdot 4\sqrt{2c{d}^3}$

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

  $\begin{array}{l}\text{For any real numbers a and b}\ne \text{0 and any integer}\\ \text{n >1,}\\ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}},if\text{ all roots are defined}\text{.}\end{array}$

• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

Calculation:

• Given $6\sqrt{8c^3{d}^5}\cdot 4\sqrt{2c{d}^3}$
• $\begin{array}{l}6\sqrt{8c^3{d}^5}\cdot 4\sqrt{2c{d}^3}\\ =24\sqrt{8c^3{d}^5}\cdot \sqrt{2c{d}^3}\\ =24\sqrt{8c^3{d}^5\cdot 2c{d}^3}\mathrm{.....................}\text{[Product property of radicals]}\\ \text{=}24\sqrt{16c^4{d}^8}\mathrm{....................}[a^{x+y}=a^x\cdot a^y]\\ =24\sqrt{4^2\cdot {\left(c^2\right)}^2\cdot {\left(d^4\right)}^2}\mathrm{...................}[\text{Factor into squares]}\\ =24\cdot 4c^2{d}^4\\ =96c^2{d}^4\end{array}$
• Since , the index is 2 , which is even and the result has no odd exponents , so we need not use the absolute value symbol to find the principal root.
• So, $\begin{array}{l}96c^2{d}^4\text{ is the principal root }\text{.}\\ -96c^2{d}^4\text{ is the other root}\text{.}\end{array}$