Chapter 7, Problem 52SGR

To determine

To Simplify:

  $4\sqrt{6y}\cdot 3\sqrt{7x^{2}y}$

Answer to Problem 52SGR

  $12\left|xy\right|\sqrt{42}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $4\sqrt{6y}\cdot 3\sqrt{7x^{2}y}$

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

• 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:

  $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$

Calculation:

• Given $4\sqrt{6y}\cdot 3\sqrt{7x^{2}y}$

    $\begin{array}{l}4\sqrt{6y}\cdot 3\sqrt{7x^{2}y}\\ =12\sqrt{6y\cdot 7x^{2}y}\mathrm{...........}[\text{product property of radicals}]\\ =12\sqrt{42x^2{y}^2}\\ =12xy\sqrt{42}\\ \text{Since the index is 2, which is even and the power of 'xy' in the solution is 1, }\\ \text{which is odd , we need to apply the absolute value symbol on it }\text{.}\\ \text{The principal root is }12\left|xy\right|\sqrt{42}\text{ and the other root is }-12\left|xy\right|\sqrt{42}\end{array}$

  So, $4\sqrt{6y}\cdot 3\sqrt{7x^{2}y}=12\left|xy\right|\sqrt{42}$