Chapter 7.5, Problem 60HP

To determine

To Find :

Who among Twyla and Ben is correct.

Answer to Problem 60HP

Twyla is correct and Ben is wrong.

Explanation of Solution

Given Information:

Twyla and Ben are simplifying

  $4\sqrt{32}+6\sqrt{18}$

  

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

    $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• Like radical expressions : If the radicals have both the index and radicand identical , then they are like radicals.
• In order to add or subtract , the radicals must be like radicals.

Calculation:

• Given $4\sqrt{32}+6\sqrt{18}$
• $\begin{array}{l}4\sqrt{32}+6\sqrt{18}\\ =4\sqrt{4^2\cdot 2}+6\sqrt{3^2\cdot 2}\mathrm{..................}\text{[Factor using squares]}\\ \text{=4}\cdot \text{4}\cdot \sqrt{2}+6\cdot 3\cdot \sqrt{2}\\ =16\sqrt{2}+18\sqrt{2}\\ =(16+18)\sqrt{2}\sqrt{2}\mathrm{....................}[\text{Adding like radicals]}\\ \text{=34}\sqrt{2}\end{array}$
• So, $\text{34}\sqrt{2}$ is the solution.
• Twyla is correct.
• Ben is wrong.Ben’s mistakes was to multiply the 4 by 16 instead of 4 and multiplying the 6 by 9 instead of 3.