Chapter 7.5, Problem 37PPS

To determine

To Simplify:

  $(7\sqrt{2}-3\sqrt{3})(4\sqrt{6}+3\sqrt{12})$

Answer to Problem 37PPS

  $\text{56}\sqrt{3}+42\sqrt{6}-36\sqrt{2}-54$

Explanation of Solution

Given Information:

Given expression to simplify:

  $(7\sqrt{2}-3\sqrt{3})(4\sqrt{6}+3\sqrt{12})$

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 $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• 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 $(7\sqrt{2}-3\sqrt{3})(4\sqrt{6}+3\sqrt{12})$
• $\begin{array}{l}(7\sqrt{2}-3\sqrt{3})(4\sqrt{6}+3\sqrt{12})\\ =7\sqrt{2}\cdot (4\sqrt{6}+3\sqrt{12})-3\sqrt{3}\cdot (4\sqrt{6}+3\sqrt{12})\mathrm{..............}\text{[Distributive property ]}\\ \text{=28}\sqrt{2}\cdot \sqrt{6}+21\sqrt{2}\cdot \sqrt{12}-12\sqrt{3}\cdot \sqrt{6}-9\sqrt{3}\cdot \sqrt{12}\mathrm{.....}[\text{Distributive property]}\\ \text{=28}\sqrt{12}+21\sqrt{24}-12\sqrt{18}-9\sqrt{36}\mathrm{............................}[\text{Product property of radicals]}\\ \text{=28}\sqrt{2^2\cdot 3}+21\sqrt{2^2\cdot 6}-12\sqrt{3^2\cdot 2}-9\sqrt{6^2}\mathrm{..................}[\text{Factor using squares]}\\ \text{=56}\sqrt{3}+42\sqrt{6}-36\sqrt{2}-\mathrm{54......................................}[\text{No like radicals , no cannot be simplified further]}\end{array}$
• So, $\text{56}\sqrt{3}+42\sqrt{6}-36\sqrt{2}-54$ is the answer.