Chapter 7.5, Problem 5ACYP

To determine

To Simplify:

  $(6\sqrt{3}-5)(2\sqrt{5}+4\sqrt{2})$

Answer to Problem 5ACYP

  $\text{12}\sqrt{15}+24\sqrt{6}-10\sqrt{5}-20\sqrt{2}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $(6\sqrt{3}-5)(2\sqrt{5}+4\sqrt{2})$

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 $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• 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 $(6\sqrt{3}-5)(2\sqrt{5}+4\sqrt{2})$
• $\begin{array}{l}(6\sqrt{3}-5)(2\sqrt{5}+4\sqrt{2})\\ =6\sqrt{3}\cdot (2\sqrt{5}+4\sqrt{2})-5\cdot (2\sqrt{5}+4\sqrt{2})\mathrm{..............}\text{[Distributive property ]}\\ \text{=12}\sqrt{3}\cdot \sqrt{5}+24\sqrt{3}\cdot \sqrt{2}-10\cdot \sqrt{5}-20\cdot \sqrt{2}\mathrm{.....}[\text{Distributive property]}\\ \text{=12}\sqrt{15}+24\sqrt{6}-10\sqrt{5}-20\sqrt{2}\mathrm{............................}[\text{Product property of radicals]}\\ =\text{12}\sqrt{15}+24\sqrt{6}-10\sqrt{5}-20\sqrt{2}\mathrm{..............................}[\text{No like radicals , no cannot be simplified further]}\end{array}$
• So, $\text{12}\sqrt{15}+24\sqrt{6}-10\sqrt{5}-20\sqrt{2}$ is the answer.