Chapter 6, Problem 5CR

To determine

To Simplify:

  $\sqrt{6}\left(\frac{\sqrt{2}}{2}+\sqrt{3}\right)-\sqrt{8}$

Answer to Problem 5CR

(a)

Explanation of Solution

Given :

The expression is :

  $\sqrt{6}\left(\frac{\sqrt{2}}{2}+\sqrt{3}\right)-\sqrt{8}$

  $\begin{array}{l}a.\sqrt{3}+\sqrt{2}\\ b.\sqrt{3}+5\sqrt{2}\\ c.\sqrt{5}\\ d.2\sqrt{3}+\sqrt{2}\end{array}$

Concept Used:

Product property of Radicals:

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

Calculation:

Given:

  $\begin{array}{l}\sqrt{6}\left(\frac{\sqrt{2}}{2}+\sqrt{3}\right)-\sqrt{8}\\ =\sqrt{6}\left(\frac{\sqrt{2}+2\sqrt{3}}{2}\right)-\sqrt{8}\\ =\frac{\sqrt{6}\cdot \sqrt{2}+\sqrt{6}\cdot 2\sqrt{3}}{2}-\sqrt{8}\mathrm{........}[\text{distribute }\sqrt{6}]\\ =\frac{\sqrt{6\cdot 2}+2\sqrt{6\cdot 3}}{2}-\sqrt{8}\mathrm{...............}[\text{product property of radicals]}\\ \text{=}\frac{\sqrt{3\cdot 2^2}+2\sqrt{2\cdot 3^2}}{2}-\sqrt{8}\\ =\frac{2\sqrt{3}+6\sqrt{2}}{2}-\sqrt{8}\\ =\frac{2(\sqrt{3}+3\sqrt{2})}{2}-\sqrt{2^2\cdot 2}\\ =\sqrt{3}+3\sqrt{2}-2\sqrt{2}\\ =\sqrt{3}+\sqrt{2}\mathrm{...............................}[\text{combine like terms]}\end{array}$

So, (a) is the correct option.