Chapter CSB, Problem 2.36E

To determine

To simplify: the given square root.

Answer to Problem 2.36E

  $\frac{18}{6-\sqrt{2}}=\frac{18\sqrt{3}+3\sqrt{6}}{17}$ .

Explanation of Solution

Given:

The given square rootis $\frac{3\sqrt{3}}{\sqrt{6}-2}$ .

Calculation:

As the given square root is $\frac{3\sqrt{3}}{\sqrt{6}-2}$ .

Now, to simplify the square root, therefore,

  $\begin{array}{l}\frac{3\sqrt{3}}{\sqrt{6}-2}=\frac{3\sqrt{3}}{\sqrt{6}-2}\cdot \left(\frac{6+\sqrt{2}}{6+\sqrt{2}}\right)\left\{\text{conjugate multiplication}\right\}\\ \frac{3\sqrt{3}}{\sqrt{6}-2}=\frac{3\sqrt{3}\left(6+\sqrt{2}\right)}{\left(6-\sqrt{2}\right)\left(6+\sqrt{2}\right)}\left\{\text{simplify}\right\}\\ \{\text{expand as}\left(a+b\right)\left(a-b\right)=a^2-b^2\}\\ \therefore \frac{3\sqrt{3}}{\sqrt{6}-2}=\frac{18\sqrt{3}+3\sqrt{6}}{{\left(6\right)}^2-{\left(\sqrt{2}\right)}^2}\\ \Rightarrow \frac{3\sqrt{3}}{\sqrt{6}-2}=\frac{18\sqrt{3}+3\sqrt{6}}{36-2}\\ \Rightarrow \frac{18}{6-\sqrt{2}}=\frac{18\sqrt{3}+3\sqrt{6}}{17}\end{array}$

Hence, the simplification of the given function is $\frac{18}{6-\sqrt{2}}=\frac{18\sqrt{3}+3\sqrt{6}}{17}$ .