Chapter CSB, Problem 2.33E

To determine

To simplify: the given square root.

Answer to Problem 2.33E

  $\frac{3}{2+\sqrt{2}}=\frac{3}{2}\left(2-\sqrt{2}\right)$ .

Explanation of Solution

Given:

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

Calculation:

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

Now, to simplify the square root, therefore,

  $\begin{array}{l}\frac{3}{2+\sqrt{2}}=\frac{3}{2+\sqrt{2}}\cdot \left(\frac{2-\sqrt{2}}{2-\sqrt{2}}\right)\left\{\text{conjugate multiplication}\right\}\\ \frac{3}{2+\sqrt{2}}=\frac{3\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\left\{\text{simplify}\right\}\\ \{\text{expand as}\left(a+b\right)\left(a-b\right)=a^2-b^2\}\\ \therefore \frac{3}{2+\sqrt{2}}=\frac{3\left(2-\sqrt{2}\right)}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}\\ \Rightarrow \frac{3}{2+\sqrt{2}}=\frac{3\left(2-\sqrt{2}\right)}{4-2}\\ \Rightarrow \frac{3}{2+\sqrt{2}}=\frac{3}{2}\left(2-\sqrt{2}\right)\end{array}$

Hence, the simplification of the given function is $\frac{3}{2+\sqrt{2}}=\frac{3}{2}\left(2-\sqrt{2}\right)$ .