Chapter CSB, Problem 2.37E

To determine

To simplify: the given square root.

Answer to Problem 2.37E

  $\frac{10}{\sqrt{7}+\sqrt{2}}=2\left(\sqrt{7}-\sqrt{2}\right)$ .

Explanation of Solution

Given:

The given square rootis $\frac{10}{\sqrt{7}+\sqrt{2}}$ .

Calculation:

As the given square root is $\frac{10}{\sqrt{7}+\sqrt{2}}$ .

Now, to simplify the square root, therefore,

  $\begin{array}{l}\frac{10}{\sqrt{7}+\sqrt{2}}=\frac{10}{\sqrt{7}+\sqrt{2}}\cdot \left(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}\right)\left\{\text{conjugate multiplication}\right\}\\ \frac{10}{\sqrt{7}+\sqrt{2}}=\frac{10\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}\left\{\text{simplify}\right\}\\ \{\text{expand as}\left(a+b\right)\left(a-b\right)=a^2-b^2\}\\ \therefore \frac{10}{\sqrt{7}+\sqrt{2}}=\frac{10\left(\sqrt{7}-\sqrt{2}\right)}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{2}\right)}^2}\\ \Rightarrow \frac{10}{\sqrt{7}+\sqrt{2}}=\frac{10\left(\sqrt{7}-\sqrt{2}\right)}{7-2}\\ \Rightarrow \frac{10}{\sqrt{7}+\sqrt{2}}=\frac{10}{5}\left(\sqrt{7}-\sqrt{2}\right)\\ \therefore \frac{10}{\sqrt{7}+\sqrt{2}}=2\left(\sqrt{7}-\sqrt{2}\right)\end{array}$

Hence, the simplification of the given function is $\frac{10}{\sqrt{7}+\sqrt{2}}=2\left(\sqrt{7}-\sqrt{2}\right)$ .