Chapter 6.4, Problem 14WE

To determine

To Simplify:

  $\frac{10}{2\sqrt{3}-\sqrt{7}}$

Answer to Problem 14WE

  $4\sqrt{3}+2\sqrt{7}$

Explanation of Solution

Given:

The expression is :

  $\frac{10}{2\sqrt{3}-\sqrt{7}}$

Formula Used:

  $(a+b)(a-b)=a^2-b^2$

Calculation:

Given:

  $\begin{array}{l}\frac{10}{2\sqrt{3}-\sqrt{7}}\\ =\frac{10}{2\sqrt{3}-\sqrt{7}}\cdot \frac{2\sqrt{3}+\sqrt{7}}{2\sqrt{3}+\sqrt{7}}\mathrm{.......}\left[\begin{array}{l}\text{rationalize the denominator by multiplying and dividing by}\\ \text{the conjugate of the denominator}\text{.}\end{array}\right]\\ =\frac{10\left(2\sqrt{3}+\sqrt{7}\right)}{\left(2\sqrt{3}-\sqrt{7}\right)\left(2\sqrt{3}+\sqrt{7}\right)}\\ =\frac{10\left(2\sqrt{3}+\sqrt{7}\right)}{{\left(2\sqrt{3}\right)}^2-{\left(\sqrt{7}\right)}^2}\mathrm{.................}[\left(a-b\right)\left(a+b\right)=a^2-b^2]\\ =\frac{10\left(2\sqrt{3}+\sqrt{7}\right)}{12-7}\\ =\frac{10\left(2\sqrt{3}+\sqrt{7}\right)}{5}\\ =2\left(2\sqrt{3}+\sqrt{7}\right)\\ =4\sqrt{3}+2\sqrt{7}\end{array}$