Chapter 6.4, Problem 32WE

To determine

To Simplify:

  $\frac{2\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$

Answer to Problem 32WE

  $\frac{17-3\sqrt{21}}{4}$

Explanation of Solution

Given:

The expression is :

  $\frac{2\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$

Formula Used:

• $(a+b)(a-b)=a^2-b^2$
• Product property of Radicals:
$\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$

Calculation:

Given:

  $\begin{array}{l}\frac{2\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\\ =\frac{2\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\cdot \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}\mathrm{.......}\left[\begin{array}{l}\text{rationalize the denominator by multiplying and dividing by}\\ \text{the conjugate of the denominator}\text{.}\end{array}\right]\\ =\frac{\left(2\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\\ =\frac{\left(2\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{3}\right)}^2}\mathrm{.................}[\left(a-b\right)\left(a+b\right)=a^2-b^2]\\ =\frac{2\sqrt{7}\cdot \sqrt{7}+2\sqrt{7}\cdot \left(-\sqrt{3}\right)+\left(-\sqrt{3}\right)\cdot \sqrt{7}+\left(-\sqrt{3}\right)\cdot \left(-\sqrt{3}\right)}{7-3}\mathrm{................}[\text{Use FOIL method to multiply the numerator]}\\ =\frac{14-2\sqrt{21}-\sqrt{21}+3}{4}\mathrm{.........................}\left[\text{Product property of radicals}\right]\\ =\frac{17-3\sqrt{21}}{4}\mathrm{.......................................}[\text{Combine like radicals]}\end{array}$