Question

Find the values of a and b if
$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+b\sqrt{5}$

Answer 1

$$\begin{gathered} \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}} \\ =\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ =\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5} \\ =\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4} \end{gathered}$$
$$\begin{gathered} =\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4} \\ =\frac{4\sqrt{5}}{4} \\ =\sqrt{5} \end{gathered}$$
$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1\times \sqrt{5}$
Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.