Question

Find the rational values of a and b from each of the following:
(a) $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+b\sqrt{15}$
(b) $\frac{\sqrt{2}-\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=a+b\sqrt{6}$
(c) $\frac{4-3\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}$

Answer 1

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\Rightarrow \frac{5+3+2\sqrt{15}}{2}\Rightarrow 4+\sqrt{15}\therefore a=4,b=1$

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}\Rightarrow \frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{18-12}\Rightarrow \frac{6+2\sqrt{6}+3\sqrt{6}+6}{6}\Rightarrow 2+\frac{5}{6}\sqrt{6}\therefore a=2,b=\frac{5}{6}$

$\frac{\sqrt{4}+3\sqrt{5}}{\sqrt{4}-3\sqrt{5}}\times \frac{\sqrt{4}+3\sqrt{5}}{\sqrt{4}+3\sqrt{5}}\Rightarrow \frac{16+45+24\sqrt{5}}{16-45}\Rightarrow \frac{61+24\sqrt{5}}{-29}\therefore a=-\frac{61}{29},b=-\frac{24}{29}$