Question

If $\frac{7a^2+2b^2}{7a^2+2b^2}=\frac{113}{13}$, find the values of $\frac{a}{b}$

Answer 1

$$\begin{gathered} \frac{7a^2+2b^2}{7a^2-2b^2}=\frac{113}{13} \\ \therefore \frac{7a^2+2b^2+7a^2-2b^2}{7a^2+2b^2-7a^2+2b^2}=\frac{113+13}{113-13}(\mathrm{Componendo}\mathrm{and}\mathrm{dividendo}) \\ =\frac{14a^2}{4b^2}=\frac{126}{100} \\ =\frac{a^2}{b^2}=\frac{126}{100}\times \frac{4}{14} \\ =\frac{a^2}{b^2}=\frac{504}{1400} \\ =\frac{a^2}{b^2}=\frac{63}{175} \\ =\frac{a}{b}=\pm \frac{\sqrt{63}}{\sqrt{175}} \end{gathered}$$