Question

If $\frac{7a^2+2b^2}{7a^2-2b^2}=\frac{113}{13}$, find the modulus value of $\frac{a}{b}$ to the nearest integer

Answer 1

Given,
$\frac{7a^2+2b^2}{7a^2-2b^2}=\frac{113}{13}$
Applying componendo and dividendo,
$\frac{(7a^2+2b^2)+(7a^2-2b^2)}{(7a^2+2b^2)-(7a^2-2b^2)}=\frac{113+13}{113-13}$
$\therefore \frac{14a^2}{4b^2}=\frac{126}{100}$
$\frac{7a^2}{2b^2}=\frac{63}{50}$
$\frac{a^2}{b^2}=\frac{9}{25}$
$\therefore \frac{a}{b}=\sqrt{\frac{9}{25}}=\frac{3}{5}$ or $-\frac{3}{5}$