Question

If $\frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}$ then find the values of following ratios.

(i) $\frac{a}{b}$ (ii) $\frac{7a-3b}{7a+3b}$ (iii) $\frac{b^2-2a^2}{b^2+2a^2}$ (iv) $\frac{b^3-2a^3}{b^3+2a^3}$

Answer 1

(i)
$\frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}$

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{\left(15a^2+4b^2\right)+\left(15a^2-4b^2\right)}{\left(15a^2+4b^2\right)-\left(15a^2-4b^2\right)}=\frac{47+7}{47-7} \\ \Rightarrow \frac{15a^2+4b^2+15a^2-4b^2}{15a^2+4b^2-15a^2+4b^2}=\frac{54}{40} \\ \Rightarrow \frac{30a^2}{8b^2}=\frac{27}{20} \\ \Rightarrow \frac{a^2}{b^2}=\frac{27\times 8}{20\times 30}=\frac{9}{25} \\ \Rightarrow \frac{a}{b}=\sqrt{\frac{9}{25}}=\frac{3}{5} \end{gathered}$$
(ii)
$$\begin{gathered} \frac{7a-3b}{7a+3b} \\ =\frac{{\displaystyle \frac{7a}{b}}-{\displaystyle \frac{3b}{b}}}{{\displaystyle \frac{7a}{b}}+{\displaystyle \frac{3b}{b}}} \\ =\frac{7\times {\displaystyle \frac{3}{5}}-3}{7\times {\displaystyle \frac{3}{5}}+3}\left(\frac{a}{b}=\frac{3}{5}\right) \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle \frac{21-15}{5}}}{{\displaystyle \frac{21+15}{5}}} \\ =\frac{6}{36} \\ =\frac{1}{6} \end{gathered}$$
(iii)
$$\begin{gathered} \frac{b^2-2a^2}{b^2+2a^2} \\ =\frac{{\displaystyle \frac{b^2}{b^2}}-{\displaystyle \frac{2a^2}{b^2}}}{{\displaystyle \frac{b^2}{b^2}}+{\displaystyle \frac{2a^2}{b^2}}} \\ =\frac{1-2\times {\displaystyle \frac{9}{25}}}{1+2\times {\displaystyle \frac{9}{25}}}\left(\frac{a}{b}=\frac{3}{5}\Rightarrow \frac{a^2}{b^2}=\frac{9}{25}\right) \\ =\frac{{\displaystyle \frac{25-18}{25}}}{{\displaystyle \frac{25+18}{25}}} \end{gathered}$$
$=\frac{7}{43}$

(iv)
$$\begin{gathered} \frac{b^3-2a^3}{b^3+2a^3} \\ =\frac{{\displaystyle \frac{b^3}{b^3}}-{\displaystyle \frac{2a^3}{b^3}}}{{\displaystyle \frac{b^3}{b^3}}+{\displaystyle \frac{2a^3}{b^3}}} \\ =\frac{1-2\times {\displaystyle \frac{27}{125}}}{1+2\times {\displaystyle \frac{27}{125}}}\left(\frac{a}{b}=\frac{3}{5}\Rightarrow \frac{a^3}{b^3}=\frac{27}{125}\right) \\ =\frac{{\displaystyle \frac{125-54}{125}}}{{\displaystyle \frac{125+54}{125}}} \end{gathered}$$
$=\frac{71}{179}$