Let $\lambda \ne 0$ be a real number. Let $\alpha,\beta$ be the roots of the equation $14{x^2} - 31x + 3\lambda = 0$ and $\alpha,\gamma$ be the roots of the equation $35{x^2} - 53x + 4\lambda = 0$. Then ${{3\alpha } \over \beta }$ and ${{4\alpha } \over \gamma }$ are the roots of the equation

$14 x^{2}-31 x+3 \lambda=0$ <br/><br/> $$ \begin{aligned} & \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\ & 35 x^{2}-53 x+4 \lambda=0 \\\\ & \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\ & \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma \end{aligned} $$ <br/><br/> (1) $-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$ <br/><br/> $\frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5}$ <br/><br/> $\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2}$ <br/><br/> $\Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$ <br/><br/> $\Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5$ <br/><br/> so, sum of roots $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$<br/><br/>$$ \begin{aligned} & =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\\\ & =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5 \end{aligned} $$<br/><br/> Product of roots <br/><br/> $$ =\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^{2}}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49} $$ <br/><br/> So, required equation is $x^{2}-5 x+\frac{250}{49}=0$ <br/><br/> $\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0$