Let α_ 1 , α_ 2 and β_ 1 , β_ 2 be the roots of a x^ 2 +b x+c=0 and...

Since $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ are the roots of ax $^{2}+b x+c=0$ and $p x^{2}+q x+r=0$ respectively therefore \alpha_{1}+\alpha_{2}=$\frac{-b}{a}, \alpha_{1} \alpha_{2}=\frac{c}{a}$and $\beta_{1}+\beta_{2}=\frac{-q}{p}, \beta_{1} \beta_{2}=\frac{r}{p}$ Since the given system of equation has a non-trivial solution \therefore\left|$\begin{array}{ll} \alpha_{1}& \alpha_{2} \\ \beta_{1} & \beta_{2} \end{array}$\right|=0\text{ i.e. } \alpha_{1} \beta_{2}-\alpha_{2} \beta_{1}=$0 or$\quad \frac{\alpha_{1}}{\beta_{1}}=\frac{\alpha_{2}}{\beta_{2}}=\frac{\alpha_{1}+\alpha_{2}}{\beta_{1}+\beta_{2}}=\sqrt{\frac{\alpha_{1} \alpha_{2}}{\beta_{1} \beta_{2}}}\Rightarrow \frac{\mathrm{pb}}{\mathrm{qa}}=\sqrt{\frac{\mathrm{pc}}{\mathrm{ra}}} \Rightarrow \frac{\mathrmb^{2}}{\mathrmq^{2}}=\frac{\mathrm{ac}}{\mathrm{pr}}