If the ratio of the roots of equation $ax^{2} + 2bx + c = 0$ is same as the ratio of the roots of eq

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If the ratio of the roots of equation $ax^{2} + 2bx + c = 0$ is same as the ratio of the roots of equation $px^{2} + 2qx + r = 0$, where $a,b,c,p,q,r$ are non zero real numbers. Then the value of $\left( \frac{b^{2}}{q^{2}} \right)\left( \frac{p}{a} \right)\left( \frac{r}{c} \right)$ is equal to:

Accepted Answer

$$\begin{matrix} & ax^{2} + 2bx + c = 0\int_{\beta}^{\alpha}\mspace{2mu}\mspace{2mu} px^{2} + 2qx + r = 0\int_{\delta}^{\gamma}\mspace{2mu}\mspace{2mu} \Rightarrow \frac{\alpha}{\gamma} = \frac{\beta}{\delta} = \frac{\alpha + \beta}{\gamma + \delta} \\ & \frac{\alpha}{\beta} = \frac{\gamma}{\delta} \\ & \text{~Also~}\ \frac{\alpha\beta}{\gamma\delta} = \left( \frac{\alpha + \beta}{\gamma + \delta} \right)^{2} \\ & \ \Rightarrow \ \frac{c/a}{r/p} = \frac{4b^{2}\left( p^{2} \right)}{a^{2}\left( 4q^{2} \right)}\ \Rightarrow \ \left( \frac{b^{2}}{q^{2}} \right)\left( \frac{p}{a} \right)\left( \frac{r}{c} \right) = 1 \end{matrix}$$

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