Let $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ be the roots of $a x^2+b x+c=0$ and $p x^2+q x+r=0$ respectively. If the system of equations $\alpha_1 y+\alpha_2 z=0, \beta_1 y+\beta_2 z=0$ has a non-trivial solution, then $\frac{b^2}{q^2}=$.

$\frac{a c}{p r}$

Since the system of equations possess non-trivial solution,

$\left|\begin{array}{ll} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{array}\right|=0 \Rightarrow \alpha_1 \beta_2-\alpha_2 \beta_1=0$

$\Rightarrow \frac{\alpha_1}{\beta_1}=\frac{\alpha_2}{\beta_2}=\frac{\alpha_1+\alpha_2}{\beta_1+\beta_2}=\left(\frac{\alpha_1 \alpha_2}{\beta_1 \beta_2}\right)^{1 / 2}$           .......(1)

Also $\alpha_1+\alpha_2=-\frac{b}{a}, \alpha_1 \alpha_2=\frac{c}{a}$

and $\beta_1+\beta_2=-\frac{q}{p}, \beta_1 \beta_2=\frac{r}{p}$

∴ equation (1)

$\Rightarrow \frac{-b / a}{-q / p}=\left(\frac{c / a}{r / p}\right)^{1 / 2} \Rightarrow \frac{b^2}{q^2}=\frac{a c}{p r}$

Hence (1) is the correct answer.