$\begin{aligned} & \mathrm{px} \mathrm{x}^2+\mathrm{qx}-\mathrm{r}=0<\beta \\ & \mathrm{p}=\mathrm{A}, \mathrm{q}=\mathrm{AR}, \mathrm{r}=\mathrm{AR}^2 \\ & A x^2+A R x-A R^2=0 \\ & \mathrm{x}^2+\mathrm{Rx}-\mathrm{R}^2=0<\beta \\ & \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\ & \therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-\mathrm{R}}{-\mathrm{R}^2}=\frac{3}{4} \Rightarrow \mathrm{R}=\frac{4}{3} \\ & (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=R^2-4\left(-R^2\right)=5\left(\frac{16}{9}\right) \\ & =80 / 9\end{aligned}$