$\text{ The roots }\alpha \text{ and }\beta \text{ of the quadratic equation }$${\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=0\text{ are real and of opposite signs. The roots of}$$\alpha (x-\beta {)}^2+\beta (x-\alpha {)}^2=0\text{ are }$

$\begin{array}{l}\text{ We have }\alpha +\beta =-\frac{\text{q}}{\text{p}},\alpha \beta =\frac{\text{r}}{\text{p}}\\ \text{ Now the given equation}\\ \Rightarrow (\alpha +\beta )x^2-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0\\ \Rightarrow \left(-\frac{\text{q}}{\text{p}}\right){\text{x}}^2-4\frac{\text{r}}{\text{p}}\text{x}+\frac{\text{r}}{\text{p}}\left(-\frac{\text{q}}{\text{p}}\right)=0\\ \Rightarrow {\text{pqx}}^2+4\text{prx}+\text{rq}=0\\ \text{ Since }\alpha \text{ and }\beta \text{ have opposite signs, therefore p and r}\\ \text{ must have opposite signs. }\\ \Rightarrow \text{pq and rq must have opposite signs }\Rightarrow {\text{ roots of equation\hspace{0.17em}pqx}}^2+4\text{prx}+\text{rq}=0\text{ have opposite signs }\end{array}$