Verify that roots of the quadratic equation $\displaystyle (\mathrm{p}-\mathrm{q}) x^{2}+(\mathrm{q}-\mathrm{r}) x+(\mathrm{r}-\mathrm{p})=0$ are equal when $\displaystyle \mathrm{q}+\mathrm{r}=2 \mathrm{p}$.

Discriminant \(\displaystyle (\mathrm{D})=(\mathrm{q}-\mathrm{r})^{2}-4(\mathrm{p}-\mathrm{q})(\mathrm{r}-\mathrm{p})\) \[=(\mathrm{q}+\mathrm{r}-2 \mathrm{p})^{2} \] Substituting, \(\displaystyle \mathrm{q}+\mathrm{r}=2 \mathrm{p}\) \[\Rightarrow \mathrm{D}=(2 \mathrm{p}-2 \mathrm{p})^{2}=0 \] ∴ Roots of the given equation are equal.