Mathematics - Quadratic Equation Question with Solution | TestHub

Given equation is
$(p-q)x^2+(q-r)x+(r-p)=0$
By using formula for finding the roots viz: $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we get
$\begin{array}{rl}& x=\frac{(r-q)\pm \sqrt{(q-r{)}^2-4(r-p)(p-q)}}{2(p-q)}\\ & \Rightarrow x=\frac{(r-q)\pm (q+r-2p)}{2(p-q)}=\frac{r-p}{p-q},1\end{array}$