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Q.
The roots of the given equation
$(p-q) x^{2}+(q-r) x+(r-p)=0$ are :
Complex Numbers and Quadratic Equations
Solution:
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}-4 ac}}{2a}$, we get
$x=\frac{(r-q) \pm \sqrt{(q-r)^{2}-4(r-p)(p-q)}}{2(p-q)}$
$\Rightarrow x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1$