Q. The roots of the equation $(q - r)x^{2} + (r - p)x + (p - q) = 0$ are
A) $\frac{r - p}{q - r},\ 1$
B) $\frac{p - q}{q - r},\ 1$
C) $\frac{p - r}{q - r},\ 2$
D) $\frac{q - r}{p - q},\ 2$
Explanation
(b)
$(q - r)x^{2} + (r - p)x + (p - q) = 0$
$\Rightarrow \ \ (q - r)x^{2} + (r - q + q - p)x + (p - q) = 0$
$\Rightarrow (q - r)x^{2} - (q - r)x - (p - q)x + (p - q) = 0$
$\Rightarrow (q - r)x(x - 1) - (p - q)(x - 1) = 0$
$\Rightarrow (x - 1)\{(q - r)x - (p - q)\} = 0$
$\Rightarrow x = 1,\ \frac{p - q}{q - r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.