Q. The roots of the given equation $(p - q)x^{2} + (q - r)x + (r - p) = 0$ are

A) $\frac{p - q}{r - p},1$
B) $\frac{q - r}{p - q},1$
C) $\frac{r - p}{p - q},1$
D) $1,\frac{q - r}{p - q}$
Explanation

(c)
Given equation is
$(p - q)x^{2} + (q - r)x + (r - p) = 0\$
$\Rightarrow x = \frac{(r - q) \pm \sqrt{(q - r)^{2} - 4(r - p)(p - q)}}{2(p - q)}$
$= \frac{(r - q) \pm \sqrt{q^{2} + r^{2} - 2qr - 4(rp - rq - p^{2} + pq)}}{2(p - q)}$
$\Rightarrow x = \frac{(r - q) \pm (q + r - 2p)}{2(p - q)}$
$\Rightarrow x = \frac{r - p}{p - q},1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.