Q. Suppose the quadratic equations $x^{2} + px + q = 0$ and $x^{2} + rx + s = 0$ are such that $p,\ q,\ r,\ s$ are real and $pr = 2(q + s)$. Then
A) Both the equations always have real roots
B) At least one equation always has real roots
C) Both the equation always have non-real roots
D) At least one equation always has real and equal roots
Explanation
(b)
Let the discriminant of the equation $x^{2} + px + q = 0$ is $D_{1},\ then\ D_{1} = p^{2} - 4q$
and the discriminant of the equation $x^{2} + rx + s = 0\ is\ D_{2},\ then\ D_{2} = r^{2} - 4s$
$\therefore\ D_{1} + D_{2} = p^{2} + r^{2} - 4(q + s) = p^{2} + r^{2} - 2pr$ (from the given relation)
$\Rightarrow \ D_{1} + D_{2} = (p - r)^{2} \geq 0$
Clearly, at least one of $D_{1}$ and $D_{2}$ must be non-negative, consequently at least one of the equation has real roots.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.