Q. If $a < b,$ then the solution $x^{2} + (a + b)x + ab < 0$ is given by
A) $(a,b)$
B) $( - \infty,a) \cup (b,\infty)$
C) $( - b, - a)$
D) $( - \infty, - b) \cup ( - a,\infty)$
Explanation
(c)
We have, $x^{2} + (a + b)x + ab < 0$
$\Rightarrow (x + a)(x + b) < 0$
$\Rightarrow - b < x < - a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.