Q. If the equation $ax^{2} + 2bx - 3c = 0$ has non-real roots and $(3c/4) < (a + b),$ then $c$ is
A) $< 0$
B) $> 0$
C) $\geq 0$
D) = 0
Explanation
(a)
Let $f(x) = ax^{2} + 2bx - 3c$ and $f(x) = 0$ has non-real roots, $f(x)$ will have the same sign for all values of $x$.
Also, $\frac{3c}{4} < (a + b) \Rightarrow 4a + 4b - 3c < 0$
$\Rightarrow f(2) > 0$
$\Rightarrow f(0) > 0$
$\Rightarrow c < 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.