Question

lf the equation $a{x}^2+2bx-3c=0$ has non-real roots and $\frac{3c}{4}<a+b$, then $c$ is always

• A. $<0$
• B. $>0$
• C. $\ge 0$
• D. $0$

Answer 1

Answer: (A)

$<0$

Say $f\left(x\right)=a{x}^2+2bx-3c$
The inequality given in the question is equivalent to $f\left(2\right)>0$
Since equation has non real roots, $f\left(x\right)>0$ for all $x$
Hence, $f\left(0\right)>0$
So $-3c>0$ which implies $c<0$