Question

If roots of $a{x}^2+2bx+c=0$ $(a\ne 0)$ are complex and $a+c<2b$, then

• A. $c>0$
• B. $c<0$
• C. $4a+c<4b$
• D. $4a+c>4b$

Answer 1

Answer: (B), (C)

The correct options are
B $c<0$
C $4a+c<4b$

Roots of $a{x}^2+2bx+c=0$ are non-real.
$\therefore f\left(x\right)=a{x}^2+2bx+c>0$ or $<0\left(\forall x\right)$
But, $f(-1)=a-2b+c<0$
$\therefore f\left(0\right)$ and $f(-2)$ must be less than zero.
$f\left(0\right)<0\Rightarrow c<0$
and $f(-2)<0\Rightarrow 4a+c<4b$.