Question

The roots of the equation $a{x}^2+bx+c=0,a\ne 0,a,b,c\in R$ are non-real and $a+c<b$, then

• A. $4a+c=2b$
• B. $4a+c<2b$
• C. $4a+c>2b$
• D. $4a-c<2b$

Answer 1

Answer: (B)

$4a+c<2b$

Let $f\left(x\right)=a{x}^2+bx+c$
Roots of $f\left(x\right)=0$ are non-real .
$\therefore$ The graph of $f\left(x\right)$ does not cut the X-axis .
This means that $f\left(x\right)$ has the same sign for all $x$
Also, $a+c<b\Rightarrow a-b+c<0\Rightarrow f(-1)<0$

$\therefore f\left(x\right)<0$ for all $x$

$\therefore f(-2)<0\Rightarrow 4a-2b+c<0$

Hence, Option B is correct.