Question

The roots of $a{x}^2+bx+c=0$, where $a\ne 0,b,cϵR$ are non real complex and $a+c<b$. Then

• A. $4a+c>2b$
• B. $4a+c<2b$
• C. $4a+c=2b$
• D. None of these

Answer 1

The correct option is B $4a+c<2b$

Non real roots$\Rightarrow D<0$

$a+c<b$ given

$a+c-b<0$

$\Rightarrow f(-1)<0$

So $f(-2)$ will also be $<0$

$f(-2)=4a-2b+c<0$

$4a+c<2b$