Question

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

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

Answer 1

Answer: (B)

$4a+c<2b$

There seems to be typographical error in question. It reads '....a+b<b...'. It should read as '...a+b<c..'

Given roots of $a{x}^2+bx+c=0\&(a\ne 0)$ are complex and non-real,

$\Rightarrow$ expression $a{x}^2+bx+c=0$ will be either always positive or always negative in sign

Also given that a+b<c ---(i)

& $f(-1)=a-b+c$

$\Rightarrow f(-1)<0$

as $a-b+c<0$

$a+c<b$ (from (i))

$\Rightarrow a{x}^2+bx+c=0$ will always be negative

$\Rightarrow f(-2)<0$

$\Rightarrow 4a-2b+c<0$

$\Rightarrow 4a+c<2b$

Hence the answer is B