Question

If the equation $a{x}^2+bx+c=0$, $a,b,c\in R$ have non-real roots, then

• A. $c(a-b+c)>0$
• B. $c(a+b+c)>0$
• C. $c(4a-2b+c)>0$
• D. None of the above.

Answer 1

Answer: (A), (B), (C)

The correct options are
A $c(a-b+c)>0$
B $c(a+b+c)>0$
C $c(4a-2b+c)>0$

Since the roots of ${ax}^2+bx+c=0$ are non real, so $f\left(x\right)={ax}^2+bx+c$ will have same sign for every value of $x$.

Hence,

$f\left(0\right)=cf\left(1\right)=a+b+cf(-1)=a-b+cf(-2)=4a-2b+c\therefore c(a+b+c)>0,c(a-b+c)>0,c(4a-2b+c)>0$