Question

If the equation $a{x}^2+bx+c=0,a,b,c\in R$ has 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 these

Answer 1

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

$c(a-b+c)>0$
$c(a+b+c)>0$
$c(4a-2b+c)>0$

Since, equation $f\left(x\right)=a{x}^2+bx+c=0$ has non-real roots.
Therefore, $f\left(x_1\right)f\left(x_2\right)>0$ where, $x_1,x_2\in R$
$f\left(0\right)f\left(1\right)=c(a+b+c)>0$
$f\left(0\right)f(-1)=c(a-b+c)>0$
$f\left(0\right)f(-2)=c(4a-2b+c)>0$
Ans: A,B,C