Question

If $a{x}^2+bx+c,a,b,c\in R,a<0$ has no real zeros and $a-b+c<0$, then the value of $ac$

• A. $=0$
• B. $>0$
• C. $<-1$
• D. $<0$

Answer 1

The correct option is B

$>0$

$f\left(x\right)=a{x}^2+bx+c$

$f(-1)=a-b+c<0\Rightarrow a<0$

$f\left(0\right)=0+0+c<0$

$\Rightarrow ac>0$