Question

If $a{x}^2+bx+c,a,b,cϵR$, a < 0 has no real zeros and a - b+c<0, then the

value of ac

• A. > 0
• B. < 0
• C. = 0
• D. < -1

Answer 1

The correct option is A

> 0

f(X) = $a{x}^2+bx+c$

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

f(0) = 0 + 0 + c < 0

ac > 0