Prove that if $ac + bc + c^2 < 0$ then equation (usual notation) has two roots

Question

We have $a, b, c$ real parameters, $a ≠ 0$.

Prove that $ax^2 + bx + c = 0$ has two different roots ($b^2-4ac > 0$), if $ac + bc + c^2 < 0$

Answer 1

If $P(x) = ax^2+bx+c$, from $c(a+b+c)<0$, we have $P(0)P(1) < 0$. So there is a root in $(0, 1)$. Further, note it cannot be a double root, as then the sign wouldn't have changed...

Answer 2

Hint $$ac+bc+c^2=c(a+b+c)=f(0)f(1)<0$$ where $f(x)=ax^2+bx+c$.

Answer 3

$ac+bc+c^2<0\Rightarrow -ac>bc+c^2$. Now $\Delta=b^2-4ac>b^2+4bc+4c^2=(b+2c)^2>0\Rightarrow$ there are $2$ distinct real roots.