The square trinomial $y=ax^2+ bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$ . The square trinomial   $y=ax^2 + bx + c$   has no roots and   $a + b + c > 0$. Find the sign of the coefficient $c$.
I'm having difficulties with this problem. 
What I've tried:
I realized that a quadratic equation doesn't have roots if the discriminant $b^2 - 4ac < 0$, so I've tried to combine that with the condition $a + b + c > 0 <=> a > -b -c $, but that didn't help me that much. 
I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .
 A: Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
A: That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x \in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
A: It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
A: Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
A: Short answer:
Perforce
$$\text{sgn}(y(0))=\text{sgn}(y(1)).$$
