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I decided to look at the graphs of the parabola to solve this problem. These are only two types that will fit this problem: parabolas that "open" upwards and parabolas that "open" downwards. These parabolas would never touch the $x$-axis, and this rules out the upwards opening parabola because $a+b+c>0$. However, I am not sure how to continue from here.

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Define $f(x) = ax^2 + bx + c$. If $f(0) = c \geqslant 0$, note that $f(1) = a + b + c < 0$, by continuity $f$ has a zero on $[0, 1)$.

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  • $\begingroup$ Is there a way to answer my question using the work I have so far? $\endgroup$ – Gerard L. Mar 10 '18 at 2:03
  • $\begingroup$ Now since it must be a downward-open parabola, if $c > 0$, it will cross the $x$-axis at some point by looking at the graph. $\endgroup$ – Saad Mar 10 '18 at 2:05
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By the quadratic formula a parabola has roots at $x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$ if such is a real number.

The only way for there not to be roots is if $b^2 - 4ac < 0$. Or if $b^2 < 4ac$.

If $c > 0$ then for this to happen we must have $0 < b^2 <4ac$ so $a > 0$ as well.

But $a+b+c < 0$ so $b$ must be negative. "More negative than $a,c$ are positive". In other words $b < -(a+c)<0$ or $|b| > |a+c|$.

So $b^2 > (a+c)^2 = a^2 + 2ac + c^2$

Now by AM-GM $a^2 + c^2 \ge 2 \sqrt{a^2c^2} = 2ac$

So $b^2 > a^2 + 2ac + c^2 > 4ac$ and thus real roots exist.

...

To get a grasp geometrically.

$ax^2 + bx + c = a(x^2 + \frac ba x) + c= a(x^2 + \frac ba x + \frac {b^2}{4a^2}) + c - \frac {b^2}{4a^2}) = a(x + \frac b{2a})^2 + c - \frac {b^2}{4a})$.

$a$ tells whether the parabola "goes up" or "goes down". $c - \frac {b^2}{4a})$ is the y value of the tip of the parabola.

to Not have roots either the tip is above the $x$-axis and it points up ($c > \frac {b^2}{4a}; a > 0$. Or the tip is below the $x$-axis and it points down ($c < \frac {b^2}{4a}; a < 0$).

In second case if $a < 0$ then $c< \frac {b^2}{4a} < 0$.

If the first case $c >0$ and $4ac > b^2$. ... which puts us where we were above.

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We have $ax^2+bx+c=0$ has no real roots.

Assume that $c=0$, we will have $\Delta=b^2-4ac<0$ or $b^2<0$, which is wrong.

Assume that $c>0$, we will have $\Delta=b^2-4ac<0 \Rightarrow 0\le b^2<4ac$ leads to $c>0$, $a>0$.

If $a+b+c<0$ is correct, then $0<a+c<-b \Rightarrow (-b)^2>(a+c)^2\ge 4ac$ (Cauchy inequality)

$\Rightarrow \Delta=b^2-4ac \ge 0$, contradiction.

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