If $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Say $a,b,c$ are integers, $a>0$. Suppose $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Find an example for $a=5$. 
I am struggling with this for some time with no success.  
I try Vieta's formula $0<x_1+x_2 = -{b\over a}<2$ and $x_1x_2 ={c\over a}<1$ so $c<a$ and $0<-b<2a$.
 A: Let $z=-b$. We know that the minimum is going to be at $(\frac{z}{2a},c-\frac{z^2}{4a})$. (For proof plug $\frac{z}{2a}$, the extremum, into $ax^2-zx+c$). We must find values of $a,b,c$ such that $a>c>0,z>0,a+c>z,c-\frac{z^2}{4a}<0$. Rewriting the last one, we get $z^2>4ac$. However, because everything is in integers we can change $A>B$ to $A\geq B+1$. We get $a>c\geq 1,z\geq 1,a+c\geq z+1,z^2\geq 4ac+1$. Let $w=z+1$, and we see that $c\geq 1,a+c\geq w,\frac{z^2-1}{4}=\frac{(z+1)(z-1)}{4}=\frac{w^2-2w}{4}\geq ac$.
We see that these form the boundaries of the region of solutions in the $a$-$c$ plane. In order to have a non-empty solution set, the intersection of $a+c\geq w$ and $\frac{w^2-2w}{4}\geq ac$ must be above the line $c=1$. We do the following:
$$
a=-c+w=\frac{w^2-2w}{4c}\\
-c^2+wc-\frac{w^2-2w}{4}=0\\
c=\frac{-w \mp \sqrt{w^2-4(-1)(-\frac{w^2-2w}{4})}}{-2}=\frac{w \pm \sqrt{w^2-(w^2-2w)}}{2}=\frac{w \pm \sqrt{2w}}{2}.
$$
Because the original equations were symmetric in $a$ and $c$, we know that they will share these roots. As $a>c$, $a$ will take the $+$ and $c$ will take the $-$, so $c=\frac{w-\sqrt{2w}}{2}$.
This means that $c=\frac{w-\sqrt{2w}}{2}\geq 1$, which we now simplify.
$$
\frac{w-\sqrt{2w}}{2}=1\\
w-\sqrt{2w}=2\\
w-2=\sqrt{2w}\\
z-1=\sqrt{2z+2}\\
(z-1)^2=z^2-2z+1=2z+2\\
z^2-4z-1=0\\
z\geq \frac{4+\sqrt{4^2-4*(-1)*1}}{2}=2+\sqrt{4-(-1)}=2+\sqrt{5}\approx 4.236\\
z\geq 5
$$
When $z=5$ (or $b=-5$) and $c=1$, you get that $a=5$. If you increase in z, c grows so slowly that a is forced to grow faster in order to meet up with z. This means $a$ can never get lower than $5$, and is only $5$ when $b=-5$ and $c=1$. 
