Let $P(x)=ax^2-bx+c$ be a parabola. For this parabola to have two distinct roots in the interval $(0,1)$, $a,b,c$ must satisfy the following. $$P(0)>0,P(1)>0, \Delta>0, 1>r>0$$So we have $$c>0, a-b+c>0, b^2>4ac, 2a>b>0.$$Actually what this means is when we are given by an arbitrary number $a\in\mathbb{Z^+}$, we will scan the group of numbers $\{1,2,\cdots,2a-1\}$ to determine if when $b$ is given by a value in this group, one can find a integer $c$ such that $$c>0, c>b-a, b^2>4ac, 2a>b>0.$$With this mindset we can easily find the minimal value of $a$ possible, by just trying them all one by one, starting from $a=2$. (If $a=1$, then $b=1$, so $1>4c$. No solutions.) $a=2$. $4>b>0, b^2>8c, c>b-2$. $b=3$ gives $c\leq1$, $b=\{2,1\}$ gives $c\leq0$. No solutions. $a=3$. $6>b>0, b^2>12c, c>b-3$. $b=5$ gives $c\leq2$, $b=4$ gives $c\leq1$, $b=\{3,2,1\}$ gives $c\leq0$. No solutions. $a=4$. $8>b>0, b^2>16c, c>b-4$. $b=7$ gives $c\leq3$, $b=6$ gives $c\leq2$, $b=5$ gives $c\leq1$, $b=\{4,3,2,1\}$ gives $c\leq0$. No solutions. $a=5$. $10>b>0, b^2>20c, c>b-5$. For $b=5$, we have $c\leq1$, so for $a=5$, $b=5$, $c=1$ works just fine. So minimal value of $a$ should be $5$. The parabola $5x^2-5x+1$ has two roots $\frac{+5+\sqrt{5}}{10}$, $\frac{+5-\sqrt{5}}{10}$, and both of them belongs to the interval $(0,1)$, as desired.