Inequality on Quadratic Equation Coefficients 
Let $a \in \mathbb{Z}_{>0}$ and $b, c \in \mathbb{Z}$, such that $$ax^2 + bx + c = 0$$ has two different solutions in the interval $(0, \frac{1}{2}]$. Prove that $a \geq 6$.

My work consists of just some observations:


*

*if the roots are $\alpha$ and $\beta$, then $\alpha + \beta = \frac{-b}{a} \in (0, 1)$

*from the quadratic equation, $\alpha - \beta = \frac{\sqrt{b^2 - 4ac}}{a}$ (WLOG assuming $\alpha > \beta$) which gives $\frac{\sqrt{b^2 - 4ac}}{a} \in (0, \frac{1}{2}) \implies \frac{a^2}{4} > b^2 - 4ac$

*graphically, for the roots to be close together, the curve has to be steep enough which is affected primarily by $a$ (not rigorous at all but intuitive to me). A further thought that occurred to me was that this question might be generalizable, for different intervals and bounds on $a$.


How do I solve this?

\begin{align}
&f(x):=ax^2+bx+c\\\\
&f(0)>0\quad\rightarrow\quad c>0\tag1\\
&f\left(\frac12\right)>0\quad\rightarrow\quad a+2b+4c>0\tag2\\
&0<-\frac{b}{2a}<\frac12\quad\rightarrow\quad b^2<a^2\tag3\\
&b^2-4ac>0\quad\rightarrow\quad a<\frac{b^2}{4c}\tag4\\\\
&\text{As $a$ is maximuzed when $c$ is 1, let's set $c=1$. Then, from (3) and (4),}\\
&a<\frac{b^2}4<\frac{a^2}4\\
&\text{As $a$ and $b$ are integers,}\\
&a+2\le\frac{b^2}4+1\le\frac{a^2}4\\
&a+2\le\frac{a^2}4\quad\rightarrow\quad a^2-4a-8\ge0\\
&a\ge2+2\sqrt3\approx 5.46\\
\therefore\space&a\ge6
\end{align}
You could generalize this, for example, by replacing $\frac12$ into $k$.
Also, I didn't use the condition (2), but it seems that it is not required to use it as we are only interested in the range of $a$ here.
