# 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?

• One way to do it is the following: the minimum happens at $-b/2a$. This must be in (0,1/2). So if $0 < a < 6$, this condition implies there are only finitely many possibilities for $a$ and $b$. Just check each one. [Each has only finitely many possibilities for $c$.] – Pat Devlin Nov 12 '16 at 2:08
• @PatDevlin Thanks: in an actual contest, five minutes after the above observations, I would have done that. I'm looking for more general/elegant solutions however. – shardulc says Reinstate Monica Nov 12 '16 at 2:15
• For a quicker shortcut, note that $0 \lt \alpha \beta \lt \frac{1}{4}$ implies $a \ge 4 c + 1 \ge 5$ and $a=5$ can be excluded fairly easily by brute force given the other constraints, which then leaves $a \ge 6$. – dxiv Nov 12 '16 at 8:10

\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.
• As a is maximized when c is 1, let's set c=1 "Maximized" in relation to what? The conclusion is correct, of course, but this step is not clear to me. – dxiv Nov 12 '16 at 6:37
• @dxiv I think we're making the assumption that $b$ and $c$ can be independently manipulated, and $a$ will be largest when $c$ is smallest in the mentioned relation. The independence is not true because $b^2 - 4ac > 0$, but if $c = 1$ leads to an appropriate choice of $a$ and $b$, then we're done. – shardulc says Reinstate Monica Nov 12 '16 at 16:50
• Sorry, it's still not clear to me. Actually I am not sure why one would try to maximize $a$ when in fact looking for a lowest bound on $a$. That said, the core idea still works if you leave $c$ in all the way to the end, which gives $a^2 - 4ca-8c \ge 0$ then $a \ge 2c + 2\sqrt{c^2 + 2c} \ge 2 + 2 \sqrt{3}$. – dxiv Nov 12 '16 at 18:50
• We would maximize $a$ because we know that we are comparing it to $a^2$ later; a large value of $a$ would mean a large lower bound for $a^2$, and hence for $a$. This works in part because $a^2$ has a higher degree than $a$. I like your method with $c$ slightly better, actually :) – shardulc says Reinstate Monica Nov 12 '16 at 20:39
• Guys, the reason behind I said "$a$ is maximized" pretty much aligns with what shardulc is saying. For example, $a < \frac{a^2}4$ would give us $a>4$ and $a < \frac{a^2}8$ would give us $a>8$, so actually what I wanted to say was "the lower-bound of $a$ is minimized when the upper-bound of this inequality is maximized" but somehow I ended up with just saying that way. At that time I was still struggling to put all those together to get the right answer. I agree that @dxiv's method of keeping $c$ until the end would be more clear. – Kay K. Nov 13 '16 at 1:27