1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $c$.] $\endgroup$ – Pat Devlin Nov 12 '16 at 2:08
  • $\begingroup$ @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. $\endgroup$ – shardulc says Reinstate Monica Nov 12 '16 at 2:15
  • 1
    $\begingroup$ 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 $a \ge 6$. $\endgroup$ – dxiv Nov 12 '16 at 8:10
3
$\begingroup$

\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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – dxiv Nov 12 '16 at 6:37
  • 1
    $\begingroup$ @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. $\endgroup$ – shardulc says Reinstate Monica Nov 12 '16 at 16:50
  • $\begingroup$ 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 $a \ge 2c + 2\sqrt{c^2 + 2c} \ge 2 + 2 \sqrt{3}$. $\endgroup$ – dxiv Nov 12 '16 at 18:50
  • $\begingroup$ 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 :) $\endgroup$ – shardulc says Reinstate Monica Nov 12 '16 at 20:39
  • 1
    $\begingroup$ 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. $\endgroup$ – Kay K. Nov 13 '16 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.