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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 $0<-b<2a$.

Bernard
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nonuser
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    Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Jan 24 '18 at 20:56
  • Yes, that one does give full solution to this one. Thanks. – nonuser Jan 24 '18 at 21:02
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    Using the quadratic formula: $x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$ has two roots if $b^2 - 4ac>0$. And $0<\frac {-b \pm \sqrt {b^2 - 4ac}}{2a}<1$ What is the smallest value of $a$. – Doug M Jan 24 '18 at 21:17
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    By your Vieta inequalities, for a given value of $a$ there are only $a-1$ possible values of $c$ and $2a-1$ possible values of $b$.

    So this isn't that hard to brute force: there are only $$ 1(3)+2(5)+3(7)=34 $$ polynomials with $a < 5$ you have to check.

    (Of course, the linked solution is a lot more elegant...)

     – Micah Jan 24 '18 at 21:18
  • @Micah I tried to avoid that ...:) – nonuser Jan 24 '18 at 21:22

1 Answers1

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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 $c=1$.

Marcus Luebke
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