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So I was tasked with the following problem: Given that the following equation is quadratic and has a real root: $$ax^2+bx+c=0$$

Prove that if a,b,c $\in \mathbb{Z}$ and $|a|\leq 2011$,$|b|\leq 2011$,$|c|\leq 2011$ the root is in the interval $(-2012,2012)$

I am not sure how to solve this with calc and other related theorems (which is my task) but I have come up with a solution:

For the sake of contradiction, let's assume that the root : x does not belong in the above mentioned interval.
Therefore $ax^2 = -bx-c$ and $|a||x|^2 \leq |b||x|+|c|$. But we also know that $|a||x|^2\ge 2012|x|$ From which we derive that:
$2012|x|\leq |b||x|+|c|\leq 2011|x|+2011$ contradiction.

My question is how would we do this with calculus?

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  • $\begingroup$ Why $|a|x^2\ge 2012|x|$? (btw, probable misprints in "$a|x|^2\ge 2012|x|$", and in "contra[di]ction"). + What is your question exactly? (Please don't answer with a comment, rather edit your post.) $\endgroup$ Commented Mar 13, 2023 at 16:40
  • $\begingroup$ @AnneBauval it is obvious that |a|>1 since otherwise the quadratic wouldn't exist. Edit oops I didn't add that it is quadratic my bad $\endgroup$
    – Helixglich
    Commented Mar 13, 2023 at 16:48
  • $\begingroup$ Still, $|x|\ge2011,$ not $2012.$ $\endgroup$ Commented Mar 13, 2023 at 16:57
  • $\begingroup$ I think one of the roots of $x^2-2011 x-2011=0$ is larger than $2011$ $($in fact about $2011.9995)$ $\endgroup$
    – Henry
    Commented Mar 13, 2023 at 17:02
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    $\begingroup$ Hope I didn't waste too much of your time. I have mistaken the interval we were supposed to solve this problem for $\endgroup$
    – Helixglich
    Commented Mar 13, 2023 at 17:58

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You didn't answer my question whether you allow explicit resolution of quadratic equations, so let's admit you forbid it, and only use "calculus" in a strict sense.

Let $N=2011.$ Wlog $a>0,$ so the quadratic polynomial $P(x)=ax^2+bx+c$ is $>0$ outside its two (possibly equal) real roots, and $\le0$ at $-\frac b{2a}.$ In order to prove that these roots are in $(-N-1,N+1),$ we just have to check that $\frac {|b|}{2a}<N+1$ and $P(\pm(N+1))>0.$ This is feasible, but less expeditious than your method: $$\frac {|b|}{2a}\le|b|\le N<N+1$$ $$P(\pm(N+1))\ge a(N+1)^2-|b|(N+1)-|c|\ge(N+1)^2-N(N+1)-N=1>0.$$

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