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?
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: $$\frac {|b|}{2a}\le|b|\le N<N+1$$ $$P(\pm(N+1))\ge a(N+1)^2-|b|(N+1)-|c|\ge(N+1)^2-N(N+1)-N=1>0.$$