# Prove location of roots

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?

• 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.) Mar 13 at 16:40
• @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 Mar 13 at 16:48
• Still, $|x|\ge2011,$ not $2012.$ Mar 13 at 16:57
• I think one of the roots of $x^2-2011 x-2011=0$ is larger than $2011$ $($in fact about $2011.9995)$ Mar 13 at 17:02
• Hope I didn't waste too much of your time. I have mistaken the interval we were supposed to solve this problem for Mar 13 at 17:58

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 $$\frac {|b|}{2a} and $$P(\pm(N+1))>0.$$ This is feasible, but less expeditious than your method: $$\frac {|b|}{2a}\le|b|\le N $$P(\pm(N+1))\ge a(N+1)^2-|b|(N+1)-|c|\ge(N+1)^2-N(N+1)-N=1>0.$$