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I understand that we would normally square both sides of an inequality in something like this: $|x|>|y|$, but from Andreescu & Andrica NT: SEP, there is this problem:

(Duke) What is the sum of all integers $n$ such that $n^2+2n+2$ divides $n^3 + 4n^2 + 4n − 14$?

It is solved with simple number theory and the division algorithm, but part of the solution involves this inequality: $|-2n-18| \ge |n^2+2n+2|$. Apparently, the solution range is $-4 \le n \le 4$. We could square both sides, but we would get a higher degree polynomial, and it might take a bit longer to find the solutions, something like $0 \ge n^4 +4n^3 -64n - 320$.

There are other threads like this one: double absolute value that suggest splitting into cases, but I'm not sure how I would do that for this, with the quadratic.

mpnm
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2 Answers2

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You might notice that $n^2+2n+2 = (n+1)^2+1$ which is always positive, so you can throw the right-hand absolute value away.

For inequalities, I prefer the split-point method which starts with solving the related equality. There are two cases.

Case 1: If $-2n-18$ is positive, the equation is

$$-2n-18 = n^2+2n+2$$

$$0 = n^2+4n+20 = (n+2)^2+16$$

which has no solutions.

Case 2: If $-2n-18$ is negative the equation is

$$2n+18 = n^2+2n+2$$

$$0 = n^2 - 16$$

which has solutions $n= \pm 4$. These are the split points and they cut the real line into three intervals. Each interval is either entirely in the solution set or entirely out of the solution set. So we test $n = -5$, $n=0$ and $n=5$ in the original inequality and discover that the solution set is the interval $-4<n<4.$

B. Goddard
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For a qualitative understanding of the location of the solutions, a graphical method is acceptable.

enter image description here

From the figure, it is obvious the solutions are between the roots of $2n+18=n^2+2n+2$.