Question

How many integers satisfy the inequality
∣10(x+1)/(x^2+2x+3)∣≥1?

Answer 1

since x^2+2x+3 is always positive,

|10(x+1)| >= (x^2+2x+3)

Now, x+1 is either positive or negative
If x+1 is positive, |x+1| = x+1, and
10(x+1) >= x^2+2x+3
4-√23 <= x <= 4+√23
-.8 <= x <= 8.8
We started with x+1>=0, so every integer between -.8 and 8.8 works. There are 9 of them

If x+1 < 0, |x+1| = -(x+1) and we have
-10(x+1) >= x^2+2x+3
-6-√23 <= x <= -6+√23
-10.8 <= x <= -1.2
We started with x+1 < 0, so x < -1, and every integer between -10.8 and -1.2 works. There are 9 of those.

So, there are 18 integers that satisfy the inequality.