Maths
posted by Ian .
How many integers satisfy the inequality
∣10(x+1)/(x^2+2x+3)∣≥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.