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March 2, 2015

March 2, 2015

Posted by **Ian** on Thursday, March 28, 2013 at 10:58pm.

∣10(x+1)/(x^2+2x+3)∣≥1?

- Maths -
**Steve**, Friday, March 29, 2013 at 2:17pmsince 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.

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