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

  • Maths -

    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.

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