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Math Help - [SOLVED] more inequalities

  1. #1
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    [SOLVED] more inequalities

    I'm once again having issues solving an inequality. Any assistance would be much appreciated
    x^3-2x^2-9x-2 >= -20

    I can't simplify by grouping, as that results in
    x^2(x-2) + (-9x-2) >= -20 which lacks a common term

    and as far as I can tell, this isn't a difference or sum of cubes, so how are the critical points calculated?
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  2. #2
    Newbie driegert's Avatar
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    Kingston, Ontario
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    Factoring

    So the only way I know how to do these questions is the following:

    1) Bring everything over on one side so you have:
    x^3 - 2x^2 - 9x + 18 >= 0

    2) Unfortunately, this part you just guess. Pick some numbers for x and see if one of them will cause the equation to become zero (for now we ignore the greater than).

    I found that x=2 will work just dandy.

    Use synthetic division or long division to factor x=2, that is, (x-2) out of that equation and you are left with a quadratic.

    3) Factor the quadratic.

    4) Investigate the invervals that are created and determine which ones will cause the inequality to be true.

    For this question I got:

    (x-2)(x-3)(x+3) >= 0

    View this on a number line:

    <----(-3)-----(2)---(3)------->

    Examine x < -3, does the above give a positive or negative number, examine the next interval (i.e. -3 < x < 2) again, positive or negative number... ect..

    Hope that helps
    Last edited by driegert; February 7th 2010 at 09:38 AM. Reason: Making more clear.
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  3. #3
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    Yes, that's exactly what I needed. As usual, I was being blind. Once you pointed out shifting the -20 over, everything else fell into place. Thanks a lot for your time.
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