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Math Help - Question

  1. #1
    Junior Member
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    Mar 2012
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    LA
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    Question

    2x^2-7X>15
    Let me just break it down till the end.

    (x-5)(2x+3)>0

    x-5-0 2x+3=0
    x=5 x=-3/2

    When I do my test point and let it be 0 It looks something like

    <--//////-----------------------------------------------------///////--------->
    True -3 over 2 the middle is F and 5 is also true and the interval notation is (-∞.-3/2)U(5,∞).

    My question is, how can -3/2 be greater than 0?
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  2. #2
    Senior Member
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    Sheffield England
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    Re: Question

    Go back to (x-5)(2x+3)>0
    This is true if both brackets are positive or both brackets negative.
    Both positive if x>5 and both negative if x<-3/2
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  3. #3
    Senior Member
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    Nov 2011
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    Crna Gora
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    Re: Question

    Actually :

    2\left(\frac{-3}{2}\right)^2-7\left(\frac{-3}{2}\right)-15=0
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