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Math Help - 1-D Inequality Help

  1. #1
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    1-D Inequality Help

    Solve: 6x^4 - 10x^2 - 12 > 10x^3 + 38x
    where x is a member of the real numbers.

    I rearranged it as an equality equaling zero, as:
    6x^4 - 10x^3 - 10x^2 - 38x - 12 = 0

    I then factored out a 2, giving:
    (2)(3x^4 - 5x^3 - 5x^2 -19x - 6) = 0

    Using factor theorum and synthetic division, I determined that:
    (2)(3)(x + 1/3)(x - 3)(x^2 + x + 2) = 0

    However, the quadratic equation above does not yield any real solutions (unless I'm doing it wrong). Where do I go from here?
    Last edited by Kangles; April 12th 2011 at 01:22 PM.
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  2. #2
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    "Solve:  6x^4 - 10x^2 - 12 > 10x^3 + 36x" but you wrote: 6x^4 - 10x^3 - 10x^2 - 38x - 12 . It's 36, not 38.

    Are you sure you wrote the inequality correct?
    x=2: 96 - 40 - 12 > 80+72 is false.
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  3. #3
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    It's 38, not 36. I wrote it incorrectly the first time. The bold line at the top is what was given to me on the assignment sheet.
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    Quote Originally Posted by Kangles View Post
    Solve: 6x^4 - 10x^2 - 12 > 10x^3 + 36x
    where x is a member of the real numbers.

    I rearranged it as an equality equaling zero, as:
    6x^4 - 10x^3 - 10x^2 - 38x - 12 = 0

    I then factored out a 2, giving:
    (2)(3x^4 - 5x^3 - 5x^2 -19x - 6) = 0

    Using factor theorum and synthetic division, I determined that:
    (2)(3)(x + 1/3)(x - 3)(x^2 + x + 2) = 0

    However, the quadratic equation above does not yield any real solutions (unless I'm doing it wrong). Where do I go from here?
    Hi Kangles,

    In your first line, you have 36x, but when you "rearranged" it, it changed to 38x.

    Assuming you meant 38x, and your factorization is correct, you have two real roots showing.

    If (x + 1/3) is a factor (and it is!), then x = -1/3 is a root.

    If (x - 3) is a factor (and it is, too!), then x = 3 is a root.

    Ignore the two complex roots. Your beginning statement said x is a member of the real numbers.

    Plot these two real values on a number line and check the intervals that will make your original inequality true.

    I found the solution to be r\: x > 3\}" alt="\{x|x < -\frac{1}{3} \r\: x > 3\}" />

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