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Math Help - Solving a quartic equation

  1. #1
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    Solving a quartic equation

    This problem seems a bit out of the ordinary. s^4+ s^2-6 =0
    Last edited by mr fantastic; June 1st 2009 at 01:33 AM. Reason: Moved, edited and corrected
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    Quote Originally Posted by sgonzalez90 View Post
    This problem seems a bit out of the ordinary. s^4+ s^2 -6 =0
    This is a quadratic in s^2. Let w = s^2:

    w^2 + w - 6 = 0.

    Solve for w. Hence solve for s.
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  3. #3
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    Oh my gosh I'm so sorry, i typed it wrong again. I'm sorry im really tired and had a long day. I'm just trying to brush up on my algebra before i begin my calculus course in two weeks.

    s^4+ 3s^2 -6 =0
    this is the correct problem i promise, thank you.
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    Quote Originally Posted by sgonzalez90 View Post
    Oh my gosh I'm so sorry, i typed it wrong again. I'm sorry im really tired and had a long day. I'm just trying to brush up on my algebra before i begin my calculus course in two weeks.

    s^4+ 3s^2 -6 =0
    this is the correct problem i promise, thank you.
    Just change the appropriate numbers in post #2.
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  5. #5
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    Quote Originally Posted by sgonzalez90 View Post
    s^4+ 3s^2 -6 =0
    this is the correct problem i promise, thank you.
    This one won't factor, so apply the Quadratic Formula:

    x^2\, =\, \frac{-(3)\, \pm\, \sqrt{(3)^2\, -\, 4(1)(-6)}}{2(1)}

    x^2\, =\, \frac{-3\, \pm\, \sqrt{9\, +\, 24}}{2}

    x^2\, =\, \frac{-3\, \pm\, \sqrt{33}}{2}

    Then take the square root of either side to solve for "x=". Yes, you'll have some nested radicals.
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