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Math Help - Babylonian reduction

  1. #1
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    Babylonian reduction

    This question has me stumped. My professor wants us to find out how the Babylonians may have made a reduction of the general cubic equation into the form n^3 + n^2 = c
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  2. #2
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    Quote Originally Posted by rbernadine View Post
    This question has me stumped. My professor wants us to find out how the Babylonians may have made a reduction of the general cubic equation into the form n^3 + n^2 = c
    The general cubic equation may be written:

    x^3+a x^2 + b x + c=0

    Make a change of variable y=x+u, and choose u so that the coefficient of y
    is zero (this will entail solving a quadratic). then make another change of
    variable n=ky so that the equation has the form:

    An^3+An^2+C=0

    Then divide through by A to get:

    n^3+n^2=C/A

    (the only problem that I can see is that the quadratic is not guaranteed to have real roots, and complex roots would have stopped a Babalonian in their tracks)

    RonL
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  3. #3
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    Talking Thanks

    I was trying so hard to use x+1, x-1,etc. How did you come up with x+u? the babylonian reduction is nonlinear, so I assumed it would have to be a negative somewhere. Is that where choosing u comes in to make the coefficient of y 0? I will try it now to see how it all works out.

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  4. #4
    Oli
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    Quote Originally Posted by CaptainBlack View Post
    The general cubic equation may be written:

    x^3+a x^2 + b x + c=0

    Make a change of variable y=x+u, and choose u so that the coefficient of y
    is zero (this will entail solving a quadratic). then make another change of
    variable n=ky so that the equation has the form:

    An^3+An^2+C=0

    Then divide through by A to get:

    n^3+n^2=C/A

    (the only problem that I can see is that the quadratic is not guaranteed to have real roots, and complex roots would have stopped a Babalonian in their tracks)

    RonL
    Wow... nice. Didn't know that.
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