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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- April 8th 2008, 01:30 PMrbernadineBabylonian 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

- April 8th 2008, 03:32 PMCaptainBlack
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 - April 10th 2008, 08:36 AMrbernadineThanks
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.

(Rofl) - April 10th 2008, 08:50 AMOli