# Babylonian reduction

• Apr 8th 2008, 12:30 PM
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
• Apr 8th 2008, 02:32 PM
CaptainBlack
Quote:

Originally Posted by rbernadine
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
• Apr 10th 2008, 07:36 AM
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.

(Rofl)
• Apr 10th 2008, 07:50 AM
Oli
Quote:

Originally Posted by CaptainBlack
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.