Posted by T.Gracken on October 26, 2002 at 08:05:57:
In Reply to: taking it a step further.... posted by Denis Borris on October 26, 2002 at 02:15:54:
: Change that equation to:
: x^4 - x^3 + 18x^2 - 16x - 42 = 0
: And I'll do same thing:
: k= x^4 - x^3; so:
: 18x^2 - 16x - 42 + k = 0; quadratic:
: 36x = 16 +- sqrt(3712 - 72k); simplify:
: x = (8 +- sqrt(928 - 18k)) / 18
: Looking for INTEGER solutions:
: k=46, x=1
: k=8, x=2
: k=14, x=-1
: k=48, x=0
: Clearly, only x=2 is a solution (x^4 - x^3 = 8).
: So why can't all equations of this type be solved this way ??
perhaps it is because some of us want to know all four solutions; real, complex, rational, integer, et.al.
...but I do like your approach
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