Posted by Denis Borris on October 26, 2002 at 02:15:54:
In Reply to: Re: question.... posted by Joel on October 26, 2002 at 01:46:06:
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:
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 ??
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