Posted by T.Gracken on October 26, 2002 at 08:13:28:
In Reply to: Re: taking it a step further.... posted by T.Gracken on October 26, 2002 at 08:05:57:
: : 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).
by the way, x=2 is not a solution to x^4 - x^3 + 18x^2 - 16x - 42 = 0...