# Re: question....

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Posted by Joel on October 26, 2002 at 01:46:06:

In Reply to: question.... posted by Denis Borris on October 26, 2002 at 00:03:00:

: x^4 - x^3 + 18x^2 - 16x + 96 = 0

: Say I got that equation; before solving it, I go:
: let k = x^4 - x^3; then:
: 18x^2 - 16x + 96 + k = 0 ; quadratic:
: 36x = 16 +- sqrt(-6656 - 72k)

: Because x^4 - x^3 can't be negative, can I tell whoever that
: equation belongs to to stick it up his/her sigmoid flexure ?

I think you're right. x has to be greater than 1 or less than -1 because if -1 < x < 1, then
18x^2 - 16x >= -3.555.....
and x^4 - x^3 would be a tiny negative number
so there's no way that all the x-terms could offset the 96.

And if x > 1 or x < -1 18x^2 - 16x has to be positive.

And as you said, x^4 - x^3 has to be positive.

So, there you go. (Be gentle.)
:)

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