Notice thatOriginally Posted by DenMac21
Now notice that,
Further, factor thus,
But I did not finish it, I do not know if that helps.
Thus, one solution is
x^4 -2ax^2 +x +a^2 -a = 0Originally Posted by DenMac21
x^4 -2ax^2 +a^2 = a -x
(x^2 -a)^2 = a-x
x^2 -a = +,-sqrt(a-x) ----------(1)
The RHS +,-sqrt(a-x):
----"a" cannot be zero because sqrt(-x) is not real.
----"a" cannot be negative because the radicand will be negative.
So, a > 0 ----------(i)
----for (a-x) to be positive, x < a -----------(ii)
----since "a" cannot be negative and x<a, then x cannot be negative too.
In the whole (1):
---x cannot be zero because (-a) cannot be equal to +,-sqrt(a).
Hence, x > 0
Therefore, 0 < x < a, but x cannot be zero. --------answer.
TD!Originally Posted by ticbol
I have to go to work now to put some food on the table. I will be back in about 12 hours from now. Maybe by then you could show us why you think x could be negative. Then I would comment on your posting.
Like I said before somewhere in this Forum, I do not comment on somebody's answer whether that answer is correct, wrong, lacking, too long, whatever.
But, of course, I will comment on any comment re my answers.
I haven't really looked into the problem so I haven't been able to come up with a more elegant solution, but here's a go.
We have a 4th degree polynomial without a 3rd degree term so a possible factorization would be two second degree polynomials with opposite lineair terms.
Identifying coefficients gives us a 3x3 system
So we have the following factorization
Then it's just solving two quadratics, solutions are
Of course, these solutions for x depend on a, so I don't see why a negative solution would be impossible - for certain values of the parameter a.