If you want to call x a complex number, it's a complex number. There's no need to write it in a + bi form, though you may do so if you like.
-Dan
Hi
which specific formula/method I can use to find the complex roots of a polynomial of degree 4 on form:
which has irrational coefficients
a = and b = and
where
If I write the equation as , then you will notice that it factorises as
Then by fundamental theorem of algebra I see that
where the complex roots of the original polymial p(x) is
Getting back to the original polynomial
(*)
where
I would like to prove that a complex number x makes (*) true iff
is a root of the
I see that that
So the solutions of in my equation are .
I then need to plug each of these numbers into the equation , which can be written as . Using the quadratic formula again, you get .
By the way do x^-1 then exist ?
Yes, there's no problem about that. In fact since , you can see that
I get this x value for both values of s to be:
This expression makes the equation
true
since
This proves that there exist a number x which is both a root of the original polynomial and Q(s = x + x^{-1}), since .
But one question remains on my part. Since x is supposedly a complex number. Is it a complex number in its current form? That I'm a bit unsure of. Since then it should be written in the form x = a + bi ?
Best Regards,
Billy
I know how to solve the cubic algebraically but never learned how to solve the quartic, it is too long. I find it amazing how these Italian mathematicians were so good at manipulation of radical terms and factorizations.
I read that one Italian mathematician was killed for being able to solve the quartic. Because people believed it was not withing human skills and thus he made a pact with a Devil (similar story with the violinist, Pagannini, they used the same excuse).
This is my 37th Post!!!