Re: New equation

Le 25/02/2025 à 09:21, Barry Schwarz a écrit :

On Mon, 24 Feb 25 21:11:40 +0000, Richard Hachel <r.hachel@tiscali.fr>
wrote:
 

Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <r.hachel@tiscali.fr>

>

A quartic always has four roots.

>
Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is perhaps not entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of degree 3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial intelligence, and I was given three roots, but they are incorrect, because those who answer do not seem to understand the real concept of imaginary numbers.
There is in fact only one root.
A very strange root composed of a real root and a complex root. Both placed on the same point A(1,0) and A(-i,0).
>
R.H.

 A cubic has three roots.

This is what is generally said, but is it always true?

The roots of your equation are 1, (-1+i*sqrt(15))/2, and
(-1-i*sqrt(15))/2.

That one of the roots is 1, and that it can be represented on a Cartesian coordinate system, is obvious. I then set the point A(1,0).
I then look for the other two roots of the equation, but I realize that I can't find any others, even complex ones, and that the two complex roots given are fanciful.
I then start from the principle that the complex roots are the real roots of the mirror curve, and that the real roots are the complex roots of this other curve, and I find a complex root which is x'=-1.
I therefore obtain the point A(-i,0) which is exactly the same as the point A(1,0) knowing that i=-1 and -i=+1.
It seems that this curve is its own mirror.
R.H.