Re: x²+4x+5=0

Le 22/01/2025 à 14:48, Richard Hachel  a écrit :

x²+4x+5=0
 This equation has no root, and it never will.
 We can then find two roots of its mirror curve.
 Let x'=-3 and x"=-1
 These are not roots of this curve, but the roots of the imaginary mirror curve.
 What is this imaginary mirror curve?
 It is the curve with equation y=-x²-4x-3
 Let's look for its roots, and we find x'=-3 and x'=-1
 These are the imaginary roots of x²-4x+5.
 Or x'=-3(i) and x'=-1(i)
 R.H.

I wrote all this very badly.
The equation has only one and the same root (double) which is obviously x=-2+i
This is the imaginary solution for y=x²+4x+5 which has no solution in R.
It is at the same time the solution for its mirror curve y=-x²-4x-3 (we give i the values ​​-1 and 1, and we find x=-3 and x=-1.
R.H.