Re: x²+4x+5=0

Le 24/01/2025 à 20:45, Python a écrit :

 Your "imagination" is off-topic. Another polynomial has different roots? So what?  They are not roots of x²+4x+5 anyway.

I proposed the curve y=x²+4x+5 and we showed that it had no real roots.
I then proposed to study the mirror curve centered on its vertex, and to look at where the two real roots were.
I said that this curve necessarily had the aspect of f'(x)=-x²-4x+c' with c'=c-(b²/2a)
Julien also says the same thing.
And so f'(x)=-x²-4x-3
Two roots in R: x'=-3 and x"=-1
The two roots of the first curve were x'=-2-i and x"=-2+i
It therefore seems that the imaginary roots of a curve without real roots have a resemblance to the real roots of the mirror curve.
R.H.