Sujet : Re: y=f(x)=(x²)²+2x²+3
De : dohduhdah (at) *nospam* yahoo.com (sobriquet)
Groupes : sci.mathDate : 06. Feb 2025, 20:15:40
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vo31ot$33ac7$1@dont-email.me>
References : 1
User-Agent : Mozilla Thunderbird
Op 06/02/2025 om 16:42 schreef Richard Hachel:
Bonjour les amis !
I asked for the roots of the following equation on the French forums, I only got one answer that didn't satisfy me, and the rest is just contempt and insults.
So I'm trying my luck here.
y=f(x)=(x²)²+2x²+3
Il y a pour moi, deux racines très simples pour cette équation, dont aucun n'est réelle.
Can the Anglo-Saxons find these two roots?
R.H.
Actually there are four complex roots.
https://www.wolframalpha.com/input?i=x%5E4%2B2x%5E2%2B3https://www.desmos.com/calculator/in0q2rqzc9
y=f(x)=(x²)²+2x²+3
Re: y=f(x)=(x²)²+2x²+3