Sujet : Re: x²+4x+5=0
De : invalid (at) *nospam* example.invalid (Moebius)
Groupes : sci.mathDate : 22. Jan 2025, 23:42:14
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vmrs86$185g8$1@dont-email.me>
References : 1 2
User-Agent : Mozilla Thunderbird
Am 22.01.2025 um 23:16 schrieb Chris M. Thomasson:
On 1/22/2025 5:48 AM, Richard Hachel wrote:
x² + 4x + 5 = 0
>
This equation has no root, and <bla>
It has the two "roots" (solutions):
x = -2 - i
x = -2 + i
Hint:
x² + 4x + 5 = (x + 2)² + 1.
Hence x² + 4x + 5 = 0 is equivalent with (x + 2)² + 1 = 0 or (x + 2)² = -1. So x + 2 has to be a (one or more) complex number(s) z such that z² = -1. We know such numbers, they are i and -i (and there aren't more). Hence we have x + 2 = i or x + 2 = -i resp. And hence x = -2 + i or x = -2 - i resp.
The x^2 component has two roots.
Doesn't make much sense. :-P