Re: x²+4x+5=0

On 1/24/2025 5:15 AM, FromTheRafters wrote:

Richard Hachel wrote :

Le 23/01/2025 à 22:23, Moebius a écrit :

Am 23.01.2025 um 00:58 schrieb sobriquet:

Op 22/01/2025 om 14:48 schreef Richard Hachel:

x²+4x+5=0
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This equation has no root, and it never will.
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We can then find two roots of its mirror curve.
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Let x'=-3 and x"=-1
What is this imaginary mirror curve?
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It is the curve with equation y=-x²-4x-3
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Let's look for its roots, and we find x'=-3 and x'=-1
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These are the imaginary roots of x²-4x+5.

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Wolfram Alpha tells us there are two roots:
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https://www.wolframalpha.com/input?i=solve+x%5E2%2B4x%2B5%3D0

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Wolfram Alpha must be wrong! :-P

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No, here it is fine. The two imaginary roots are x'=-2-i and x"=-2+i
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But that is not the question.

 Shouldn't imaginary roots be on the y axis?

:^) If I were to plot point 0+3i, I would place it at vector point (0, 3) on the canvas.