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That One Thread


Florete
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Nightmare, bite me, I want to be a vampire. The cool Castlevania kind, not the gay Twilight kind.

I don't bite guys. That's gay.

Dammit, I want to change back to John Morris now, but I can't. ):

And it would've been so perfect, too.

Maybe Sirius...!

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As I went home on Friday night

as drunk as drunk could be,

I saw a head upon the bed

where my old head should be.

Well, I called me wife and I said to her

"Will you kindly tell to me

Who owns that head upon the bed

where my old head should be?"

"Ah, you're drunk, you're drunk, you silly old fool

Still you cannot see

That's a baby boy that

me mother sent to me."

Well it's many a day I've travelled

a hundred miles or more

But a baby boy with his whiskers on

sure I never saw before.

Edited by Grimsworth of Natal
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But that's the thing. The substitution is not a valid step. There is no real solution to x = -1 - 1/x, and yet when he subbed it into the other equation he did not say "x = complex number". This is no different than not saying "x /= 0" after dividing by x and then using x = 0 later on. That's the problem. There are three roots to the cubic function, and if you reject the real solution you arrive at the correct answer.

He's creating a third answer from an equation that only has two. The first section is quadratic (two roots). The second is cubic (three roots). Obviously you can't have three roots to the first equation. You have to reject one of them. It's pretty obvious which one.

The fact is that x does not equal -1 - 1/x in the real number system and so the substitution is not valid when dealing only with real numbers. Maybe you don't know that without trying to solve it, but the fact remains it is not a true statement.

But with the perfect square method, all of the steps are valid even if the equation has no real roots[/s]. In fact, we wouldn't have even known about whether an equation has real roots or not without checking b2 - 4ac, which is derived from that process. So is there something exceptional about substituion that makes it only valid for real solutions? That actually makes perfect sense (since subsitution is under the assumption that the equation being substituted in is true) but it makes me wonder if there any other stuff that you can't do when an equation has no real roots.

Oh, yeah, and thanks a lot for making me finally figure it out, Narga!

Edited by Proto
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