It's straightforward to see that there's no real solution. If you happen to remember that the minimum point of an upward sloping parabola is at x = -b/2a, then you know that the minimum is at (-1/2, 3/4). So there's no solution. Or you can note that |x^2| > |x| anytime |x| > 1, while a solution requires that x^2 - x < -1.

I actually don't think any of his algebraic manipulations are at fault. It's just garbage in, garbage out. If an equation has two solutions, you can substitute just one for a variable, so long as you don't subsequently use the other. Moreover, if a variable occurs twice, you can substitute for one instance of it and not the other, so long as you don't later perform an inconsistent substitution on the other.

Oh, the last post had a huge brain fart--every polynomial has at least one complex solution, though I spoke the truth so far as I indicated that I didn't know how to check what they were...

The quadratic equation works just fine for the complex numbers.

The original equation had two solutions, both of them solutions to the equation x^3 = 1. I am not a mathematician, but increasing the degree of a polynomial expression often introduces a spurious solution.

I fail at physics. Reading the newspaper in front of the mirror I did not notice any flipping whatsoever. After some thought I decided to try to read the headline in the mirror instead. After turning the front page toward the mirror I noticed that the up-down directions had been flipped but the left-right directions remained the same! I am very confused now. Is it possible that this could be because I am in Europe and you are in the US, so that our points of view are rotated 90 degrees compared to each other?

as a formal mathmatician myself...let me try..

first..you should not divide both side of quadradic equation by X because you will lose one of the value.
second... you should not multiply both side of the equation by X because you will create false answer which created paradox in your example.

i would like to look at this more..but i am too busy and dont have time..
after i come back next month i will find exactly what is wrong.

The original quadratic is fine, and has 2 roots, which are both of the complex cube roots of 1 (e^2pi*i/3 and e^-2pi*i/3). Both are also roots of x^3=1, of course, but that equation has a third root, 1. You can get from the original equation x^2+x+1=0 to x^3=1 just by multiplying it by x-1 on both sides. The fatal flaw is in going from x^3=1 to x=1 - this only works if you assume x has to be real, which isn't the case.

I believe the original quadratic answer is flawed... to get x from x^2 you should take squareroot of both sides... that gives you:

x = +/- the squareroot of (-x-1)


unlike in the post which divides x into both sides...

The dividing by x is perfectly valid, as x is nonzero. It's not standard for dealing with quadratics, but this does tell you that all solutions are of the form x = -1-1/x.

This is why I never claimed to be a mathmatician

You can't just multiply with (x-1) on both sides to get x^3=1, because then x=1 is a solution, making the previous step a multiplication by zero. That introduces more solutions.

As for the last step: I don't see a problem there. The only flaw it introduces is that it ignores the other solutions. If you interpret the original question as "find some x such that ..." instead of "find all x such that ...", that's a perfectly valid step.
Just assume the last step would have been from x^3=1 to x=(-1/2 + i sqrt(3)/2). That would be a correct solution to the original equation. Not all solutions, but a correct solution.
The actual error must be at an earlier point, where the equation changes from two solutions to three solutions.

The error must obviously be in the transition from the fourth line to the fifth. That's where the third solution (x=1) appears.

In that line, *one* x is substituted by another formula that contains another x. That in itself is not a problem. But now, you suddenly have two different xs in the equation, the one from the substitution and one "leftover" x. In the remaining steps, these different xs are mixed up. You have to substitute *both* the x and the x^2, then it all works out correctly

The preceding lines imply line 5, and line 5 is true. The error is that one-way implication is not equivalence.

Can you give me a precise definition of when substitution is safe and when it is unsafe? You claim that partial substitution is risky, can you prove that "full" substitution never introduces extraneous solutions?

Thanks guys, I didn't see that there's no problem until you actually take a step (x^3 = 1 => x = 1) that depends on the solution being real.