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Equation development

What's the error in this equation/solution?

The correct answer should be $x = 1 ⇒ x = ±1$, right?

Here's my attempt:
For $x=-1$, I can't "come back" from the first line to $x=1$. I get $4=0$ as a result.
However, for $x=1$, I can do this procedure and find $0=0$.

All help is appreciated.

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  • $\begingroup$ $x^2-2x+1=0$ does not imply $x=1$? What other solution exists? $\endgroup$ – David C. Ullrich Mar 30 at 15:50
  • $\begingroup$ My mistake, Sorry $\endgroup$ – Ray LittleRock Mar 30 at 15:54
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$(1)$ and $(2)$ are not equivalent.

As you can see, $(2)$ also holds for $x=-1$.

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  • $\begingroup$ In this case, x^2 also holds x = -1, making the equation 0=0, but the "-2x" doesn't hold x = -1, right? Sorry, but I don't know how to use MathJax. $\endgroup$ – Daniel Sehn Colao Mar 30 at 18:43
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    $\begingroup$ @DanielSehnColao If I understand you correctly then it looks like you get the idea. It's true that $x^2 = 1$ whenever $x=1$ or $x=-1$. Indeed the important part is that once they insert $x=1$ in $-2x$ they change everything because it has different values for $x=1$ and $x=-1$. $\endgroup$ – Yanko Mar 30 at 19:38
  • $\begingroup$ Yes, if I insert x = -1 in -2x the equation will change at all. That's why (1) doesn't hold x = -1. And in the (2) I just have x^2, then it's okay to insert whether x = 1 or x = -1, right? $\endgroup$ – Daniel Sehn Colao Mar 30 at 20:02
  • $\begingroup$ @DanielSehnColao Yes, exactly. $\endgroup$ – Yanko Mar 30 at 20:29
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The answer given is right. The second logical connective (i.e., $\iff$) should be $\implies$.

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