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If I want to solve $x^2=x$ then obviously $x(x-1)=0$ so $x=0$ or $x=1$.

But, what if I try to solve it differently:

$$x^2=x$$ $$\frac{x^2}{x}=1$$ $$x=1$$

Then for some reason the solution $x=0$ vanishes. Apparently I am missing something. Question is: what am I missing and how to also get $x=0$ as a solution in the second method?

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    $\begingroup$ You can do it if you suppose $x\neq 0$, otherwise it's wrong. $\endgroup$
    – user301068
    Commented Jan 2, 2016 at 9:45
  • $\begingroup$ Cancelling out a factor eliminates the possibility of you solving that factor. $\endgroup$
    – Airdish
    Commented Jan 4, 2016 at 17:48

2 Answers 2

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Division by $x$ is only valid if $x \neq 0$ here, that's why you cannot have $x=0$ as a solution with the second method.

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If we set $x=0$, we get $0^2=0$, which is true. OK, $x=0$ is a solution of our equation; let's move on supposing $x\ne0$. Now we can divide both sides by $x$, getting $x=1$. Fine, we have two solutions.

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