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Probably a dumb question, but solving this:

$$x^2=2x$$ $$x=2$$

But also (or using the quadratic formula);

$$x^2-2x=0$$

$$x(x-2)=0$$ $$x=0/x=2$$

Is the first computation wrong, and why is it limiting?

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    $\begingroup$ In the first part of the question, you are dividing by $x$ to simplify. This is valid if $x \ne 0$, and $x = 0$ is a solution. $\endgroup$ – T. Bongers Nov 2 '17 at 17:05
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In the first part, you can «cancel» the $x$ therm if $x\not = 0$. If you cancel that term, you are asuming implicitly that $x\neq 0$.

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The first simplification only works under the assumption that x is nonzero. If you divide both sides of an equation by an unknown quantity, the result is not a valid representation of the same equation if that unknown quantity were actually zero.

So, an equation that results when you divide both sides by x does not eliminate the possibility that x could be zero. You need to separately check the case where the simplified equation is invalid. In an equation of one variable, this is easy to do.

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    $\begingroup$ Alright, thanks a lot for a clear explanation. Understood $\endgroup$ – Axel Nov 2 '17 at 17:16
  • $\begingroup$ This is actually a common error in beginning algebra. Google "2 = 1 proof" for the most common example used to show what happens when you forget that division by zero is invalid. $\endgroup$ – Trixie Wolf Nov 2 '17 at 18:16

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