Probably a dumb question, but solving this:

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

But also (or using the quadratic formula);


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

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

  • 4
    $\begingroup$ 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. $\endgroup$ – user296602 Nov 2 '17 at 17:05

In the first part, you can «cancel» the $x$ therm if $x\not = 0$. If you cancel that term, you are asuming implicitly that $x\neq 0$.

| cite | improve this answer | |

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.