How do I prove $\frac{x-1}{x}=0$ has two real solutions?

I know it has one solution at $x=1$, but the question said "find the 2 real solutions", so I was confused if I was overlooking something. Maybe it's an error in the question.

Does anybody have any ideas to this problem?


closed as unclear what you're asking by user137731, Jack D'Aurizio, ajotatxe, qbert, Jean Marie Oct 2 '16 at 21:43

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    $\begingroup$ Usually one talk about solutions when there is an equation (or inequation) involve, but in your question there was only one function. Are you sure you should not include some $=$, or some inequality? $\endgroup$ – Darío G Oct 2 '16 at 21:18
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    $\begingroup$ Shouldn't this be equal to something if it is to have "solutions?" I don't see an equation. Unless if you're asking if $f(x) = \frac{x-1}{x}$ has exactly two real roots. $\endgroup$ – Sean Roberson Oct 2 '16 at 21:19
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    $\begingroup$ This question is not well posed just yet. To what is it supposed to be equal? $\endgroup$ – B. Pasternak Oct 2 '16 at 21:19
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    $\begingroup$ If roots are meant, there is only one. At the root of the denominator , the function has a pole. Maybe, the OP means "asymptotics". In this case, we actually have two , namely $x=0$ and $y=1$ $\endgroup$ – Peter Oct 2 '16 at 21:22
  • $\begingroup$ Ah yes I meant (x-1)/x = 0 , I know it has one solution but the question said find the 2 real solutions so I was confused if I was overlooking something. Maybe it's an error in the question. $\endgroup$ – Curtis Cleary Oct 2 '16 at 21:39

Well, assuming you want to find the roots of $f(x)=\frac{x-1}{x}$: $$ f(x)=1-\frac{1}{x} $$ The roots are given when $f(x)=0$, so $$ 1=\frac{1}{x}\Rightarrow x=1 $$

You could even plot $f(x)$ to see it:

  • $\begingroup$ (+1) for showing the graph of the function. Whatever is meant : This answer should be helpful. $\endgroup$ – Peter Oct 2 '16 at 21:34

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