Why does basic algebra provide one value for $x$ when there should be two? I have the equation $x^2=x$.
If I divide $x$ from both sides I get $x=1$.
Yet clearly $x$ can also equal $0$.
What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?
I hope this isn't a stupid question because I feel dumb asking about something so basic.
EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.
 A: When you "divided both sides by $x$", you tacitly assumed that $x \neq 0$. It does not make sense to divide by zero.
In general, when confronted with problems like this, you can try to substitute in your solution from the beginning and go through the steps to see what goes awry. Here, starting with $x = 0$, the equation $x^2 = x$ is $0 = 0$. The next step is to divide both sides by $x$... whoops!
A: $x^2=x\Rightarrow x^2-x=0\Rightarrow x(x-1)=0\Rightarrow $either $x=0$ or $x=1$
A: The given quadratic equation $x^2=x$ will have two real roots given as follows $$x^2-x=0$$ $$x(x-1)=0$$ $$x=0\ \ \ \ \textrm{or}\ \ \ x-1=0$$
$$x=0\ \ \ \ \textrm{or}\ \ \ x=1$$ Hence, we get $x=0$ or $x=1$
A: You cannot divide by $x$ if $x=0$ in the first place. That's why if you want to divide by $x$, you take cases. If $x=0$, we have a solution. If $x\neq0$, then we get $1$.
A: Your methodology isn't wrong. You just need to modify it a bit. Before you divide both sides by $x$, you must check what  happens when $x = 0$; that is, you must check whether or not $x=0$ is a solution. Once you have taken care of that, you are free to consider what happens when $x \ne 0$. In particular, now you can divide both sides by $x$.
(1)    $x^2 = x$
(2)    Clearly $x = 0$ is a solution. 
(3)    Now supose that $x \ne 0$. Then we can multiply both sides by $\dfrac 1x$.
(4)    We get $x = 1$.
(5)    The solution set is $x \in \{0, 1\}$.
