We all know the solutions of

$$\sqrt{x}=x$$

are $x=0$ and $x=1$, which are easily checked. If I wanted to solve for the latter solution, I would proceed as follows:

\begin{eqnarray}
\sqrt{x}&=&x\\
\left(\sqrt{x}\right)^2&=&\left(x\right)^2 \\
x &=& x^2 \\
1 &=& x
\end{eqnarray}

However, I am at a loss to solve the solution $x=0$ in a similar algebraic manner. Can anyone show me where to begin?