Determine all solutions of the equation $\sqrt{x}=x-2$.

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By squaring both sides of the equation, we obtain $x=x^{2}-4 x+4$, which is equivalent to $x^{2}-5 x+4=0$. The roots of this quadratic equation are $x=1$ and $x=4$. However, when substituting $x=1$ into the original equation $\sqrt{x}=x-2$, we get $\sqrt{1}=-1$, which is false. However, when substituting $x=4$, we get $\sqrt{4}=2$, which is true. Therefore, the given equation has $x=4$ as its only solution.

Note: The appearance of the "strange solution" $x=1$ is due to the fact that the implication

$$
a^{2}=b^{2} \Rightarrow a=b
$$

is not generally true. The correct statement is

$$
a^{2}=b^{2} \Rightarrow a= \pm b
$$

Thus, when we square both sides of an equation, we obtain a new equation that may, eventually, contain more solutions than the original equation. You can also see this clearly, for example, in the equations: $x=1 \mathrm{and}$ $x^{2}=1^{2}$.