$2 \cdot 11$ The equation $x=k^{2}(x-1)(x-2)$ has real roots, the value of $k$ is
(A) does not exist.
(B) $-21$ or $k<-2$.
(E) all real numbers.
(13th American High School Mathematics Examination, 1962)
［Solution］From the given equation, it can be transformed into
$$
k^{2} x^{2}-\left(3 k^{2}+1\right) x+2 k^{2}=0,
$$

If $k \neq 0$, for the equation to have real roots, then $\Delta \geqslant 0$, i.e., $k^{4}+6 k^{2}+1 \geqslant 0$. This holds for any real number $k$.
If $k=0$, there is still a real root $x=0$.
Therefore, the answer is $(E)$.