$2 \cdot 26$ If two real numbers $x$ and $y$ satisfy the equation $\frac{x}{y}=x-y$. Then
(A) $x \geqslant 4$ and $x \leqslant 0$.
(B) $y$ can equal 1.
(C) $x$ and $y$ are both irrational numbers.
(D) $x$ and $y$ cannot both be integers.
(E) $x$ and $y$ should both be rational numbers.
(8th American High School Mathematics Examination, 1957)
[Solution] From $\frac{x}{y}=x-y$, we transform it into $y^{2}-x y+x=0$. Since $y$ is a real number, when the above equation is regarded as a quadratic trinomial in $y$, $\Delta \geqslant 0$. That is, $x^{2}-4 x \geqslant 0$, which gives $x \geqslant 4$ or $x \leqslant 0$. Therefore, the answer is $(A)$.