4. The solution situation of the equation $x^{2}-x=\frac{1}{x}$ is ( ).
(A) Only one positive root
(B) Only one negative root
(C) One positive root and one negative root
(D) No real roots
4. (A).

Method 1. Let $y=x^{2}-x$ $=\left(x-\frac{1}{2}\right)-\frac{1}{4}, y=$ $\frac{1}{x}$, their graphs intersect at only one point in the first quadrant.

Method 2. If $x0$, so $x>0$, when $x \leqslant 1$, $x^{2}-x$ $\leqslant 0$, when $x \geqslant 2$, $x^{2}-x>1, \frac{1}{x}0$. Therefore, the equation $x^{2}-x-\frac{1}{x}=0$ must have a real root in the interval $1<x<2$.