China Mathematical Competition

We have
$$kx > 0, \qquad \textcircled{1}$$
$$x + 1 > 0, \qquad \textcircled{2}$$
$$kx = (x + 1)^2. \qquad \textcircled{3}$$
The expression ③ can be standardized as
$$x^2 + (2 - k)x + 1 = 0. \qquad \textcircled{4}$$
The two roots of ④ are
$$x_1, x_2 = \frac{1}{2}[k - 2 \pm \sqrt{k^2 - 4k}], \quad ⑤$$
where
$$\Delta = k^2 - 4k \ge 0 \Leftrightarrow k \le 0 \text{ or } k \ge 4.$$
(i) When $k < 0$, it is easy to see from ⑤ that $x_1 + 1 > 0$, $x_2 + 1 < 0$, and $kx_1 > 0$. Then the equation has one real root, $x_1 = \frac{1}{2}[k - 2 + \sqrt{k^2 - 4k}]$.

(ii) When $k = 4$, the equation has one real root, $x = \frac{4}{2} - 1 = 1$.

(iii) When $k > 4$, the two roots $x_1, x_2$ are both positive, as well as $x_1 \ne x_2$. Discarded.

Therefore, the range of $k$ is $k < 0$, $k = 4$.