China Mathematical Competition

We have $$kx>0,\stackrel{◯}{1}$$ $$x+1>0,\stackrel{◯}{2}$$ $$kx=(x+1{)}^2.\stackrel{◯}{3}$$ The expression ③ can be standardized as $$x^2+(2-k)x+1=0.\stackrel{◯}{4}$$ The two roots of ④ are $$x_1,x_2=\frac{1}{2}[k-2\pm \sqrt{k^2-4k}],⑤$$ where $$\Delta =k^2-4k\ge 0\iff k\le 0\text{ or }k\ge 4.$$ (i) When $k<0$, it is easy to see from ⑤ that $x_1+1>0$, $x_2+1<0$, and $k{x}_1>0$. 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 $k<0$, $k=4$.