value of k

[man111]

Number of real values of $k$ for which the equation $\displaystyle\frac{x+2}{kx-1} = x$ has exactly one real solution

[Руст]

$k=0$ is not solution.
Consider $x(kx-1)=x+2$. If $x=-2$ solution of quadratic equation and $k=-\frac 12$, we must except this solution and had only another solution solution. But in this case another solution is $x=-2$. Diskriminant $D=4(1+2k)=0\to k=-\frac 12$. Therefore not exist suth k. Number of real values of k is 0.

[ewert]

Руст в сообщении #519187 писал(а):

$k=0$ is not solution.

Вполне себе решение.

[Dave]

This problem contains two pitfalls. The first one is the assumption that the equation $x+2=x(kx-1)$ is equivalent to the initial one. And the second is the conviction that the latter equation is always quadtratic. After avoiding both piftals we conclude that the answer is 1, which corresponds to the case of $k=0$ and $x=-1$.