jmc

Multiplying both sides by $(x+1)(x+2),$ we get $$x(x+2)+x(x+1)=kx(x+1)(x+2),$$or $$2x^2+3x=k{x}^3+3k{x}^2+2kx.$$This rearranges to the equation $$0=k{x}^3+(3k-2)x^2+(2k-3)x,$$or $$0=x(k{x}^2+(3k-2)x+(2k-3)).$$Clearly $x=0$ is a root of this equation. All the other roots must satisfy the equation $$0=k{x}^2+(3k-2)x+(2k-3).$$If $k=0,$ then the equation becomes $-2x-3=0,$ so $x=-\frac{3}{2}.$ Thus, $k=0$ works.

Otherwise, the $x^2$ coefficient of the right-hand side is nonzero, so the equation is a proper quadratic equation. For the given equation to have exactly two roots, one of the following must be true:

The quadratic has $0$ as a root, and the other root is nonzero. Setting $x=0,$ we get $0=2k-3,$ so $k=\frac{3}{2}.$ This is a valid solution, because then the equation becomes $0=\frac{3}{2}{x}^2+\frac{5}{2}x,$ which has roots $x=0$ and $x=-\frac{5}{3}.$

The quadratic has two equal, nonzero roots. In this case, the discriminant must be zero: $$(3k-2{)}^2-4k(2k-3)=0,$$which simplifies to just $k^2+4=0.$ Thus, $k=\pm 2i.$ These are both valid solutions, because we learned in the first case that $k=\frac{3}{2}$ is the only value of $k$ which makes $0$ a root of the quadratic; thus, the quadratic has two equal, nonzero roots for $k=\pm 2i.$

The possible values for $k$ are $k=\boxed{0,\frac{3}{2},2i,-2i}.$