Complex Roots in a Complex Equation 
Problem: The equation $$\frac{x}{x+1} + \frac{x}{x+2} = kx$$ has exactly two complex roots. Find all possible complex values for $k.$


Progress: I was online today, and I saw this problem. Heres what I tried: $$(x+1)x+(x+2)x=kx(x+1)(x+2),$$ and I immediately got stuck :(. I'm quite new at complex numbers. Help? Thanks.
 A: Hint: A polynomial equation of degree $n$ has exactly $n$ roots (counted with multiplicity). The simplified expression has degree at most $3$. Therefore, it either has a double root or is of degree $2$ with simple roots. Note that $x=0$ is a root.
A: While @lhf has provided the right hint, some careful case work is needed, hence elaborating a bit.
First, one must note the denominators can't be zero, so the domain of $x$ does not contain $\{-1, -2\}$.  Given this, you can multiply by $(x+1)(x+2)$ to get your equation as
$$(x+1)x + (x+2)x = kx(x+1)(x+2)$$
$$\iff kx^3 + (3k-2)x^2+(2k-3)x=0$$
Clearly $x=0$ is a root.  If in all there are exactly two roots, and $a$ is the other root, then we could have the roots as $(0,0,a), (0, a, a), (0,a)$ (of course without considering ordering).
For $x=0$ to be a double root, it is evident $\color{red}{k=\frac32}$ must be true.  So this is one value for $k$, which is indeed a complex number. (You should verify that this value indeed gives the last root in the allowable domain.)
If $x=0$ is not a multiple root, then we have what remains after cancellation of $x$, viz. $kx^2+(3k-2)x+(2k-3)=0$ should have exactly one more root.  This can happen in two ways,:
EITHER
the quadratic $kx^2+(3k-2)x+(2k-3)=0$, has a double root.  Thus it's discriminant is zero, hence
$$(3k-2)^2-4k(2k-3)=0 \iff k^2+4=0$$
which gives $\color{red}{k=\pm2i}$, which are the only other possible values.  (Once again you should verify this leads to roots for the quadratic which are allowable in the domain.)
OR
we have essentially only one root, i.e. $\color{red}{k=0}$ so that what remains after cancellation is really $-2x-3=0 \implies x =-\frac32$.
Collecting all cases, the values possible are $\color{red}{k \in \{\frac32,2i, -2i, 0\}}$
A: Good!
Yes, multiplying both sides by $(x+ 1)(x+ 2)$ "reduces" the equation to $(x+ 1)x+ (x+ 2)x= kx(x+ 1)(x+ 2)$.  Now, I would do the indicated multiplications to get $x^2+ x+ x^2+ 2x= kX*(x^2+ 3x+ 2)= kx^3+ 3kx^2+ 2kx$ which can be written as $kx^3+ (3k- 1)x^2+ (2k- 1)x= 0$.  The left side can be factored as $x(kx^2+ (3k- 1)x+ (2k- 1))= 0$.  One root is $x= 0$ and, if $x$ is not $0$, $kx^2+ (3k- 1)x+ (2k-1)= 0$ can be solved by the "quadratic formula".
