How do I factorize equations of the form $x^2 + Bxy + Cy^2 = 0$ Given equation
$$ x^2 + Bxy + Cy^2 = 0. $$
I want to factorize it in the form
$$ (x + my)(x + ny) = 0. $$
What are the values of $m$ and $n$ in terms of $B$ and $C$?

I tried writing the equation as
$$ x^2 + Bxy + Cy^2 = (x + kCy)(x + \dfrac{1}{k}y) = 0. $$
In this case the following equation must be satisfied:
$$ kC + \dfrac{1}{k} = B $$
I solved it for $k$ to find
$$ k = \dfrac{B \mp \sqrt{B^2 - 4C}}{2C}. $$
That makes
$$ x^2 + Bxy + Cy^2 = (x + \dfrac{B \mp \sqrt{B^2 - 4C}}{2}y)(x + \dfrac{2C}{B \mp \sqrt{B^2 - 4C}}y) = 0. $$
Which implies that
$$ \begin{array}{rcl}
m &=& \dfrac{B \mp \sqrt{B^2 - 4C}}{2} \\
n &=& \dfrac{2C}{B \mp \sqrt{B^2 - 4C}},
\end{array} $$
and there are at most four $(m,n)$ pairs.
Is my solution correct? Are there any (other) solution methods for this factorization?
 A: Your approach is quite satisfactory. It just skips a formal step by assuming the solution form rather than deriving it.
$$
(x+my)(x+ny) = x^2 +(m+n)xy + mny^2
$$
Therefore
$$
B = m+n\\
C = mn
$$
From the equation for $C$, we can write
$$
n = \frac{C}m
$$
which turns the first equation into
$$
B = m + \frac{C}m
$$
Now, if you let $k=\frac1m$, then you can write these as
$$
n = Ck\\
m = \frac1k\\
B = \frac1k + Ck
$$
And this is the same as the assumed form that you started with. As you can see, this derivation requires no additional assumptions (except, perhaps, that $m\neq 0$, which should be fine because $m=0$ would make $C=0$ anyway), and so there are no other solutions than the one you provided.
A: Nice solution! Please note that Glen O has answered your question very well, and what comes below does not. Its purpose is to give a slightly different view of the matter, and to connect it with other things that you know.
Let us factor $4x^2+4Bxy+4Cy^2$. Completing the square, we get $(2x+By)^2-(B^2-4C)y^2$. If $B^2-4C\lt 0$, there is no factorization over the reals. But if we are willing to work with complex non-real numbers, we can go on. If we are unwilling, assume that $B^2-4C\ge 0$. 
Let $D=B^2-4C$. We are looking at $(2x+By)^2-(\sqrt{D}y)^2$. This is a difference of squares, and factors as 
$$(2x+By+\sqrt{D}y)(2x+By-\sqrt{D}y).$$
But we had multiplied by $4$, so let's undo that by dividing each term by $2$. We get
$$\left(x-\frac{B+\sqrt{D}}{2}y\right)\left(x+\frac{B+\sqrt{D}}{2}y\right).$$
Remark: Note that in particular we have derived the Quadratic Formula. For an alternate approach, let's use the Quadratic Formula. Divide our quadratic by $y^2$, not worrying about possible division by $0$, and temporarily let $w=\frac{x}{y}$. Then our quadratic is $y^2(w^2+Bw+C)$. The quadratic equation $w^2+Bw+C=0$ has the roots $\frac{-B\pm\sqrt{B^2-4C}}{2}$, and therefore the polynomial $w^2+Bw+C$ factors as 
$$w^2+Bw+C=\left(w-\frac{-B+\sqrt{B^2-4C}}{2}\right) \left(w-\frac{-B-\sqrt{B^2-4C}}{2}\right) .$$
Now replace $w$ by $\frac{x}{y}$, and multiply each term by $y$.
