TITLE: Can the quadratic formula be used with variable coefficients?
QUESTION [11 upvotes]: Could we use the quadratic formula on an expression such as $z^2xy - zx^2y+y = 0$ to find $z$ in terms of $x$ and $y$?

REPLY [4 votes]: Write $$z^2xy - zx^2y + y = 0$$ 
as $$(xy)z^2 - (x^2y)z + (y) = 0$$ 
so that you can see $a = xy$, $b=-x^2y$ and $c=y$.
Then 
\begin{align}
   z  &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\
   z  &= \dfrac{x^2y \pm \sqrt{x^4y^2-4xy^2}}{2xy} \\
   z  &= \dfrac{x^2y \pm |y|\sqrt{x^4-4x}}{2xy} \\
   z  &= \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x} \\
\end{align}
Notice that $y$ has disappeared. That is because 
$(xy)z^2 - (x^2y)z + (y) = 0 \iff y(xz^2 - x^2z + 1) = 0$.
So we need to notice that $y=0$ is a solution. If $y\ne 0$ then we get
$z = \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x}$; where we are going to need to require that $x \ne 0$ and, if we only want real-valued solutions, $x^4-4x \ge 0$.