TITLE: Solution af a system of 2 quadratic equations
QUESTION [5 upvotes]: I have a system of two quadratic equations with unknowns $x$ and $y$:
$$a_{1 1} x y + a_{1 2} x^2 + a_{1 3} y^2 + a_{1 4} x + a_{1 5} y + a_{1 6} = 0,\\
a_{2 1} x y + a_{2 2} x^2 + a_{2 3} y^2 + a_{2 4} x + a_{2 5} y + a_{2 6} = 0,$$
where $a_{i j}$ are arbitrary scalars.
Is there an algebraic solution of the above system?

REPLY [8 votes]: From Intersecting two conics:

The solutions to a two second degree equations system in two variables may be seen as the coordinates of the intersections of two generic conic sections. In particular two conics may possess none, two or four possibly coincident intersection points. The best method of locating these solutions exploits the homogeneous matrix representation of conic sections, i.e. a 3x3 symmetric matrix which depends on six parameters.
  The procedure to locate the intersection points follows these steps:

given the two conics $C_1$ and $ C_2$ consider the pencil of conics given by their linear combination $\lambda C_1 + \mu C_2$
identify the homogeneous parameters $(\lambda,\mu)$ which corresponds to the degenerate conic of the pencil. This can be done by imposing that $\det(\lambda C_1 + \mu C_2) = 0$, which turns out to be the solution to a third degree equation.
given the degenerate conic $C_0$, identify the two, possibly coincident, lines constituting it
intersects each identified line with one of the two original conic; this step can be done efficiently using the dual conic representation of $C_0$
the points of intersection will represent the solution to the initial equation system