# system of two quadratic equations with two variables

Is there a general way to solve exactly a system of this shape (the $a_i$ are constants):

$$\begin{array}{cc}a_1x^2+a_2x+a_3y^2+a_4y+a_5=0\\ a_6xy+a_7x+a_8y+a_9=0 \end{array}$$

It comes from a geometrical problem: the first equation states two vectors have same length, and the second ones state that they are perpendicular.

It can be reduced to one equation of degree four with one variable, but is it the best we can do?

• In principle, the second equation can be used to solve for $x$ as a rational linear function in $y$. Substituting into the first equation and clearing denominators gives a quartic in $y$, and all quartics can be solved exactly. – davidlowryduda Jun 23 '14 at 19:49
• You ask whether reduction to a quartic is "the best we can do." We know exactly how to solve every quartic exactly, so I think that's pretty good. – davidlowryduda Jun 23 '14 at 20:06
• yes but I was looking for a simpler one, the general solution of quartic equations seems painful to program (I'm looking to implement this). – Denis Jun 23 '14 at 20:41

Assume your $a_1\neq0,\,a_6\neq0$. Dividing by these, you can have the slightly simpler system with two fewer constants,

$$x^2 + b_1 x + b_2 y^2 + b_3 y + b_4 = 0$$

$$x y + b_5 x + b_6 y + b_7 = 0$$

Eliminating $y$ results in a monic quartic of form,

$$x^4+ax^3+bx^2+cx+d=0\tag1$$

The solution of the general quartic is not that bad. It is just,

$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$

$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag3$$

where,

$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$

and the $v_i$ are the two roots of the quadratic,

$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$

P.S. However, one has to be take note that some older CAS have $(-n)^{1/3} = \text{complex}$, even for real $n$.