Given a system of two bivariate quadratic polynomials:

\begin{eqnarray} a_0 + a_1 x + a_2 y + a_3 xy+a_4 x^2 + a_5 y^2 &= 0 \\ b_0 + b_1 x + b_2 y + b_3 xy+b_4 x^2 + b_5 y^2 &= 0 \end{eqnarray}

is there a closed-form formula for finding all of its roots?

I'm asking becase according to Bézout's theorem there are 4 solutions and there is a closed-form formula for quartic polynomial.

  • $\begingroup$ Here's an ugly solution: Solve both equations for $y$ in terms of $x$. Then equate the solutions of these. You should get a quartic in $x$ after eliminating radicals. $\endgroup$ – KReiser Jul 1 '16 at 22:39

Yes. To reduce it to one equation in one unknown, the command for Walpha is

Resultant[a0 + a1 x + a2 y + a3 x y + a4 x^2 + a5 y^2, b0 + b1 x + b2 y + b3 x y + b4 x^2 + b5 y^2, y]

which eliminates $y$ and will give you a quartic solely in $x$,


where the $P_i$ are in the $a_i$ and $b_i$. Of course, with symbolic coefficients, it is pretty ugly.

P.S. There is a maximum of four solutions but if your $a_i$ and $b_i$ obey certain constraints such that $P_4=0$, you will end up with a cubic. Or, if $P_4=P_3=0$, then even a quadratic in $x$. For ex. if, $$a_5=\frac{-a_4b_3^2+2a_3b_3b_4}{4b_4^2}$$ $$b_5=\frac{b_3^2}{4b_4}$$

then this constraint causes $P_4=0$ and $(1)$ is just a cubic.

  • $\begingroup$ "obey certain constraints" - do you happen to know more details about it? $\endgroup$ – Ecir Hana Nov 2 '16 at 9:23
  • 1
    $\begingroup$ @EcirHana: I've edited and given an example. $\endgroup$ – Tito Piezas III Nov 2 '16 at 13:33

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