# Analytic solution of a system of four second order polynomials

Can I systematically solve in $\mathbb{R}^4$ the following system without using Grobner basis algorithm ? If not, can I find the exact number of solutions ?

\begin{equation*} \left\{ \begin{aligned} X_1^2+X_2^2=a\\ X_1X_3+X_2X_4=b\\ X_3^2+X_4^2+eX_1+fX_2=c\\ eX_3+fX_4=d \end{aligned} \right. \end{equation*}

$a,b,c,d,e,f$ are known reals constants

One can use resultants. Given two equations of degrees $k_1, k_2$,

$$F_1(x_1,x_2,\dots,x_m) = 0$$

$$F_2(x_1,x_2,\dots,x_n) = 0$$

the resultant can eliminate one variable between them. By applying it repeatedly to a system, you can reduce it to just one equation. (It is computationally expensive, but if the degrees $k_i$ of your equations are small enough, then it is feasible.)

For your system, I find one can resolve it into a sextic in $x_4$. (I used Mathematica). Thus, in general, you have six solutions. However, the transformation,

$$x_4 = \frac{ez+df}{e^2+f^2}$$

simplifies it to the form,

$$z^6+c_1z^4+c_2z^2+c_3=0$$

so the sextic has a solvable Galois group and your $x_i$ in fact can be expressed in radicals.

• The command in Mathematica is, for example, $Resultant[x^2+y^2-z^2,x^3+y^3+z^3,z]$ which eliminates the variable $z$. You can also use the Eliminate command like this. – Tito Piezas III Mar 10 '15 at 10:03
• The transformation worked wonders – Antoine Bassoul Mar 11 '15 at 13:12
• @AntoineBassoul: What is the background of this problem? A system of non-linear equations is not always solvable in radicals, so I'm curious why it is so in this particular case. – Tito Piezas III Mar 12 '15 at 4:29
• We're trying to estimate the kinematic parameter $x(t=t_0),y(t=t_0),v_x,v_y$ of a target with a rectilinear uniform motion (MRU) in the plane. We're measuring the distance between the target and an observer which is manoeuvring (think of boats), the observer's manoeuver is known. If the observer doesn't move, you can't estimate the target's motion only from distance measurements, same problem if the observer is in MRU. It is however observable if the observer has a uniformly accelerated trajectory. In fact in the above equations $e$ and $f$ are the x and y coordinate of the acceleration vecto – Antoine Bassoul Mar 12 '15 at 9:35
• How did you find the transformation to simplify the polynomial ? Is this mathematician intuition or did you asked mathematica ? – Antoine Bassoul Mar 18 '15 at 16:40