Solve the following system. $$\left\{\begin{aligned} x^2+4 x y+6 y^2 &= 11 \\ y^2+3 y z+2 z^2 &= 0 \\ z^2+4 z x+12 x^2 &= 9 \end{aligned} .\right.$$
I have tried several tricks with this system but I am getting nowhere. Can someone please help me with just an idea?
On
I do not know what methods you are allowed to use, but the task here is to calculate a so called variety, that is the set of all points where an Ideal of a multivariate polynomial ring vanishes.
This can be tackled using a groebner basis of the same ideal with regards to the lexicographic term order.
Using sage we can quickly calculate this groebner basis like this https://sagecell.sagemath.org/?z=eJw1jcEKwjAQRO-F_sPSU7rGQKIIgnrvrc1VRBIaSiA2kF6SfL0rxdMb3jC7WtwyL7w-4A5jDGWNH2-C9uvCpsmaxCGm2SVqu-By17fNSPmZERUc4IwZC_GC5SeOICWHsncncpWosJLgUP-TipkoFeZ9c321zUBHtfCzM4GN9GQQS4rOri69rdn8xvovAVcsBQ==&lang=sage&interacts=eJyLjgUAARUAuQ==
although we could also do this with some effort by hand. This gives us the basis made from $$ x + \left(-\frac{199740516965191}{20000267695200}\right) z^{7} + \frac{157285576503190573}{4380058625248800} z^{5} + \left(-\frac{11168712229508887}{336927586557600}\right) z^{3} + \frac{7144738116409}{868542261600} z $$ $$ y + \frac{126612178716625}{50571783823968} z^{7} + \left(-\frac{23070511777829797}{2768805164362248}\right) z^{5} + \frac{64521464738498479}{11075220657448992} z^{3} + \frac{367100796197}{366026196624} z $$ $$ z^{8} + \left(-\frac{2540288788}{602720193}\right) z^{6} + \frac{3279495766}{602720193} z^{4} + \left(-\frac{523604604}{200906731}\right) z^{2} + \frac{76295547}{200906731} $$
Note that due to using the lexicographic term order these polynomials descend in number of variables. So to calculate the common roots you can start with the last polynomial to get all possible solutions for $z$, then for each use the second polynomial to get all corresponding solutions for $y,z$ and then for each of these insert into the first polynomial to get all solutions for $x,y,z$.
The trick is noticing that the second equation can be written as $$(y+z)(y+2x)=0$$ $$\Rightarrow y=-z \text{ or } y=-2z$$
Case 1: If $y=-z$, the system of equation becomes $$\begin{cases} x^2-4xz+6z^2=11\\ 12x^2+4xz+x^2=9\\ \end{cases}$$
which simplifies to $$\Rightarrow 13x^2+7z^2=20$$ $$\Rightarrow x=\frac{\sqrt{13(20-7z^2)}}{13}\text{ or } x=-\frac{\sqrt{13(20-7z^2)}}{13}$$
Case 2: If $y=-2z$, the system of equation becomes $$\begin{cases} x^2-4xz+6z^2=11\\ 12x^2+4xz+x^2=9\\ \end{cases}$$ $$\Rightarrow\begin{cases} x^2-8xz+24z^2=11\\ 24x^2+8xz+2x^2=18\\ \end{cases}$$ which simplifies to $$\Rightarrow 25x^2+26z^2=29$$ $$\Rightarrow x=\frac{\sqrt{29-26z^2}}{5}\text{ or } x=-\frac{\sqrt{29-26z^2}}{5}$$ and you can take it from there.