How to solve this system of quadratic equations?

Question

I have the following system of quadratic equations:

\begin{align*} v_2x_1^2 + v_2x_1 - v_1x_2^2 - v_1x_2 & = 0\\ v_3x_1^2 + v_3x_1 - v_1x_3^2 - v_1x_3 & = 0\\ v_3x_2^2 + v_3x_2 - v_2x_3^2 - v_2x_3 & = 0 \end{align*}

How can I solve them for $x_1$, $x_2$, and $x_3$ in terms of $v_1$, $v_2$, and $v_3$ only?

Answer 1

It is a hidden linear system. Put $y_i={x_i}^2+x_i$.

Answer 2

As suggested before, you can substitute

$y_1={x_1}^2+x_1$

$y_2={x_2}^2+x_2$

$y_3={x_3}^2+x_3$

Then the new equations are:

$v_2y_1-v_1y_2=0$

$v_3y_1-v_1y_3=0$

$v_3y_2-v_2y_3=0$

Updated:

The solution can be of form $y_1=a_1*v_1, y_2=a_2*v_2, y_3=a_3*v_3$. where $a_1, a_2, a_3$ are real numbers.

Therefore, The following equations need to be solved:

${x_1}^2+x_1=a_1*v_1$

${x_2}^2+x_2=a_2*v_2$

${x_3}^2+x_3=a_3*v_3$

You get two solutions for each $x_i$ and therefore 8 solutions overall (symbolic solutions since $a_i$ can be any real number).