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?

  • $\begingroup$ $x_i = -1$ for any $v_j$ seems to do the trick. Do you need something more? $\endgroup$
    – Chinny84
    Nov 10, 2015 at 15:43
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. Typing the question is preferable to including an image since images cannot be searched. $\endgroup$ Nov 10, 2015 at 15:49
  • $\begingroup$ @Chinny84 Would you please explain your answer in a bit more details? I have no idea of how to solve the question. Even if there are some limitations to put on $x_i$ values in order to solve them, it is fine. $\endgroup$
    – MM Khan
    Nov 11, 2015 at 10:55

2 Answers 2


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


As suggested before, you can substitute




Then the new equations are:





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:




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

  • $\begingroup$ $y_i=0$ is not the unique solution. $\endgroup$
    – user91684
    Sep 2, 2017 at 14:25
  • 1
    $\begingroup$ $Y=\alpha V$ where $\alpha$ is a scalar. $\endgroup$
    – user91684
    Sep 2, 2017 at 16:20
  • $\begingroup$ right. didn't notice that $\endgroup$
    – David
    Sep 2, 2017 at 16:26
  • $\begingroup$ I fixed the solution above. $\endgroup$
    – David
    Sep 11, 2017 at 15:24

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