# How to solve this system of quadratic equations?

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?

• $x_i = -1$ for any $v_j$ seems to do the trick. Do you need something more? Nov 10, 2015 at 15:43
• 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. Nov 10, 2015 at 15:49
• @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. Nov 11, 2015 at 10:55

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

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).

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