I've got a school system of equations:


and i know it's roots are (2;2) (-2;-2) but how i could simplify it? I've tried many different ways like multiplying and raising to the power of 2 or 3 but the only useful result i got is x = -y.

  • $\begingroup$ What if you let $r = x^2, s = y^2$? You would throw out extraneous solutions. $\endgroup$ – Moo Aug 24 '17 at 19:53
  • $\begingroup$ You are missing $(2,2)$ and $(-2,-2)$. $\endgroup$ – Yves Daoust Aug 24 '17 at 20:03
  • $\begingroup$ oh sure, that's right. $\endgroup$ – Dmitrii Aug 24 '17 at 20:23

First use intermediate variables $u=x^2,v=y^2$ to make the degree less scary.


Then subtracting one from the other,

$$u-v+v^2-u^2=0=(u-v)(1-v-u)$$ so that $$u=v\lor u+v=1.$$

For the first case,

$$u+u^2=20$$ is a quadratic equation with solutions $u=-5,4$. The negative root must be rejected.

For the second case, eliminate $v$ by $v=1-u$ and

$$u+(1-u)^2=20$$ so that

$$u=\frac{1\pm\sqrt{77}}2.$$ The negative root must also be rejected. But with the positive one $v=1-u$ is also negative and this must be rejected.

Finally, we just have

$$\pm x=\pm y=2.$$


Let $a=x^2$ and $b=y^2$. Then you have the system:

$$ \left\{\begin{array}{l} a+b^2=20\\ b+a^2=20 \end{array}\right.$$

These imply $0\leq a,b\leq 2\sqrt{5}\,$.
Moreover, the second equation yields $b=20-a^2$ which we can substitute into the first to obtain

$$a+(400-40a^2+a^4)=20\iff a^4-40a^2+a+380=0$$

Any root $0\leq a \leq 2\sqrt{5}$ of the polynomial above corresponds to a solution of the system, and these in turn correspond to four solutions via choices of sign for $x=\pm \sqrt{a}$ and $y=\pm\sqrt{b}$.

At this point you can use your favorite method to show that $a=4$ is the only root in the desired interval.


Rearrange the second equation to $y^2=20-x^\color{red}{4}$ and substitute it into the the first equation \begin{eqnarray*} x^2+(20-x^4)^2=20 \\ x^8-40x^4+x^2+380=0 \\ (x^4-x^2-19)(x^2+5)(x+2)(x-2)=0 \end{eqnarray*} Is that enough ?

  • $\begingroup$ The factorization of the octic polynomial is by... magic. $\endgroup$ – Yves Daoust Aug 24 '17 at 20:05
  • 1
    $\begingroup$ @Moo Thanks ... edited $\endgroup$ – Donald Splutterwit Aug 24 '17 at 20:07
  • $\begingroup$ @YvesDaoust We knew $(x^2-4)$ would be a factor & it is a function of $x^2$ so we only need to factor the cubic $X^3+4X^2 +24X-95$ ... Hey we got lucky with $(X+5)$ ... $ \ddot \smile$ $\endgroup$ – Donald Splutterwit Aug 24 '17 at 20:11

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