2
$\begingroup$

I've got a school system of equations:

equation

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$$\begin{cases}u+v^2=20,\\v+u^2=20.\end{cases}$$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.