How would you solve the following system of equations: $$ x^2 + y = 4 \\ x + y^2 = 10 $$ Thanks very much!

I tried defining y in terms of x and then inserting in to the second equation:

$$ y = 4 - x^2 \\ x + (4 - x^2)^2 = 10 $$

Expand the second equation:

$$ x + 16 - 8x^2 + x^4 = 10 $$

Rearrange the terms:

$$ x^4 - 8x^2 +x + 6 = 0 $$

I tried factoring out this polynomial to simplify it for solving, but didn't succeed :(


2 Answers 2


$$y=4-x^2$$ put it in 2nd equation

$$x+(4-x^2)^2=10$$ $$x+16+x^4-8x^2=10$$ $$x^4-8x^2+x+6=0$$ $$x^4-x^3+x^3-7x^2-x^2+7x-6x+6=0$$ $$x^4-x^3+x^3-x^2-7x^2+7x-6x+6=0$$

$$x^3(x-1)+x^2(x-1)-7x(x-1)-6(x-1)=0$$ $$(x-1)(x^3+x^2-7x-6)=0\implies x-1=0\implies x=1$$

solve the above equation one of the value of x=1 and y=3

  • $\begingroup$ I should have posted your last equation as my question, because that is my problem. Could you please lay out steps for solving that (4th degree) equation? $\endgroup$
    – Davit
    Commented Aug 14, 2013 at 9:00
  • 1
    $\begingroup$ If you have to, and you don't have the formula with you, try to substitute factors of 6 into x and see if that fits, including negative factors. Because if $a$ is a root, then $x-a$ is a factor of $x^4 - 8x^2 + x+6$, so when you match the constant term, $a$ must be a factor of $6$. If all roots are irrational or complex, then there are not much to do with that with bare hand. $\endgroup$
    – peterwhy
    Commented Aug 14, 2013 at 9:16
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    $\begingroup$ This at least used to be called the Theory of Equations. Descartes' Rule of Signs tells us that the last equation has two or no positive solutions and two or no negative solutions (hence no, two or four nonreal solutions). The rational solutions must be in {-6, -3, -2, -1, 1, 2, 3, 6}. Checking these (e.g., via synthetic division) shows that 1 is the only rational solution. Dividing by $x-1$, we get that the other solutions are the zeroes of $x^3+x^2-7x-6$. There is one positive, irrational solution to this. I think this gives you enough terms to research. $\endgroup$ Commented Aug 14, 2013 at 9:23
  • $\begingroup$ Indeed, continuing, one gets three other real, irrational roots. There is one in each of the intervals (-3, -2), (-1,0) and (2, 3). You need Cardano's formula to express these exactly. $\endgroup$ Commented Aug 14, 2013 at 9:32
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    $\begingroup$ @iostream007 Thanks a lot. Beautiful factoring :) $\endgroup$
    – Davit
    Commented Aug 14, 2013 at 9:47

To solve general equations of the form $ax^4+bx^3+cx^2+dx+e=0$ requires quartic formula.

  • 4
    $\begingroup$ This is true in general, but in this case it's quite easy to find at least one root by inspection. Using the quartic formula in this case is a bit overkill imo. $\endgroup$
    – Daniel R
    Commented Aug 14, 2013 at 9:14
  • $\begingroup$ Though one might want to resort to Cardano's formula (for the cubic). $\endgroup$ Commented Aug 14, 2013 at 9:29

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