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What would be the easiest way to solve the following set of equations:$$ x + y^2 = 7 $$$$ x^2 + y = 11$$

I've been trying substitution method but end up in a $4$th degree bi-quadratic equation. Thanks for your time!

EDIT: Final equation: $y^4 - 14y^2 + y + 38 = 0$. I'm only concerned about the real solutions. Is it possible to write it as a product of two quadratic equations?

Cookie
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Vinit Shandilya
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  • Perhaps you should give more details of what quartic equation you got. That is what one would expect elimination to produce, but maybe an algebraic error caused the problem to become too hard. – hardmath Mar 10 '15 at 15:40
  • Aren't there four real solutions? – Casteels Mar 10 '15 at 15:52
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    @VinitShandilya It doesn't seem possible to write your final equation as a product of two quadratic equations, but rather you can write it as a product of a linear equation and a cubic equation. Namely, after long division or synthetic division, you can write your equation as $$(y-2)(y^3+2y^2-10y-19)=0.$$ From this, you get $y=2$ as your only rational root. Then you can substitute that in and get $x=3$. – Cookie Mar 10 '15 at 16:46

2 Answers2

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In finding real solutions, consider substituting your first equation into your second, that is, you will have $$x^2\pm\sqrt{7-x}=11,$$ and so $$x^2-11=\pm\sqrt{7-x}.$$ Square both sides: $$x^4-22x^2+121=7-x.$$ Algebraically rearranging, $$x^4-22x^2+x+114=0.$$ Rational root test will tell you that $x=3$ is a root, and hence a solution. WolframAlpha verifies this. Long division or synthetic division reduces our equation to $$(x-3)(x^3+3x^2-13x-38)=0.$$ We have $x=3=0$, which gives us $x=3$, and $$x^3+3x^2-13x-38=0.$$ The latter equation can be solved using Cardano's formula and find the remaining real roots.

Anyway, what we can do for now: substitute the $x=3$ in into either of your original equations to find your $y$, which should be $y=2$.

NB: There are three more real roots of $x$ that solve the fourth-degree polynomial, but those roots are irrational since they are not found in any possible roots of the rational root test.

Cookie
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First note that your two original equations represents two parabolas with orthogonal axes, so they have four intersection points, i.e. the systems has four solutions.

Your equation $y^4-14y^2+y+38=0$ has the obvious solution $y=2$ so it factorize in: $$ (y-2)(y^3+2y^2-10y-19)=0 $$ Since a cubic has always a real solution there are at least two real solution of your quartic.

Now you can study the function $f(y)=y^3+2y^2-10y-19$ . The derivative is $f'=3y^2+4y-10 $ and you can solve the second degree equation for stationary points, finding two real solutions $y=\dfrac{-2\pm\sqrt{34}}{3}$.

With a bit of work you can see that the function has a positive maximum and a negative minimum, so it must have three real solutions as we expected.

Emilio Novati
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