Ask Uncle Colin: A Quartic

Dear Uncle Colin,

I’m trying to solve the system of equations $2x^2+y^2=18$ and $xy=4$. I’ve substituted for $y$ and ended up with a quartic. How do I solve that?

- Regarding A Polynomial, Having Solutions Or Not?

Hi, RAPHSON, and thanks for your message!

Your approach seems pretty sensible – let $y=\frac{4}{x}$, which makes the first equation $2x^2+\frac{16}{x^2}=18$, or $2x^4-18x^2+16=0$.

That simplifies to $x^4-9x^2+8=0$. Can we solve a quartic?

In this form¹, we can! let $z=x^2$, so we have $z^2-9z+8=0$. That factorises as $z=8$ or $z=1$.

Now we need to solve for $x$:

• $x^2=1$, so $x=\pm 1$
• $x^2=8$, so $x=\pm 2\sqrt{2}$

… and those are the four solutions for $x$ in the original quartic. The solutions to the system are $(1,4),(-1,-4),(2\sqrt{2},\sqrt{2})$ and $(-2\sqrt{2},-\sqrt{2})$.

Another approach

• $2x^2+y^2=18$ so $4x^2+2y^2=36$ [1]
• $xy=4$ so $9xy=36$ [2]
• [1]-[2] give $4x^2-9xy+2y^2=0$
• This factors as $(4x-y)(x-2y)=0$
• So either $4x=y$ or $x=2y$

These can be solved simultaneously with either original equation to get the same four answers as before.

Hope that helps!

- Uncle Colin

Footnotes:

1. Where there are no odd-degree terms