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 form((Where there are no odd-degree terms)), 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 $(1,4), (-1,-4), (2\sqrt2, \sqrt{2})$ and $(-2\sqrt{2},-\sqrt{2})$.

### Another approach

• $2x^2 + y^2 = 18$ so $4x^2 + 2y^2 = 36$ 
• $xy = 4$ so $9xy = 36$ 
• - 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