Dear Uncle Colin,

The question tells me that $5x^2 + 4x + 4 + 9y^2 +12xy =0$ and that I have to find $xy$. I’ve tried plotting it in Desmos, but nothing shows up! What should I do?

Dramatically Overthinking Things

Hi, DOT, and thanks for your message!

I think there are at least a couple of methods here, and I’ll give them both so you can take your pick.

The first, and the one that jumps out at me, is to complete the square on $y$; the second is to mess around with a quadratic formula, which I suspect comes out much the same way.

Completing the square on $y$

If we rewrite it as $9y^2 + 12xy + 5x^2 + 4x + 4 = 0$, we can replace the first pair of terms with $(3y + 2x)^2 - 4x^2$. This makes the whole thing $(3y+2x)^2 + x^2 + 4x + 4 = 0$.

The last three terms are a perfect square, $(x+2)^2$, so the equation can be rewritten as $(3y+2x)^2 + (x+2)^2 = 0$. This is only true when both of the terms are 0, so $3y=-2x$ and $x=-2$. That means $y = \frac{4}{3}$, and $xy = -\frac{8}{3}$.

Note that there is only point on the curve – that’s why Desmos didn’t plot it!


A little less subtly, you can choose to pretend $x$ or $y$ is a constant. This time, I’ll fix $y$ and let $x$ be the variable.

Rewriting the equation as $5x^2 + (4+12y)x + (4 + 9y^2) = 0$ and applying the quadratic formula gives:

$x = \frac{-(4+12y)\pm \sqrt{(4+12y)^2 - 20(4+9y^2)}}{10}$.


Let’s look at the inside of that square root: it’s $(16 + 96y + 144y^2) - (80-180y^2)$ or $-36y^2 + 96y - 64$. That factorises as $-(6y-8)^2$.

For us to take the square root of it, it can’t be negative – and the only value of $y$ that gives a non-negative value is $y=\frac{4}{3}$ to make the square root 0.

Putting this into the formula gives $x = \frac{-20}{10}= -2$, as we got in the first part.

Hope that helps!

- Uncle Colin

* Edited 2021-06-16 to fix some LaTeX and a typo. Thanks, Sam!