TITLE: Solving a non-zero quadratic equation of two variables
QUESTION [8 upvotes]: I have a problem where I am to find the intersection between the two curves
$$
\begin{cases}
x^2 - y^2 = 3 \\
xy = 2,
\end{cases}
$$
which I can easily see that the two points $\pm(2,1)$ are the real solutions to this problem, but I don't know how to solve for this systematically. I tried two approaches to solve for it as a quadratic equation:
approach 1 (insert the second equation in the first):
$$
4x^2 - 4y^2 + 4xy = 20 \Leftrightarrow (2x + y)^2 - 5y^2 = 20 \Leftrightarrow x=\dfrac{\pm\sqrt{5y^2 + 20} - y}{2}
$$
approach 2 (substitute $y$ for $x$):
$$
x=\frac{2}{y} \Rightarrow x^2 - \frac{4}{x^2} = 3 \Leftrightarrow x^4 - 3x^2 - 4 = 0
$$
But I don't know how to continue from here. I found this math.stackexchange question where they solve a similar equation using these two approaches, but they don't end up with a constant under the root since they have $0$ on the right-hand side. Also, since that equation don't have any real solutions, I don't get how to apply it to this equation.
How do I solve this kind of equation systematically? (not 100% sure what type of the equation it is)
Edits:
Corrected substitution from incorrectly substituting $y$ for $y$. I had arrived at this step earlier, but I still don't know how to solve the equation.

REPLY [3 votes]: Once you get to $x^4-3x^2-4=0$, it follows quite easily. You can factor this as $(x^2-4)(x^2+1)$, or if you want to do it more systematically, you can do a substitution $u= x^2$. Then you have $u^2-3u-4=0$ and now you can apply the quadratic formula to solve for $u$.