Solving a non-zero quadratic equation of two variables 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.
 A: $$\begin{cases}
x^2 - y^2 = 3 \\
xy = 2,
\end{cases}$$
Substitution is as follows, you substitute $y$ in first equation with it's value according to second equation so in this case. 
$$x^2-\frac4{x^2}=3$$
That is $y=2/x$ according to second equation. Substitution should always lead to reduction in number of variables. 
Solving $x^2-\frac4{x^2}=3$ for $x$ yields $$x^4 -3x^2-4=0.$$ This equation has two real solutions $x=2$ and $x=-2$ (there are also two complex solutions at $x=i$ and $x=-i$). Plugging $x= \pm 2$ into the initial equations yields $y=\pm 1$. Likewise, you can obtain the complex solutions of $y$ by plugging $x=\pm i$ into the initial equations which yields $y=\pm 2i$.
A: An idea for you:
Multiply second equation by two: $\;2xy=4\;$ , and subtract it from first equation, getting:
$$x^2-2xy-(y^2-1)=0$$
The solution to this quadratic in $\;x\;$ is:
$$\Delta=4y^2+4y^2-4=4(2y^2-1)\implies x_{1,2}=y\pm\sqrt{2y^2-1}$$
and now substitute in first equation, say:
$$\overbrace{y^2\pm2y\sqrt{2y^2-1}+2y^2-1}^{=x^2}-y^2=3\implies y^2\pm y\sqrt{2y^2-1}=2\implies$$
$$y^4-4y^2+4=2y^4-y^2\implies y^4+3y^2-4=0\iff (y^2+4)(y^2-1)=0\implies$$
$$y=\pm1\implies x=\pm1\pm\sqrt{2-1}=\pm1\pm1$$
and since the solutions $\;(x,y)\;$ must have the same sign and both are clearly different from zero, we finally get the candidates:
$$(2,1)\,,\,\,(-2,-1)$$
Finally, check whether the above two really are solutions (since we squared stuff here...)
A: Your second approach would have been my choice as well. Once you've found that
$$x^4-3x^2-4=0,$$
you can solve this as a quadratic in $x$ to find that $x^2=-1$ or $x^2=4$, as others have explained.
Your first approach is also fine, though your choice of substitution is not. You can see that
$$y^2=x^2-3,$$
and squaring the second equation yields
$$4=x^2y^2=x^2(x^2-3)=x^4-3x^2,$$
leading to the same equation for $x$ as your first approach.
A: You made a mistake during calculation. What you should have done during substitution is to get the value $x=\frac2y$ from the second equation and substitute $x$ with that value in the first equation which should result in 
$$\left(\frac{2}{y}\right)^2 - y^2=3$$ which is an equation that is easy to solve.
A: Another approach, based on (complex) completing of the square.
Notice that if the first equation would have the term $x^2+y^2$, it would be much easier because then by taking the sum and the difference of the two equation, you would get squares. But the minus sign in front of the $y$ is ''wrong'', so let's change it using complex numbers!
$$(x+iy)^2=x^2-y^2+2ixy=3+4i$$
$$(x-iy)^2=x^2-y^2-2ixy=3-4i$$
Write the numbers in the polar form,
$$(x+iy)^2=5 e^{i\phi}$$
$$(x-iy)^2=5 e^{-i\phi}$$
Here $\phi$ is such that $\tan(\phi)=4/3$.
So, take the square roots:
$$x+iy=\sqrt{5} e^{i\phi/2}$$
$$x-iy=\sqrt{5} e^{-i\phi/2}$$
Take the sum and the difference:
$$x=\sqrt{5} \frac{e^{i\phi/2}+e^{-i\phi/2}}{2}$$
$$y=\sqrt{5} \frac{e^{i\phi/2}-e^{-i\phi/2}}{2i}$$
Remember the complex equations for sine and cosine:
$$x=\sqrt{5} \cos(\phi/2)$$
$$y=\sqrt{5} \sin(\phi/2)$$
To find $\cos(\phi/2)$ (thanks to hints from Michael Seifert in the comments): It is easy to see that $\cos(\phi)=3/5$. The double-angle formula for cosine says:
$$\cos(\phi)=2 \cos^2(\phi/2)-1$$
so
$$\cos(\phi/2)=\pm \sqrt{\frac{\cos(\phi)+1}{2}}=\pm \frac{2}{\sqrt{5}}$$
And likewise $\sin(\phi/2)=\pm\frac{1}{\sqrt{5}}$.
Altogether, this leads to the solution given by others! This might not be the easiest method to discover, especially not the last step, but I always like it if complex numbers are used to solve a problem in the reals.
A: 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 $u$.
A: Multiply the first equation by $x^2$ and substitute $xy$ from the second:
$$x^4-x^2y^2-3x^2=x^4-4-3x^2=0.$$
This is a biquadratic equation, which gives the roots $x^2=4$ and $x^2=-1$. If you are only interested in the real solutions,
$$x=\pm2,y=\frac2x.$$
A: If you only need to find real solutions, here is an approach using trigonometry: 
The equation is equivalent to $z^{2} = 3+4i$ where $z=x+yi$. Hence, there exists two solutions $z=\pm (3+4i)^{1/2}$. Using a polar coordinate, if $z=re^{i\theta}$, we should have
$$
r= \sqrt{|3+4i|} = \sqrt{5}, \quad \cos 2\theta = \frac{3}{5},\quad  \sin 2\theta = \frac{4}{5}
$$
From this, we have 
$$
\tan 2\theta = \frac{4}{3} = \frac{2\tan\theta}{1-\tan^{2}\theta}, \quad \tan\theta = \frac{1}{2}
$$
and
$$
z = \pm \sqrt{5}(\cos\theta + i\sin \theta) = \pm (2+i),
$$
which gives two solutions $\pm(2, 1)$. 
A: With symmetric functions:
First it's enough to determine $x^2$ and $y^2$ since
$$\begin{cases}
x^2-y^2=3\\ xy=2
\end{cases}\iff\begin{cases}
x^2-y^2=3\\x^2y^2=4&\text{and $x$, $y$ have the same sign.}
\end{cases}$$
So we  set $X=x^2$,$\;Y=y^2$. We have to solve the system
$\;\begin{cases}
X-Y=3,\\ XY=4,\end{cases}$ in positive numbers.
This system implies 
$\; (X+Y)^2=(X-Y)^2+4XY=9+16=25 $ , so we obtain the system
$$\begin{cases}
X+Y=5,\\ XY=4,\end{cases}\iff \begin{gathered}\{X,Y\}\;\text{ are the (positive) solutions of the quadratic equation }\\[1ex] T^2-5T+4=0
\end{gathered}$$
Now  the solutions of this equation are $\{1,4\}$, and we have $X=Y+3>Y$, so the solution are $x^2=X=4,\;y^2=Y=1$, whence
$$(x,y)=(2,1)\;\text{or}\;(x,y)=(-2,-1).$$
A: I thought about removing the constant terms.
$$2x^2-2y^2 = 6 = 3xy$$
$$ 2x^2 - 3xy -2y^2 = 0$$
$$(2x+y)(x-2y) = 0$$
$$\text{$y = -2x$ or $y=\frac 12x$}$$
Substituting into $xy=2$, I get
$$\text{$(x,y)\in \{(i,-2i), (-i,2i), (-2,-1),(2,1) \}$}$$
