I don't know of any simpler method, however there is a formula for this which I wouldn't recommend trying to memorize if you don't answer questions like these often.
There is another way to approach the same problem, with the same method. Who knows, you might find it interesting:
[tex]a^2+b^2=3[/tex] (1) , [tex]2ab=-2\sqrt{2}[/tex] (2)
[tex](a^2+b^2)^2-(2ab)^2=(a^2-b^2)^2[/tex]
Sub (1) and (2) into equation: [tex]3^2-(-2\sqrt{2})^2=(a^2-b^2)^2[/tex]
Therefore, [tex](a^2-b^2)^2=1[/tex] (3)
but from (1): [tex]a^2=3-b^2[/tex]
Sub (1) into (3): [tex](3-2b^2)^2=1[/tex]
Solving for b: [tex]3-2b^2=\pm 1 \Rightarrow b^2=\frac{3\pm 1}{2} \Rightarrow b=\pm \sqrt{\frac{3\pm 1}{2}}[/tex]
Take each case for b, and substitute back into (2) to solve for a.
This other approach doesn't really simplify things, but I only use it just to avoid trying to factorize the quadratic in a2 (yes, I don't enjoy factorizing due to the trail and error nature of it).