Advanced Algebra Square Roots

$a + b\sqrt{c} = \sqrt{4 - 2\sqrt{3}}$

We can easily speculate the c = 3 (what else could c be equal to, 17!?).

$a + b\sqrt{3} = \sqrt{4 - 2\sqrt{3}}$

We can safely square both sides of the equation...

$(a + b\sqrt{3})^2 = 4 - 2\sqrt{3}$

Open up parenthesis: $a^2 + 2ab\sqrt{3} + 3b^2 = 4 - 2\sqrt{3}$

Then we can see that the root 3 part corresponds to the root 3 part in the equation since a and b are integers:

$2ab\sqrt{3} = -2\sqrt{3}$

$ab = -1$

From earlier, we also obtained:

$a^2 + 3b^2 = 4$

How can you take it from here to find a and b? Good luck...