Since 3 is the only quantity in the radical in the RHS, I surmised that c = 3, and thus 2ab = 4. Because a, b, and c are all rational, neither a^2 nor (b^2)c can contribute to the sqrt(3) term in the RHS. Then, equating the rational terms, I solved for a using the quadratic formula:
As I required a to be rational, a = +/- 2, and thus b = +/- 1 from (5), with the signs on each the same. So, since the convention is to take the positive square root,
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