$(\sqrt{3}+2)^2 + (\sqrt{3}-2)^2 = (3+4\sqrt{3}+4)+(3-4\sqrt{3}+4) = 14$, which is a positive rational number.
Answer: A
Using $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$, the irrational terms ($4\sqrt{3}$) cancel out, leaving $2(3+4)=14$. Examiners expect the calculation shown briefly to justify the choice.