Problem: After rationalizing the numerator of 3−23\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}33−2, the denominator in simplest form is:
Answer Choices:
A. 3(3+2)\sqrt{3}(\sqrt{3}+\sqrt{2})3(3+2)
B. 3(3−2)\sqrt{3}(\sqrt{3}-\sqrt{2})3(3−2)
C. 3−323-\sqrt{3} \sqrt{2}3−32
D. 3+63+\sqrt{6}3+6
E. none of these answers. Solution:
3−23⋅3+23+2=13+6\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} \cdot \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{1}{3+\sqrt{6}}33−2⋅3+23+2=3+61.