Question

By how much does $\sqrt{12}+\sqrt{18}exceed\sqrt{3}+\sqrt{2}?$

• A. (2) ^(1/2) – 4(3) ^(1/2)
• B. (3) ^(1/2) + 2 (2) ^(1/2)
• C. 2 ( (3) ^(1/2) –(2) ^(1/2) )
• D. 3 ((3) ^(1/2) – (2) ^(1/2))
• E. None of these

Answer 1

The correct option is B

(3) ^(1/2) + 2 (2) ^(1/2)

$\begin{array}{c}\left(\sqrt{12}+\sqrt{18}\right)-\left(\sqrt{3}+\sqrt{2}\right)=\left(\sqrt{4\times 3}+\sqrt{9\times 2}\right)-\left(\sqrt{3}+\sqrt{2}\right)=\left(2\sqrt{3}+3\sqrt{2}\right)-\left(\sqrt{3}+\sqrt{2}\right)\\ =2\sqrt{3}-\sqrt{3}+3\sqrt{2}-\sqrt{2}=\sqrt{3}+2\sqrt{2}.\end{array}$