Question

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

• A. $\sqrt{3}+2\sqrt{2}$
• B. $\sqrt{2}+2\sqrt{3}$
• C. $\sqrt{3}+\sqrt{2}$
• D. $\sqrt{3}-2\sqrt{2}$

Answer 1

The correct option is A $\sqrt{3}+2\sqrt{2}$

Required difference $=(\sqrt{12}+\sqrt{18})-(\sqrt{3}+\sqrt{2})$
$=(\sqrt{4\times 3}+\sqrt{9\times 2})-(\sqrt{3}+\sqrt{2})=2\sqrt{3}+3\sqrt{2}-\sqrt{3}-\sqrt{2}$
$=(2\sqrt{3}-\sqrt{3})+(3\sqrt{2}-\sqrt{2})=\sqrt{3}+2\sqrt{2}$