Question

Simplify: $(\sqrt{75}-\sqrt{45})(\sqrt{20}+\sqrt{12})$.

Answer 1

The given surds can be simplified as follows:

$(\sqrt{75}-\sqrt{45})(\sqrt{20}+\sqrt{12})=(\sqrt{25\times 3}-\sqrt{9\times 5})(\sqrt{4\times 5}+\sqrt{4\times 3})=(5\sqrt{3}-3\sqrt{5})(2\sqrt{5}+2\sqrt{3})$

$=5\sqrt{3}(2\sqrt{5}+2\sqrt{3})-3\sqrt{5}(2\sqrt{5}+2\sqrt{3})=(5\times 2\sqrt{3\times 5})+(5\times 2\sqrt{3\times 3})-(3\times 2\sqrt{5\times 5})-(3\times 2\sqrt{5\times 3})$

$=10\sqrt{15}+10\sqrt{9}-6\sqrt{25}-6\sqrt{15}=(10\sqrt{15}-6\sqrt{15})+10\sqrt{9}-6\sqrt{25}=4\sqrt{15}+(10\times 3)-(6\times 5)$

$=4\sqrt{15}+30-30=4\sqrt{15}$.

Hence, $(\sqrt{75}-\sqrt{45})(\sqrt{20}+\sqrt{12})=4\sqrt{15}$.