Question

The value of $\frac{\sqrt{(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})}}{5+\sqrt{24}}$

• A. $\sqrt{6}-\sqrt{2}$
• B. $\sqrt{6}+\sqrt{2}$
• C. $\sqrt{6}-2$
• D. $2-\sqrt{6}$

Answer 1

The correct option is C $\sqrt{6}-2$

$\sqrt{\frac{(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})}{5+\sqrt{24}}}$

$=\sqrt{\frac{(2\sqrt{3}-2\sqrt{2})(\sqrt{3}+\sqrt{2})}{5+2\sqrt{6}}}$

$=\sqrt{\frac{2(\sqrt{3}{)}^2-2\sqrt{6}+2\sqrt{6}-2(\sqrt{2}{)}^2}{5+2\sqrt{6}}}$

$=\sqrt{\frac{6-4}{5+2\sqrt{6}}}=\sqrt{\frac{2(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}}$

$=\sqrt{\frac{2(5-2\sqrt{6})}{25-24}}=\sqrt{10-4\sqrt{6}}$

$=\sqrt{6+4-4\sqrt{6}}=\sqrt{(\sqrt{6}{)}^2+(2{)}^2-2\times 2\times \sqrt{6}}$

$=\sqrt{(\sqrt{6}-2{)}^2}=\sqrt{6}-2$