Question

After rationalising the denominator of $\frac{22}{2+\sqrt{3}+\sqrt{5}}.$ We get

• A.
  $3\sqrt{3}+\sqrt{5}-2\sqrt{15}-4$
• B.
  $3\sqrt{3}+\sqrt{15}$
• C.
  $3\sqrt{3}+\sqrt{5}-2\sqrt{15}+4$
• D.
  $3\sqrt{3}+\sqrt{5}$

Answer 1

The correct option is C

$3\sqrt{3}+\sqrt{5}-2\sqrt{15}+4$
$\frac{22}{(2+\sqrt{3})+\sqrt{5}}\times \frac{(2+\sqrt{3})-\sqrt{5}}{(2+\sqrt{3})-\sqrt{5}}=\frac{22(2+\sqrt{3}-\sqrt{5})}{(2+\sqrt{3}{)}^2-(\sqrt{5}{)}^2}=\frac{22+(2+\sqrt{3}-\sqrt{5})}{\left(2^2\right)+(\sqrt{3}{)}^2+2\times 2\times \sqrt{3}-5}=\frac{22(2+\sqrt{3}-\sqrt{5})}{2+4\sqrt{3}}\times \frac{2-4\sqrt{3}}{2-4\sqrt{3}}=\frac{22(2+\sqrt{3}-\sqrt{5})(2-4\sqrt{3})}{(2{)}^2-(4\sqrt{3}{)}^2}=\frac{22(2+\sqrt{3}-\sqrt{5})(2-4\sqrt{3})}{4-48}=\frac{22(2+\sqrt{3}-\sqrt{5})\times 2(1-2\sqrt{3})}{-44}=\frac{22(2+\sqrt{3}-\sqrt{5})\times 2(1-2\sqrt{3})}{-44}=-1(2-4\sqrt{3}+\sqrt{3}-6-\sqrt{5}+2\sqrt{15})=-1(-4-3\sqrt{3}-\sqrt{5}+2\sqrt{15})=4+3\sqrt{3}+\sqrt{5}-2\sqrt{15}$

Hence, option c is correct.