Question

If
$x=5+2\sqrt{6},\text{then find the value of}\sqrt{x}+\frac{1}{\sqrt{x}}.$

• A.
  $2\sqrt{3}$
• B.
  $5\sqrt{3}$
• C.
  $3\sqrt{5}$
• D.
  $2\sqrt{5}$

Answer 1

The correct option is A

$2\sqrt{3}$
$x=5+2\sqrt{6}\to a+2\sqrt{b}=3+2+2\sqrt{3\times 2}(6=1\times 6=2\times 3)=(\sqrt{3}{)}^2+(\sqrt{2}{)}^2+2\times \sqrt{3}\times \sqrt{2}\therefore x=(\sqrt{3}+\sqrt{2}{)}^2\sqrt{x}=\sqrt{(\sqrt{3}+\sqrt{2}{)}^2}=\sqrt{3}+\sqrt{2}\sqrt{x}+\frac{1}{\sqrt{x}}=(\sqrt{3}+\sqrt{2})+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=(\sqrt{3}+\sqrt{2})+\left(\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}\right)=(\sqrt{3}+\sqrt{2})+\left(\frac{\sqrt{3}-\sqrt{2}}{3-2}\right)=\sqrt{3}+\sqrt{2}+\sqrt{3}+\sqrt{2}=\sqrt{3}+\sqrt{3}=2\sqrt{3}\therefore \sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{3}$

Hence, option A is the right answer.