Question

If $x=6+2\sqrt{6}$, then what is the value of $\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$?

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

Answer 1

Answer: (A)

$2\sqrt{3}$

$x=6+2\sqrt{6}$
$\Rightarrow x-1=5+2\sqrt{6}={\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\sqrt{2}\times \sqrt{3}={(\sqrt{3}+\sqrt{2})}^2$
$\sqrt{x-1}=\pm (\sqrt{3}+\sqrt{2})$
$=\sqrt{x-1}+\frac{1}{\sqrt{x-1}}=\sqrt{3}+\sqrt{2}+\frac{1}{(\sqrt{3}+\sqrt{2})}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}$