Question

If $x=2\sqrt{3}+2\sqrt{2},$ then ${(x+\frac{1}{x})}^2$ =$\frac{245+98\sqrt{6}}{p}$ then $p=$

Answer 1

$x=2\sqrt{3}+3\sqrt{2}$

$\frac{1}{x}=\frac{1}{2\sqrt{3}+3\sqrt{2}}$

$=\frac{1}{2\sqrt{3}+3\sqrt{2}}\times \frac{2\sqrt{3}-3\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}$

$=\frac{2\sqrt{3}+3\sqrt{2}}{12-18}$

$=\frac{-2\sqrt{3}-3\sqrt{2}}{6}$

$\therefore x+\frac{1}{x}=2\sqrt{3}+3\sqrt{2}-\left(\frac{-2\sqrt{3}-3\sqrt{2}}{6}\right)$

$=2\sqrt{3}+3\sqrt{2}+\frac{2\sqrt{3}+3\sqrt{2}}{6}$

$=\frac{12\sqrt{3}+18\sqrt{2}+2\sqrt{3}+3\sqrt{2}}{6}$

$\therefore x+\frac{1}{x}=\frac{14\sqrt{3}+21\sqrt{2}}{6}$

$\therefore {(x+\frac{1}{x})}^2={\left(\frac{14\sqrt{3}+21\sqrt{2}}{6}\right)}^2$

$=\frac{1}{36}(588‬+882+588‬\sqrt{6})$

$=\frac{1470+588‬\sqrt{6}‬}{36}$

$=\frac{245+98\sqrt{6}‬}{6}$

By comparing $p=6$