Question

If $x=2\sqrt{6}+5$ find $x+\frac{1}{x}$

Answer 1

If $x=2\sqrt{6}+5$ then $\frac{1}{x}=\frac{1}{2\sqrt{6}+5}$.

We rationalise the denominator of $\frac{1}{x}=\frac{1}{2\sqrt{6}+5}$ and simplify as follows:

$\frac{1}{2\sqrt{6}+5}\times \frac{2\sqrt{6}-5}{2\sqrt{6}-5}=\frac{2\sqrt{6}-5}{(2\sqrt{6}+5)(2\sqrt{6}-5)}=\frac{2\sqrt{6}-5}{(2\sqrt{6}{)}^2-(5{)}^2}=\frac{2\sqrt{6}-5}{(4\times 6)-25}=\frac{2\sqrt{6}-5}{24-25}=\frac{2\sqrt{6}-5}{-1}=5-2\sqrt{6}$

Now,

$x+\frac{1}{x}=2\sqrt{6}+5+5-2\sqrt{6}=10$

Hence, $x+\frac{1}{x}=10$