Question

If $x=9+4\sqrt{5},$ then the value of $\sqrt{x}-\frac{1}{\sqrt{x}}$ is ____.

Answer 1

Given, $x=9+4\sqrt{5}$.
$\therefore \frac{1}{x}=\frac{1}{9+4\sqrt{5}}=\frac{1}{9+4\sqrt{5}}\times \frac{9-4\sqrt{5}}{9-4\sqrt{5}}$ ....(Rationalising the denominator)
$=\frac{9-4\sqrt{5}}{(9{)}^2-(4\sqrt{5}{)}^2}$

$=\frac{9-4\sqrt{5}}{81-80}$

$=9-4\sqrt{5}$.
$\therefore x+\frac{1}{x}=(9+4\sqrt{5})+(9-4\sqrt{5})$ $=18$.
Now, ${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=x+\frac{1}{x}-2$
$=18-2=16$.
$\therefore \sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{16}=4$.