If $x = 3 + 2\sqrt{2}$, then the value of $\sqrt{x} -\frac{1}{\sqrt{x}}$ is:
Answer & explanation
Correct answer: option 1
When a fraction is in the form of $x = a + b$ and the difference between the square of a and square of b is equal to 1 then we can say, \(\frac{1}{x}\) = a - b
If $x = 3 + 2\sqrt{2}$,
then the value of $\sqrt{x} -\frac{1}{\sqrt{x}}$
If $x = 3 + 2\sqrt{2}$
then, \(\frac{1}{x}\) = 3 - $2\sqrt{2}$
So, x + \(\frac{1}{x}\) = 3 + $2\sqrt{2}$ + 3 - $2\sqrt{2}$ = 6
and, $\sqrt{x} -\frac{1}{\sqrt{x}}$ = \(\sqrt {6 - 2}\) = 2