If $\sqrt{x} -\frac{1}{\sqrt{x}}=\sqrt{3}$, then what is the value of $x^4 +\frac{1}{x^4}$?
Answer & explanation
Correct answer: option 4
If x - \(\frac{1}{x}\) = n
then, $x^2 +\frac{1}{x^2}$ = n2 + 2
If $\sqrt{x} -\frac{1}{\sqrt{x}}=\sqrt{3}$
x + \(\frac{1}{x}\) = (\(\sqrt {3}\))2 + 2 = 5
$x^2 +\frac{1}{x^2}$ = 52 - 2 = 23
$x^4 +\frac{1}{x^4}$ = 232 - 2 = 527