If $\sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0$, then $x^2(x^2 - 47) = ?$

-1

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If $\sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0$

then $x^2(x^2 - 47) = ?$

If $\sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0$

Then, x + \(\frac{1}{x}\) = 3² – 2 = 7

x² + \(\frac{1}{x^2}\) = 7² – 2 = 47

Put the value of 47 in $x^2(x^2 - 47) $

= $x^2(x^2 - x^2 - \frac{1}{x^2}) $ = -1