If $x^2 + (4 - \sqrt{3}) x - 1 = 0, $ then what is the value of $x^2 +\frac{1}{x^2}$ ?

$21-8\sqrt{3}$

We know that,

If x - \(\frac{1}{x}\)  = n

then, x² + \(\frac{1}{x^2}\)  = \(\sqrt {n^2 + 2}\)

If $x^2 + (4 - \sqrt{3}) x - 1 = 0, $

then what is the value of $x^2 +\frac{1}{x^2}$

Divide the given equation by x on the both sides, then we get,

x - \(\frac{1}{x}\) = $(\sqrt{3} - 4)$

then, x² + \(\frac{1}{x^2}\)  =  \(\sqrt {(\sqrt{3} - 4)^2 + 2}\)

x² + \(\frac{1}{x^2}\)  = \(\sqrt {3 + 16 - 2×4×\sqrt{3} + 2}\) = $21-8\sqrt{3}$