If $x^{2}-x sqrt{68}+1=0$, then what is the value of $left(x-frac{1}{x} ight) ?$

$x^{2}-x sqrt{68}+1=0$

$Rightarrow x^{2}+1=x sqrt{68}$

$Rightarrow x+frac{1}{x}=sqrt{68}$

$hereforeleft(x-frac{1}{x} ight)^{2}=left(x+frac{1}{x} ight)^{2}-4=68-4=64$

$hereforeleft(x-frac{1}{x} ight)=sqrt{64}=8$