Answer
$x=\pm \sqrt{2-\sqrt{3}}$ or $x=\pm \sqrt{2+\sqrt{3}}$
Work Step by Step
$4x^{-4}-16x^{-2}+4=0$ (Multiply by $x^4$)
$4-16x^2+4x^4=0$ (Divide by $4$)
$x^4-4x^2+1=0$ (Move the constant and change the sign)
$x^4-4x^2=-1$ (Add with $4$ to complete the square)
$x^4-4x^2+4=3$
$(x^2-2)^2=3$ (Take the square root)
$x^2-2=\pm \sqrt{3}$
$x^2-2=-\sqrt{3}$ or $x^2-2=\sqrt{3}$
$x^2=2-\sqrt{3}$ or $x^2=2+\sqrt{3}$
$x=\pm \sqrt{2-\sqrt{3}}$ or $x=\pm \sqrt{2+\sqrt{3}}$
