Answer
The solutions are $x=\pm\sqrt{2-\sqrt{3}}$ and $x=\pm\sqrt{2+\sqrt{3}}$
Work Step by Step
$4x^{-4}-16x^{-2}+4=0$
Rewrite this equation:
$\dfrac{4}{x^{4}}-\dfrac{16}{x^{2}}+4=0$
Multiply the whole equation by $x^{4}$:
$x^{4}\Big(\dfrac{4}{x^{4}}-\dfrac{16}{x^{2}}+4=0\Big)$
$4-16x^{2}+4x^{4}=0$
Rearrange:
$4x^{4}-16x^{2}+4=0$
Let $u$ be equal to $x^{2}$:
$u=x^{2}$
$u^{2}=x^{4}$
Rewrite the equation using the new variable $u$:
$4u^{2}-16u+4=0$
Solve this equation using the quadratic formula, which is $u=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=4$, $b=-16$ and $c=4$.
Substitute the known values into the formula and simplify:
$u=\dfrac{-(-16)\pm\sqrt{(-16)^{2}-4(4)(4)}}{2(4)}=\dfrac{16\pm\sqrt{256-64}}{8}=...$
$...=\dfrac{16\pm\sqrt{192}}{8}=\dfrac{16\pm\sqrt{64\cdot3}}{8}=\dfrac{16\pm8\sqrt{3}}{8}=2\pm\sqrt{3}$
The solutions found are $u=2+\sqrt{3}$ and $u=2-\sqrt{3}$
Substitute $u$ back to $x^{2}$ and solve for $x$:
$x^{2}=2+\sqrt{3}$
$x=\pm\sqrt{2+\sqrt{3}}$
$x^{2}=2-\sqrt{3}$
$x=\pm\sqrt{2-\sqrt{3}}$
The solutions are $x=\pm\sqrt{2-\sqrt{3}}$ and $x=\pm\sqrt{2+\sqrt{3}}$
