x^4-4x^2+4=0 equation

Given the equation:
$$\left(x^{4} - 4 x^{2}\right) + 4 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 4 v + 4 = 0$$
This equation is of the form

a*v^2 + b*v + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -4$$
$$c = 4$$
, then

D = b^2 - 4 * a * c = 

(-4)^2 - 4 * (1) * (4) = 0

Because D = 0, then the equation has one root.

v = -b/2a = --4/2/(1)

$$v_{1} = 2$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$