x^4-8x^2+16=0 equation

Given the equation:
$$x^{4} - 8 x^{2} + 16 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 8 v + 16 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -8$$
$$c = 16$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 16 + \left(-8\right)^{2} = 0$$
Because D = 0, then the equation has one root.

v = -b/2a = --8/2/(1)

$$v_{1} = 4$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$