Given the equation:
$$\left(x^{4} + 2 x^{2}\right) + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 2 v + 1 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (1) * (1) = 0
Because D = 0, then the equation has one root.
v = -b/2a = -2/2/(1)
$$v_{1} = -1$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{\left(-1\right)^{\frac{1}{2}}}{1} = i$$
$$x_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = - i$$