4x^4-5x^2+1=0 equation

Given the equation:
$$4 x^{4} - 5 x^{2} + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$4 v^{2} - 5 v + 1 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 4$$
$$b = -5$$
$$c = 1$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 4 \cdot 4 \cdot 1 + \left(-5\right)^{2} = 9$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 1$$
Simplify
$$v_{2} = \frac{1}{4}$$
Simplify
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(\frac{1}{4}\right)^{\frac{1}{2}}}{1} = \frac{1}{2}$$
$$x_{4} = \frac{\left(-1\right) \left(\frac{1}{4}\right)^{\frac{1}{2}}}{1} + \frac{0}{1} = - \frac{1}{2}$$