Given the equation:
$$\frac{x^{2} + 4}{x} = 0$$
Multiply the equation sides by the denominators:
x
we get:
$$x^{2} + 4 = 0$$
$$x^{2} + 4 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (4) = -16
Because D<0, then the equation
has no real roots,
but complex roots is exists.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2 i$$
$$x_{2} = - 2 i$$