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x^3+x=0

x^3+x=0 equation

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Numerical solution:

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The solution

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x  + x = 0
$$x^{3} + x = 0$$
Detail solution
Given the equation:
$$x^{3} + x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} + 1 = 0$$
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:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 1$$
, then
D = b^2 - 4 * a * c = 

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

Because D<0, then the equation
has no real roots,
but complex roots is exists.
x2 = (-b + sqrt(D)) / (2*a)

x3 = (-b - sqrt(D)) / (2*a)

or
$$x_{2} = i$$
$$x_{3} = - i$$
The final answer for x^3 + x = 0:
$$x_{1} = 0$$
$$x_{2} = i$$
$$x_{3} = - i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = 0$$
The graph
Sum and product of roots [src]
sum
-I + I
$$- i + i$$
=
0
$$0$$
product
0*(-I)*I
$$i 0 \left(- i\right)$$
=
0
$$0$$
0
Rapid solution [src]
x1 = 0
$$x_{1} = 0$$
x2 = -I
$$x_{2} = - i$$
x3 = I
$$x_{3} = i$$
x3 = i
Numerical answer [src]
x1 = 0.0
x2 = 1.0*i
x3 = -1.0*i
x3 = -1.0*i
The graph
x^3+x=0 equation