Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^2+9=(x+9)^2
-
Equation cos(x)=-1
-
Equation 3*x+50=77
-
Equation 5-2(x-1)=4-x
- Express {x} in terms of y where:
- 10*x+5*y=-17
- 14*x+16*y=-3
- -19*x+1*y=0
- -15*x-17*y=13
- Graphing y =:
-
x^3+x
- Integral of d{x}:
-
x^3+x
- Derivative of:
-
x^3+x
-
Identical expressions
- x^ three +x= zero
- x cubed plus x equally 0
- x to the power of three plus x equally zero
- x3+x=0
- x³+x=0
- x to the power of 3+x=0
- x^3+x=O
-
Similar expressions
- x^3-x=0
x^3+x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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 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
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{2} = i$$
$$x_{3} = - i$$
The final answer for x^3 + x = 0:
$$x_{1} = 0$$
$$x_{2} = i$$
$$x_{3} = - i$$
$$x^{3} + x = 0$$
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
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = 0$$
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
The graph