Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 8+x^3=0
- Equation -4/9y=16/27
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Equation (89cos^2+80cosx)/(80tgx-39)=0
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- Express {x} in terms of y where:
- -10*x+18*y=6
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- Factor polynomial:
- 8+x^3
- Limit of the function:
-
8+x^3
-
Identical expressions
- eight +x^ three = zero
- 8 plus x cubed equally 0
- eight plus x to the power of three equally zero
- 8+x3=0
- 8+x³=0
- 8+x to the power of 3=0
- 8+x^3=O
-
Similar expressions
- 8-x^3=0
8+x^3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} + 8 = 0$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{x^{3}} = \sqrt[3]{-8}$$
or
$$x = 2 \sqrt[3]{-1}$$
Expand brackets in the right part
We get the answer: x = 2*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = -8$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -2$$
$$z_{2} = 1 - \sqrt{3} i$$
$$z_{3} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = -2$$
$$x_{2} = 1 - \sqrt{3} i$$
$$x_{3} = 1 + \sqrt{3} i$$
$$x^{3} + 8 = 0$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{x^{3}} = \sqrt[3]{-8}$$
or
$$x = 2 \sqrt[3]{-1}$$
Expand brackets in the right part
x = -2*1^1/3
We get the answer: x = 2*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = -8$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -2$$
$$z_{2} = 1 - \sqrt{3} i$$
$$z_{3} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = -2$$
$$x_{2} = 1 - \sqrt{3} i$$
$$x_{3} = 1 + \sqrt{3} i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 8$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 8$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 8$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 8$$
Rapid solution
[src]
x1 = -2
$$x_{1} = -2$$
___ x2 = 1 - I*\/ 3
$$x_{2} = 1 - \sqrt{3} i$$
___ x3 = 1 + I*\/ 3
$$x_{3} = 1 + \sqrt{3} i$$
x3 = 1 + sqrt(3)*i
Sum and product of roots
[src]
sum
___ ___ -2 + 1 - I*\/ 3 + 1 + I*\/ 3
$$\left(-2 + \left(1 - \sqrt{3} i\right)\right) + \left(1 + \sqrt{3} i\right)$$
=
0
$$0$$
product
/ ___\ / ___\ -2*\1 - I*\/ 3 /*\1 + I*\/ 3 /
$$- 2 \left(1 - \sqrt{3} i\right) \left(1 + \sqrt{3} i\right)$$
=
-8
$$-8$$
-8
Numerical answer
[src]
x1 = 1.0 + 1.73205080756888*i
x2 = -2.0
x3 = 1.0 - 1.73205080756888*i
x3 = 1.0 - 1.73205080756888*i
The graph