Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^3=2
-
Equation x-32/4=50
- Equation ax=b
- Equation a*x=b
- Express {x} in terms of y where:
- -4*x+7*y=-16
- 12*x-20*y=9
- -1*x-20*y=1
- -5*x-12*y=13
- Derivative of:
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x^3
- Integral of d{x}:
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x^3
- x^3
-
Identical expressions
- x^ three = two
- x cubed equally 2
- x to the power of three equally two
- x3=2
- x³=2
- x to the power of 3=2
-
Similar expressions
- x^3+a*x+b=0
- x^3=0.34
x^3=2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = 2$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{2}$$
or
$$x = \sqrt[3]{2}$$
Expand brackets in the right part
We get the answer: x = 2^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 2$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 2$$
where
$$r = \sqrt[3]{2}$$
- 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
$$p = \frac{2 \pi N}{3}$$
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:
$$z_{1} = \sqrt[3]{2}$$
$$z_{2} = - \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
$$z_{3} = - \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \sqrt[3]{2}$$
$$x_{2} = - \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
$$x_{3} = - \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
$$x^{3} = 2$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{2}$$
or
$$x = \sqrt[3]{2}$$
Expand brackets in the right part
x = 2^1/3
We get the answer: x = 2^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 2$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 2$$
where
$$r = \sqrt[3]{2}$$
- 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
$$p = \frac{2 \pi N}{3}$$
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:
$$z_{1} = \sqrt[3]{2}$$
$$z_{2} = - \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
$$z_{3} = - \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \sqrt[3]{2}$$
$$x_{2} = - \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
$$x_{3} = - \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -2$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -2$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -2$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -2$$
Sum and product of roots
[src]
sum
3 ___ 3 ___ ___ 3 ___ 3 ___ ___
3 ___ \/ 2 I*\/ 2 *\/ 3 \/ 2 I*\/ 2 *\/ 3
\/ 2 + - ----- - ------------- + - ----- + -------------
2 2 2 2
$$\left(\sqrt[3]{2} + \left(- \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}\right)\right) + \left(- \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ 3 ___ 3 ___ ___\ / 3 ___ 3 ___ ___\
3 ___ | \/ 2 I*\/ 2 *\/ 3 | | \/ 2 I*\/ 2 *\/ 3 |
\/ 2 *|- ----- - -------------|*|- ----- + -------------|
\ 2 2 / \ 2 2 /
$$\sqrt[3]{2} \left(- \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}\right) \left(- \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}\right)$$
=
2
$$2$$
2
Rapid solution
[src]
3 ___ x1 = \/ 2
$$x_{1} = \sqrt[3]{2}$$
3 ___ 3 ___ ___
\/ 2 I*\/ 2 *\/ 3
x2 = - ----- - -------------
2 2
$$x_{2} = - \frac{\sqrt[3]{2}}{2} - \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
3 ___ 3 ___ ___
\/ 2 I*\/ 2 *\/ 3
x3 = - ----- + -------------
2 2
$$x_{3} = - \frac{\sqrt[3]{2}}{2} + \frac{\sqrt[3]{2} \sqrt{3} i}{2}$$
x3 = -2^(1/3)/2 + 2^(1/3)*sqrt(3)*i/2
Numerical answer
[src]
x1 = 1.25992104989487
x2 = -0.629960524947437 + 1.09112363597172*i
x3 = -0.629960524947437 - 1.09112363597172*i
x3 = -0.629960524947437 - 1.09112363597172*i
The graph