Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2+11*x+30=0
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Equation k^2+3*k-4=0
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- Integral of d{x}:
- x^4+x
-
Identical expressions
- x^ four +x= zero
- x to the power of 4 plus x equally 0
- x to the power of four plus x equally zero
- x4+x=0
- x⁴+x=0
- x^4+x=O
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Similar expressions
- x^4-x=0
x^4+x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{4} + x = 0$$
Obviously:
next,
transform
$$\frac{1}{x^{3}} = -1$$
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:
$$\frac{1}{\sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-1}}$$
or
$$x = - \left(-1\right)^{\frac{2}{3}}$$
Expand brackets in the right part
We get the answer: x = -(-1)^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{3}} = -1$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 3 i p}}{r^{3}} = -1$$
where
$$r = 1$$
- 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:
$$z_{1} = -1$$
$$z_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$z_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = -1$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$x_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$x^{4} + x = 0$$
Obviously:
x0 = 0
next,
transform
$$\frac{1}{x^{3}} = -1$$
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:
$$\frac{1}{\sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-1}}$$
or
$$x = - \left(-1\right)^{\frac{2}{3}}$$
Expand brackets in the right part
x = 1^2/3
We get the answer: x = -(-1)^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{3}} = -1$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 3 i p}}{r^{3}} = -1$$
where
$$r = 1$$
- 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:
$$z_{1} = -1$$
$$z_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$z_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
x0 = 0
$$x_{1} = -1$$
$$x_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$x_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
Rapid solution
[src]
x1 = -1
$$x_{1} = -1$$
x2 = 0
$$x_{2} = 0$$
___
1 I*\/ 3
x3 = - - -------
2 2
$$x_{3} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
___
1 I*\/ 3
x4 = - + -------
2 2
$$x_{4} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
x4 = 1/2 + sqrt(3)*i/2
Sum and product of roots
[src]
sum
___ ___
1 I*\/ 3 1 I*\/ 3
-1 + - - ------- + - + -------
2 2 2 2
$$\left(-1 + \left(\frac{1}{2} - \frac{\sqrt{3} i}{2}\right)\right) + \left(\frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ |1 I*\/ 3 | |1 I*\/ 3 | -0*|- - -------|*|- + -------| \2 2 / \2 2 /
$$- 0 \left(\frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
0
$$0$$
0
Numerical answer
[src]
x1 = 0.5 + 0.866025403784439*i
x2 = 0.0
x3 = 0.5 - 0.866025403784439*i
x4 = -1.0
x4 = -1.0