Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation sin(x)=0
-
Equation x+x/9=-10/3
- Equation z^2-(2-i)z+3-i=0
-
Equation x^6=-64
- Express {x} in terms of y where:
- 10*x-10*y=-6
- 16*x-7*y=-12
- 10*x+10*y=6
- -14*x+6*y=16
- Graphing y =:
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x^6
- Integral of d{x}:
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x^6
- Derivative of:
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x^6
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Identical expressions
- x^ six =- sixty-four
- x to the power of 6 equally minus 64
- x to the power of six equally minus sixty minus four
- x6=-64
- x⁶=-64
-
Similar expressions
- -6-((-6*sqrt(2)+ln(1-6^2)+arctan(-6*4)+4))/(sqrt(2)+5)=x
- x^6=+64
x^6=-64 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{6} = -64$$
Because equation degree is equal to = 6 and the free term = -64 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{6} = -64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -64$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$z_{1} = - 2 i$$
$$z_{2} = 2 i$$
$$z_{3} = - \sqrt{3} - i$$
$$z_{4} = - \sqrt{3} + i$$
$$z_{5} = \sqrt{3} - i$$
$$z_{6} = \sqrt{3} + i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - 2 i$$
$$x_{2} = 2 i$$
$$x_{3} = - \sqrt{3} - i$$
$$x_{4} = - \sqrt{3} + i$$
$$x_{5} = \sqrt{3} - i$$
$$x_{6} = \sqrt{3} + i$$
$$x^{6} = -64$$
Because equation degree is equal to = 6 and the free term = -64 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{6} = -64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -64$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$z_{1} = - 2 i$$
$$z_{2} = 2 i$$
$$z_{3} = - \sqrt{3} - i$$
$$z_{4} = - \sqrt{3} + i$$
$$z_{5} = \sqrt{3} - i$$
$$z_{6} = \sqrt{3} + i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - 2 i$$
$$x_{2} = 2 i$$
$$x_{3} = - \sqrt{3} - i$$
$$x_{4} = - \sqrt{3} + i$$
$$x_{5} = \sqrt{3} - i$$
$$x_{6} = \sqrt{3} + i$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ -2*I + 2*I + -I - \/ 3 + I - \/ 3 + \/ 3 - I + I + \/ 3
$$\left(\left(\sqrt{3} - i\right) + \left(\left(\left(- \sqrt{3} - i\right) + \left(- 2 i + 2 i\right)\right) + \left(- \sqrt{3} + i\right)\right)\right) + \left(\sqrt{3} + i\right)$$
=
0
$$0$$
product
/ ___\ / ___\ / ___ \ / ___\ -2*I*2*I*\-I - \/ 3 /*\I - \/ 3 /*\\/ 3 - I/*\I + \/ 3 /
$$- 2 i 2 i \left(- \sqrt{3} - i\right) \left(- \sqrt{3} + i\right) \left(\sqrt{3} - i\right) \left(\sqrt{3} + i\right)$$
=
64
$$64$$
64
Rapid solution
[src]
x1 = -2*I
$$x_{1} = - 2 i$$
x2 = 2*I
$$x_{2} = 2 i$$
___ x3 = -I - \/ 3
$$x_{3} = - \sqrt{3} - i$$
___ x4 = I - \/ 3
$$x_{4} = - \sqrt{3} + i$$
___ x5 = \/ 3 - I
$$x_{5} = \sqrt{3} - i$$
___ x6 = I + \/ 3
$$x_{6} = \sqrt{3} + i$$
x6 = sqrt(3) + i
Numerical answer
[src]
x1 = -1.73205080756888 + 1.0*i
x2 = -1.73205080756888 - 1.0*i
x3 = 2.0*i
x4 = -2.0*i
x5 = 1.73205080756888 + 1.0*i
x6 = 1.73205080756888 - 1.0*i
x6 = 1.73205080756888 - 1.0*i
The graph