Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x*8=25+15
-
Equation cos(x)=5
- Equation (x-4)^4-4*(x-4)^2-21=0
-
Equation ((1/5))^(x)^(2)*5^(2*x+2)=(1/5)^(6)
- Express {x} in terms of y where:
- 15*x-5*y=-5
- 15*x+7*y=2
- 12*x-20*y=10
- -11*x-1*y=15
-
Identical expressions
- two *(four +x^(- three))= zero
- 2 multiply by (4 plus x to the power of ( minus 3)) equally 0
- two multiply by (four plus x to the power of ( minus three)) equally zero
- 2*(4+x(-3))=0
- 2*4+x-3=0
- 2(4+x^(-3))=0
- 2(4+x(-3))=0
- 24+x-3=0
- 24+x^-3=0
- 2*(4+x^(-3))=O
-
Similar expressions
- 2*(4+x^(3))=0
- 2*(4-x^(-3))=0
2*(4+x^(-3))=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$2 \left(4 + \frac{1}{x^{3}}\right) = 0$$
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:
$$\frac{1}{\sqrt[3]{2} \sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-8}}$$
or
$$\frac{2^{\frac{2}{3}} x}{2} = - \frac{\left(-1\right)^{\frac{2}{3}}}{2}$$
Expand brackets in the left part
Expand brackets in the right part
Divide both parts of the equation by 2^(2/3)/2
We get the answer: x = -(-1)^(2/3)*2^(1/3)/2
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{3}} = -4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 3 i p}}{r^{3}} = -4$$
where
$$r = \frac{\sqrt[3]{2}}{2}$$
- 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:
$$z_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$z_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$z_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$x_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$x_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$2 \left(4 + \frac{1}{x^{3}}\right) = 0$$
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:
$$\frac{1}{\sqrt[3]{2} \sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-8}}$$
or
$$\frac{2^{\frac{2}{3}} x}{2} = - \frac{\left(-1\right)^{\frac{2}{3}}}{2}$$
Expand brackets in the left part
x*2^2/3/2 = -(-1)^(2/3)/2
Expand brackets in the right part
x*2^2/3/2 = 1^2/3/2
Divide both parts of the equation by 2^(2/3)/2
x = -(-1)^(2/3)/2 / (2^(2/3)/2)
We get the answer: x = -(-1)^(2/3)*2^(1/3)/2
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{3}} = -4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 3 i p}}{r^{3}} = -4$$
where
$$r = \frac{\sqrt[3]{2}}{2}$$
- 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:
$$z_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$z_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$z_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$x_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$x_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
Sum and product of roots
[src]
sum
3 ___ 3 ___ 3 ___ ___ 3 ___ 3 ___ ___
\/ 2 \/ 2 I*\/ 2 *\/ 3 \/ 2 I*\/ 2 *\/ 3
- ----- + ----- - ------------- + ----- + -------------
2 4 4 4 4
$$\left(- \frac{\sqrt[3]{2}}{2} + \left(\frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}\right)\right) + \left(\frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}\right)$$
=
0
$$0$$
product
3 ___ /3 ___ 3 ___ ___\ /3 ___ 3 ___ ___\ -\/ 2 |\/ 2 I*\/ 2 *\/ 3 | |\/ 2 I*\/ 2 *\/ 3 | -------*|----- - -------------|*|----- + -------------| 2 \ 4 4 / \ 4 4 /
$$- \frac{\sqrt[3]{2}}{2} \left(\frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}\right) \left(\frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}\right)$$
=
-1/4
$$- \frac{1}{4}$$
-1/4
Rapid solution
[src]
3 ___
-\/ 2
x1 = -------
2
$$x_{1} = - \frac{\sqrt[3]{2}}{2}$$
3 ___ 3 ___ ___
\/ 2 I*\/ 2 *\/ 3
x2 = ----- - -------------
4 4
$$x_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
3 ___ 3 ___ ___
\/ 2 I*\/ 2 *\/ 3
x3 = ----- + -------------
4 4
$$x_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
x3 = 2^(1/3)/4 + 2^(1/3)*sqrt(3)*i/4
Numerical answer
[src]
x1 = -0.629960524947437
x2 = 0.314980262473718 - 0.545561817985861*i
x3 = 0.314980262473718 + 0.545561817985861*i
x3 = 0.314980262473718 + 0.545561817985861*i