Mister Exam
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Similar expressions
- 4*x^2-1/x=0
4*x^2+1/x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$4 x^{2} + \frac{1}{x} = 0$$
transform
$$\frac{1}{x^{3}} = -4$$
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:
$$\frac{1}{\sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-4}}$$
or
$$x = - \frac{\left(-1\right)^{\frac{2}{3}} \sqrt[3]{2}}{2}$$
Expand brackets in the right part
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:
$$x_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$x_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$x_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$4 x^{2} + \frac{1}{x} = 0$$
transform
$$\frac{1}{x^{3}} = -4$$
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:
$$\frac{1}{\sqrt[3]{\frac{1}{x^{3}}}} = \frac{1}{\sqrt[3]{-4}}$$
or
$$x = - \frac{\left(-1\right)^{\frac{2}{3}} \sqrt[3]{2}}{2}$$
Expand brackets in the right part
x = 1^2/3*2^1/3/2
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:
$$x_{1} = - \frac{\sqrt[3]{2}}{2}$$
$$x_{2} = \frac{\sqrt[3]{2}}{4} - \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
$$x_{3} = \frac{\sqrt[3]{2}}{4} + \frac{\sqrt[3]{2} \sqrt{3} i}{4}$$
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
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
Numerical answer
[src]
x1 = -0.629960524947437
x2 = 0.314980262473718 - 0.545561817985861*i
x3 = 0.314980262473718 + 0.545561817985861*i
x3 = 0.314980262473718 + 0.545561817985861*i
The graph