Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^3=2
-
Equation 3x^2-2x=0
- Equation x^2+y^2=1
-
Equation x^3=3
- 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:
-
x^3
- Integral of d{x}:
-
x^3
- x^3
-
Identical expressions
- x^ three = zero . thirty-four
- x cubed equally 0.34
- x to the power of three equally zero . thirty minus four
- x3=0.34
- x³=0.34
- x to the power of 3=0.34
- x^3=O.34
-
Similar expressions
- x^3+a*x+b=0
- x^3=0.34
- 1.2*(3-n)+0.3*(4n+1)=0.6*(n+11)
x^3=0.34 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = \frac{17}{50}$$
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:
$$\sqrt[3]{\left(1 x + 0\right)^{3}} = \sqrt[3]{\frac{17}{50}}$$
or
$$x = \frac{\sqrt[3]{340}}{10}$$
Expand brackets in the right part
We get the answer: x = 340^(1/3)/10
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = \frac{17}{50}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = \frac{17}{50}$$
where
$$r = \frac{\sqrt[3]{340}}{10}$$
- 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:
$$z_{1} = \frac{\sqrt[3]{340}}{10}$$
$$z_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
$$z_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \frac{\sqrt[3]{340}}{10}$$
$$x_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
$$x_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
$$x^{3} = \frac{17}{50}$$
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:
$$\sqrt[3]{\left(1 x + 0\right)^{3}} = \sqrt[3]{\frac{17}{50}}$$
or
$$x = \frac{\sqrt[3]{340}}{10}$$
Expand brackets in the right part
x = 340^1/3/10
We get the answer: x = 340^(1/3)/10
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = \frac{17}{50}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = \frac{17}{50}$$
where
$$r = \frac{\sqrt[3]{340}}{10}$$
- 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:
$$z_{1} = \frac{\sqrt[3]{340}}{10}$$
$$z_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
$$z_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \frac{\sqrt[3]{340}}{10}$$
$$x_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
$$x_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - \frac{17}{50}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = - \frac{17}{50}$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - \frac{17}{50}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = - \frac{17}{50}$$
Sum and product of roots
[src]
sum
3 _____ 3 _____ ___ 3 _____ 3 _____ ___ 3 _____
\/ 340 \/ 340 I*\/ 3 *\/ 340 \/ 340 I*\/ 3 *\/ 340
0 + ------- + - ------- - --------------- + - ------- + ---------------
10 20 20 20 20
$$\left(\left(0 + \frac{\sqrt[3]{340}}{10}\right) - \left(\frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right)\right) - \left(\frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right)$$
=
0
$$0$$
product
3 _____ / 3 _____ ___ 3 _____\ / 3 _____ ___ 3 _____\
\/ 340 | \/ 340 I*\/ 3 *\/ 340 | | \/ 340 I*\/ 3 *\/ 340 |
1*-------*|- ------- - ---------------|*|- ------- + ---------------|
10 \ 20 20 / \ 20 20 /
$$1 \cdot \frac{\sqrt[3]{340}}{10} \left(- \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right) \left(- \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right)$$
=
17 -- 50
$$\frac{17}{50}$$
17/50
Rapid solution
[src]
3 _____
\/ 340
x1 = -------
10
$$x_{1} = \frac{\sqrt[3]{340}}{10}$$
3 _____ ___ 3 _____
\/ 340 I*\/ 3 *\/ 340
x2 = - ------- - ---------------
20 20
$$x_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
3 _____ ___ 3 _____
\/ 340 I*\/ 3 *\/ 340
x3 = - ------- + ---------------
20 20
$$x_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}$$
Numerical answer
[src]
x1 = -0.348976602345444 - 0.60444520591507*i
x2 = -0.348976602345444 + 0.60444520591507*i
x3 = 0.697953204690889
x3 = 0.697953204690889
The graph