Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 2*x^2+5*x-12=0
-
Equation x^2+2x-8=0
-
Equation 6x+2=-1
- Equation (x^2-25)^2+(x^2+3x-10)^2=0
- Express {x} in terms of y where:
- 7*x-2*y=18
- 18*x+19*y=7
- 8*x-13*y=-12
- -10*x+13*y=5
- Derivative of:
-
x^3-4
- Factor polynomial:
- x^3-4
- Graphing y =:
- x^3-4
-
Identical expressions
- x^ three - four = zero
- x cubed minus 4 equally 0
- x to the power of three minus four equally zero
- x3-4=0
- x³-4=0
- x to the power of 3-4=0
- x^3-4=O
-
Similar expressions
- x^3+4=0
x^3-4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} - 4 = 0$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{4}$$
or
$$x = 2^{\frac{2}{3}}$$
Expand brackets in the right part
We get the answer: x = 2^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 4$$
where
$$r = 2^{\frac{2}{3}}$$
- 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:
$$z_{1} = 2^{\frac{2}{3}}$$
$$z_{2} = - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
$$z_{3} = - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 2^{\frac{2}{3}}$$
$$x_{2} = - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
$$x_{3} = - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
$$x^{3} - 4 = 0$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{4}$$
or
$$x = 2^{\frac{2}{3}}$$
Expand brackets in the right part
x = 2^2/3
We get the answer: x = 2^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 4$$
where
$$r = 2^{\frac{2}{3}}$$
- 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:
$$z_{1} = 2^{\frac{2}{3}}$$
$$z_{2} = - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
$$z_{3} = - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 2^{\frac{2}{3}}$$
$$x_{2} = - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
$$x_{3} = - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -4$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -4$$
Sum and product of roots
[src]
sum
2/3 2/3 ___ 2/3 2/3 ___
2/3 2 I*2 *\/ 3 2 I*2 *\/ 3
2 + - ---- - ------------ + - ---- + ------------
2 2 2 2
$$\left(2^{\frac{2}{3}} + \left(- \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}\right)\right) + \left(- \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ 2/3 2/3 ___\ / 2/3 2/3 ___\
2/3 | 2 I*2 *\/ 3 | | 2 I*2 *\/ 3 |
2 *|- ---- - ------------|*|- ---- + ------------|
\ 2 2 / \ 2 2 /
$$2^{\frac{2}{3}} \left(- \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}\right) \left(- \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}\right)$$
=
4
$$4$$
4
Rapid solution
[src]
2/3 x1 = 2
$$x_{1} = 2^{\frac{2}{3}}$$
2/3 2/3 ___
2 I*2 *\/ 3
x2 = - ---- - ------------
2 2
$$x_{2} = - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
2/3 2/3 ___
2 I*2 *\/ 3
x3 = - ---- + ------------
2 2
$$x_{3} = - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2}$$
x3 = -2^(2/3)/2 + 2^(2/3)*sqrt(3)*i/2
Numerical answer
[src]
x1 = 1.5874010519682
x2 = -0.7937005259841 - 1.3747296369986*i
x3 = -0.7937005259841 + 1.3747296369986*i
x3 = -0.7937005259841 + 1.3747296369986*i
The graph