Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation z^3+27=0
-
Equation e^x
- Equation 4+(2x-(4x-1))+x=-5-x
-
Equation x^4+3x^2-10=0
- Express {x} in terms of y where:
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
- Factor polynomial:
- z^3+27
-
Identical expressions
- z^ three + twenty-seven = zero
- z cubed plus 27 equally 0
- z to the power of three plus twenty minus seven equally zero
- z3+27=0
- z³+27=0
- z to the power of 3+27=0
- z^3+27=O
-
Similar expressions
- z^3-27=0
z^3+27=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{3} + 27 = 0$$
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:
$$\sqrt[3]{z^{3}} = \sqrt[3]{-27}$$
or
$$z = 3 \sqrt[3]{-1}$$
Expand brackets in the right part
We get the answer: z = 3*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = -27$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -27$$
where
$$r = 3$$
- 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
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = -3$$
$$w_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$w_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -3$$
$$z_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$z_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
$$z^{3} + 27 = 0$$
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:
$$\sqrt[3]{z^{3}} = \sqrt[3]{-27}$$
or
$$z = 3 \sqrt[3]{-1}$$
Expand brackets in the right part
z = -3*1^1/3
We get the answer: z = 3*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = -27$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -27$$
where
$$r = 3$$
- 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
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = -3$$
$$w_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$w_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -3$$
$$z_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$z_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 27$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = 27$$
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 27$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = 27$$
Rapid solution
[src]
z1 = -3
$$z_{1} = -3$$
___
3 3*I*\/ 3
z2 = - - ---------
2 2
$$z_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
___
3 3*I*\/ 3
z3 = - + ---------
2 2
$$z_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
z3 = 3/2 + 3*sqrt(3)*i/2
Sum and product of roots
[src]
sum
___ ___
3 3*I*\/ 3 3 3*I*\/ 3
-3 + - - --------- + - + ---------
2 2 2 2
$$\left(-3 + \left(\frac{3}{2} - \frac{3 \sqrt{3} i}{2}\right)\right) + \left(\frac{3}{2} + \frac{3 \sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ |3 3*I*\/ 3 | |3 3*I*\/ 3 | -3*|- - ---------|*|- + ---------| \2 2 / \2 2 /
$$- 3 \left(\frac{3}{2} - \frac{3 \sqrt{3} i}{2}\right) \left(\frac{3}{2} + \frac{3 \sqrt{3} i}{2}\right)$$
=
-27
$$-27$$
-27
Numerical answer
[src]
z1 = -3.0
z2 = 1.5 - 2.59807621135332*i
z3 = 1.5 + 2.59807621135332*i
z3 = 1.5 + 2.59807621135332*i
The graph