Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^4=(4*x-21)^2
-
Equation 14/(x-3)=-7/9
-
Equation 4^x-3*2^x+2=0
-
Equation x(x^2+2x+1)=6(x+1)
- Express {x} in terms of y where:
- 1*x-6*y=-19
- -20*x-13*y=15
- -20*x-14*y=5
- 18*x-12*y=-20
-
Identical expressions
- three x^3- twenty-seven = zero
- 3x cubed minus 27 equally 0
- three x cubed minus twenty minus seven equally zero
- 3x3-27=0
- 3x³-27=0
- 3x to the power of 3-27=0
- 3x^3-27=O
-
Similar expressions
- 3x^3+27=0
3x^3-27=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$3 x^{3} - 27 = 0$$
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:
$$\sqrt[3]{3} \sqrt[3]{x^{3}} = \sqrt[3]{27}$$
or
$$\sqrt[3]{3} x = 3$$
Expand brackets in the left part
Divide both parts of the equation by 3^(1/3)
We get the answer: x = 3^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 9$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 9$$
where
$$r = 3^{\frac{2}{3}}$$
- 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:
$$z_{1} = 3^{\frac{2}{3}}$$
$$z_{2} = - \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}$$
$$z_{3} = - \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 3^{\frac{2}{3}}$$
$$x_{2} = - \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}$$
$$x_{3} = - \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}$$
$$3 x^{3} - 27 = 0$$
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:
$$\sqrt[3]{3} \sqrt[3]{x^{3}} = \sqrt[3]{27}$$
or
$$\sqrt[3]{3} x = 3$$
Expand brackets in the left part
x*3^1/3 = 3
Divide both parts of the equation by 3^(1/3)
x = 3 / (3^(1/3))
We get the answer: x = 3^(2/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 9$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 9$$
where
$$r = 3^{\frac{2}{3}}$$
- 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:
$$z_{1} = 3^{\frac{2}{3}}$$
$$z_{2} = - \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}$$
$$z_{3} = - \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 3^{\frac{2}{3}}$$
$$x_{2} = - \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}$$
$$x_{3} = - \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}$$
Vieta's Theorem
rewrite the equation
$$3 x^{3} - 27 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 9 = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -9$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -9$$
$$3 x^{3} - 27 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 9 = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -9$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -9$$
Rapid solution
[src]
2/3 x1 = 3
$$x_{1} = 3^{\frac{2}{3}}$$
2/3 6 ___
3 3*I*\/ 3
x2 = - ---- - ---------
2 2
$$x_{2} = - \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}$$
2/3 6 ___
3 3*I*\/ 3
x3 = - ---- + ---------
2 2
$$x_{3} = - \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}$$
x3 = -3^(2/3)/2 + 3*3^(1/6)*i/2
Sum and product of roots
[src]
sum
2/3 6 ___ 2/3 6 ___
2/3 3 3*I*\/ 3 3 3*I*\/ 3
3 + - ---- - --------- + - ---- + ---------
2 2 2 2
$$\left(3^{\frac{2}{3}} + \left(- \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}\right)\right) + \left(- \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}\right)$$
=
0
$$0$$
product
/ 2/3 6 ___\ / 2/3 6 ___\
2/3 | 3 3*I*\/ 3 | | 3 3*I*\/ 3 |
3 *|- ---- - ---------|*|- ---- + ---------|
\ 2 2 / \ 2 2 /
$$3^{\frac{2}{3}} \left(- \frac{3^{\frac{2}{3}}}{2} - \frac{3 \sqrt[6]{3} i}{2}\right) \left(- \frac{3^{\frac{2}{3}}}{2} + \frac{3 \sqrt[6]{3} i}{2}\right)$$
=
9
$$9$$
9
Numerical answer
[src]
x1 = 2.0800838230519
x2 = -1.04004191152595 + 1.801405432764*i
x3 = -1.04004191152595 - 1.801405432764*i
x3 = -1.04004191152595 - 1.801405432764*i