Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^4=(3x-10)^2
-
Equation 7x-3=2x+1
-
Equation x^2-9*x=0
-
Equation (5x-4)(6-x)=0
- Express {x} in terms of y where:
- -14*x+8*y=6
- 6*x-2*y=17
- 12*x-18*y=-14
- -20*x+12*y=2
- Integral of d{x}:
-
2x^3+1
- Graphing y =:
- 2x^3+1
- Derivative of:
- 2x^3+1
-
Identical expressions
- 2x^ three + one
- 2x cubed plus 1
- 2x to the power of three plus one
- 2x3+1
- 2x³+1
- 2x to the power of 3+1
-
Similar expressions
- 2x^3-1
2x^3+1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$2 x^{3} + 1 = 0$$
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:
$$\sqrt[3]{2} \sqrt[3]{x^{3}} = \sqrt[3]{-1}$$
or
$$\sqrt[3]{2} x = \sqrt[3]{-1}$$
Expand brackets in the left part
Expand brackets in the right part
Divide both parts of the equation by 2^(1/3)
We get the answer: x = (-1)^(1/3)*2^(2/3)/2
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = - \frac{1}{2}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = - \frac{1}{2}$$
where
$$r = \frac{2^{\frac{2}{3}}}{2}$$
- 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:
$$z_{1} = - \frac{2^{\frac{2}{3}}}{2}$$
$$z_{2} = \frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
$$z_{3} = \frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{2^{\frac{2}{3}}}{2}$$
$$x_{2} = \frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
$$x_{3} = \frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
$$2 x^{3} + 1 = 0$$
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:
$$\sqrt[3]{2} \sqrt[3]{x^{3}} = \sqrt[3]{-1}$$
or
$$\sqrt[3]{2} x = \sqrt[3]{-1}$$
Expand brackets in the left part
x*2^1/3 = (-1)^(1/3)
Expand brackets in the right part
x*2^1/3 = -1^1/3
Divide both parts of the equation by 2^(1/3)
x = (-1)^(1/3) / (2^(1/3))
We get the answer: x = (-1)^(1/3)*2^(2/3)/2
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = - \frac{1}{2}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = - \frac{1}{2}$$
where
$$r = \frac{2^{\frac{2}{3}}}{2}$$
- 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:
$$z_{1} = - \frac{2^{\frac{2}{3}}}{2}$$
$$z_{2} = \frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
$$z_{3} = \frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{2^{\frac{2}{3}}}{2}$$
$$x_{2} = \frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
$$x_{3} = \frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
Vieta's Theorem
rewrite the equation
$$2 x^{3} + 1 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{1}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = \frac{1}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = \frac{1}{2}$$
$$2 x^{3} + 1 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{1}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = \frac{1}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = \frac{1}{2}$$
Sum and product of roots
[src]
sum
2/3 2/3 2/3 ___ 2/3 2/3 ___ 2 2 I*2 *\/ 3 2 I*2 *\/ 3 - ---- + ---- - ------------ + ---- + ------------ 2 4 4 4 4
$$\left(- \frac{2^{\frac{2}{3}}}{2} + \left(\frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}\right)\right) + \left(\frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}\right)$$
=
0
$$0$$
product
2/3 / 2/3 2/3 ___\ / 2/3 2/3 ___\ -2 |2 I*2 *\/ 3 | |2 I*2 *\/ 3 | ------*|---- - ------------|*|---- + ------------| 2 \ 4 4 / \ 4 4 /
$$- \frac{2^{\frac{2}{3}}}{2} \left(\frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}\right) \left(\frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}\right)$$
=
-1/2
$$- \frac{1}{2}$$
-1/2
Rapid solution
[src]
2/3
-2
x1 = ------
2
$$x_{1} = - \frac{2^{\frac{2}{3}}}{2}$$
2/3 2/3 ___
2 I*2 *\/ 3
x2 = ---- - ------------
4 4
$$x_{2} = \frac{2^{\frac{2}{3}}}{4} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
2/3 2/3 ___
2 I*2 *\/ 3
x3 = ---- + ------------
4 4
$$x_{3} = \frac{2^{\frac{2}{3}}}{4} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{4}$$
x3 = 2^(2/3)/4 + 2^(2/3)*sqrt(3)*i/4
Numerical answer
[src]
x1 = 0.39685026299205 - 0.687364818499301*i
x2 = 0.39685026299205 + 0.687364818499301*i
x3 = -0.7937005259841
x3 = -0.7937005259841