Mister Exam
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How to use it?
- Equation:
-
Equation x^3=125
- Equation (x^2-16)^2+(x^2-5x-36)^2=0
- Equation (2y)/(y-3)=(3y+3)/y
- Equation (7)/(1-cos^2(x))+(9)/(sin(x))=10
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
- 5*x+5*y=20
- Derivative of:
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x^3
- Integral of d{x}:
-
x^3
- x^3
- Sum of series:
-
125
-
Identical expressions
- x^ three = one hundred and twenty-five
- x cubed equally 125
- x to the power of three equally one hundred and twenty minus five
- x3=125
- x³=125
- x to the power of 3=125
-
Similar expressions
- x^3+4x^2-9x-36=0
x^3=125 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = 125$$
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:
$$\sqrt[3]{\left(1 x + 0\right)^{3}} = \sqrt[3]{125}$$
or
$$x = 5$$
We get the answer: x = 5
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 125$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 125$$
where
$$r = 5$$
- 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:
$$z_{1} = 5$$
$$z_{2} = - \frac{5}{2} - \frac{5 \sqrt{3} i}{2}$$
$$z_{3} = - \frac{5}{2} + \frac{5 \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 5$$
$$x_{2} = - \frac{5}{2} - \frac{5 \sqrt{3} i}{2}$$
$$x_{3} = - \frac{5}{2} + \frac{5 \sqrt{3} i}{2}$$
$$x^{3} = 125$$
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:
$$\sqrt[3]{\left(1 x + 0\right)^{3}} = \sqrt[3]{125}$$
or
$$x = 5$$
We get the answer: x = 5
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 125$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 125$$
where
$$r = 5$$
- 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:
$$z_{1} = 5$$
$$z_{2} = - \frac{5}{2} - \frac{5 \sqrt{3} i}{2}$$
$$z_{3} = - \frac{5}{2} + \frac{5 \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 5$$
$$x_{2} = - \frac{5}{2} - \frac{5 \sqrt{3} i}{2}$$
$$x_{3} = - \frac{5}{2} + \frac{5 \sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -125$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -125$$
$$p x^{2} + x^{3} + q x + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -125$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -125$$
Rapid solution
[src]
x_1 = 5
$$x_{1} = 5$$
___
5 5*I*\/ 3
x_2 = - - - ---------
2 2
$$x_{2} = - \frac{5}{2} - \frac{5 \sqrt{3} i}{2}$$
___
5 5*I*\/ 3
x_3 = - - + ---------
2 2
$$x_{3} = - \frac{5}{2} + \frac{5 \sqrt{3} i}{2}$$
Sum and product of roots
[src]
sum
___ ___
5 5*I*\/ 3 5 5*I*\/ 3
5 + - - - --------- + - - + ---------
2 2 2 2
$$\left(5\right) + \left(- \frac{5}{2} - \frac{5 \sqrt{3} i}{2}\right) + \left(- \frac{5}{2} + \frac{5 \sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
___ ___
5 5*I*\/ 3 5 5*I*\/ 3
5 * - - - --------- * - - + ---------
2 2 2 2
$$\left(5\right) * \left(- \frac{5}{2} - \frac{5 \sqrt{3} i}{2}\right) * \left(- \frac{5}{2} + \frac{5 \sqrt{3} i}{2}\right)$$
=
125
$$125$$
Numerical answer
[src]
x1 = 5.0
x2 = -2.5 - 4.33012701892219*i
x3 = -2.5 + 4.33012701892219*i
x3 = -2.5 + 4.33012701892219*i
The graph