Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^2+2*x+10=0
-
Equation x^2+3*x+9=0
- Equation x^2-x+30=0
-
Equation 3*x^2-2*x-8=0
- Express {x} in terms of y where:
- -15*x+1*y=10
- 3*x-1*y=5
- -6*x-9*y=-20
- -8*x-8*y=14
- Graphing y =:
- x+1/x
- Canonical form:
- x+1/x
- Limit of the function:
-
x+1/x
-
Identical expressions
- x+ one /x= zero
- x plus 1 divide by x equally 0
- x plus one divide by x equally zero
- x+1/x=O
- x+1 divide by x=0
-
Similar expressions
- x-1/x=0
- (x^4-4x+1)/(x-4)=0
- sqrt(3x+1)/x+4(sqrt(3x+1)/3-sqrt(4x-3))=13/3-4x-1/x
- (x(x-2)-(x+2)(x+1))/((x+1)(x-2))=0
x+1/x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x + \frac{1}{x} = 0$$
transform
$$x^{2} = -1$$
Because equation degree is equal to = 2 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{2} = -1$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{2} e^{2 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{2 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(2 p \right)} + \cos{\left(2 p \right)} = -1$$
so
$$\cos{\left(2 p \right)} = -1$$
and
$$\sin{\left(2 p \right)} = 0$$
then
$$p = \pi N + \frac{\pi}{2}$$
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:
$$z_{1} = - i$$
$$z_{2} = i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - i$$
$$x_{2} = i$$
$$x + \frac{1}{x} = 0$$
transform
$$x^{2} = -1$$
Because equation degree is equal to = 2 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{2} = -1$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{2} e^{2 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{2 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(2 p \right)} + \cos{\left(2 p \right)} = -1$$
so
$$\cos{\left(2 p \right)} = -1$$
and
$$\sin{\left(2 p \right)} = 0$$
then
$$p = \pi N + \frac{\pi}{2}$$
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:
$$z_{1} = - i$$
$$z_{2} = i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - i$$
$$x_{2} = i$$
The graph