Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 9^x-8*3^x-9=0
- Equation x+y=4
-
Equation 5x+8=-12
-
Equation x/9+x=-3
- Express {x} in terms of y where:
- 18*x+17*y=-14
- 16*x-17*y=-15
- -4*x-4*y=19
- 15*x+1*y=19
-
Identical expressions
- x^-5x+ three = zero
- x to the power of minus 5x plus 3 equally 0
- x to the power of minus 5x plus three equally zero
- x-5x+3=0
- x^-5x+3=O
-
Similar expressions
- x^-5x-3=0
- x^+5x+3=0
x^-5x+3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\frac{x}{x^{5}} + 3 = 0$$
Because equation degree is equal to = -4 and the free term = -3 < 0,
so the real solutions of the equation d'not exist
All other 4 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{4}} = -3$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 4 i p}}{r^{4}} = -3$$
where
$$r = \frac{3^{\frac{3}{4}}}{3}$$
- the magnitude of the complex number
Substitute r:
$$e^{- 4 i p} = -1$$
Using Euler’s formula, we find roots for p
$$- i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = -1$$
so
$$\cos{\left(4 p \right)} = -1$$
and
$$- \sin{\left(4 p \right)} = 0$$
then
$$p = - \frac{\pi N}{2} - \frac{\pi}{4}$$
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:
$$z_{1} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{2} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{3} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{4} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{2} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{3} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{4} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$\frac{x}{x^{5}} + 3 = 0$$
Because equation degree is equal to = -4 and the free term = -3 < 0,
so the real solutions of the equation d'not exist
All other 4 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{4}} = -3$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 4 i p}}{r^{4}} = -3$$
where
$$r = \frac{3^{\frac{3}{4}}}{3}$$
- the magnitude of the complex number
Substitute r:
$$e^{- 4 i p} = -1$$
Using Euler’s formula, we find roots for p
$$- i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = -1$$
so
$$\cos{\left(4 p \right)} = -1$$
and
$$- \sin{\left(4 p \right)} = 0$$
then
$$p = - \frac{\pi N}{2} - \frac{\pi}{4}$$
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:
$$z_{1} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{2} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{3} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$z_{4} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{2} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{3} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
$$x_{4} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
Rapid solution
[src]
___ 3/4 ___ 3/4
\/ 2 *3 I*\/ 2 *3
x1 = - ---------- - ------------
6 6
$$x_{1} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
___ 3/4 ___ 3/4
\/ 2 *3 I*\/ 2 *3
x2 = - ---------- + ------------
6 6
$$x_{2} = - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
___ 3/4 ___ 3/4
\/ 2 *3 I*\/ 2 *3
x3 = ---------- - ------------
6 6
$$x_{3} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
___ 3/4 ___ 3/4
\/ 2 *3 I*\/ 2 *3
x4 = ---------- + ------------
6 6
$$x_{4} = \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}$$
x4 = sqrt(2)*3^(3/4)/6 + sqrt(2)*3^(3/4)*i/6
Sum and product of roots
[src]
sum
___ 3/4 ___ 3/4 ___ 3/4 ___ 3/4 ___ 3/4 ___ 3/4 ___ 3/4 ___ 3/4
\/ 2 *3 I*\/ 2 *3 \/ 2 *3 I*\/ 2 *3 \/ 2 *3 I*\/ 2 *3 \/ 2 *3 I*\/ 2 *3
- ---------- - ------------ + - ---------- + ------------ + ---------- - ------------ + ---------- + ------------
6 6 6 6 6 6 6 6
$$\left(\left(\frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right) + \left(\left(- \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right) + \left(- \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right)\right)\right) + \left(\frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right)$$
=
0
$$0$$
product
/ ___ 3/4 ___ 3/4\ / ___ 3/4 ___ 3/4\ / ___ 3/4 ___ 3/4\ / ___ 3/4 ___ 3/4\ | \/ 2 *3 I*\/ 2 *3 | | \/ 2 *3 I*\/ 2 *3 | |\/ 2 *3 I*\/ 2 *3 | |\/ 2 *3 I*\/ 2 *3 | |- ---------- - ------------|*|- ---------- + ------------|*|---------- - ------------|*|---------- + ------------| \ 6 6 / \ 6 6 / \ 6 6 / \ 6 6 /
$$\left(- \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right) \left(- \frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right) \left(\frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right) \left(\frac{\sqrt{2} \cdot 3^{\frac{3}{4}}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} i}{6}\right)$$
=
1/3
$$\frac{1}{3}$$
1/3
Numerical answer
[src]
x1 = -0.537284965911771 - 0.537284965911771*i
x2 = -0.537284965911771 + 0.537284965911771*i
x3 = 0.537284965911771 - 0.537284965911771*i
x4 = 0.537284965911771 + 0.537284965911771*i
x4 = 0.537284965911771 + 0.537284965911771*i