Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation -x^2-4*x-3=0
-
Equation 3x+3=-2-7x
-
Equation cos(x)=-3
-
Equation -x^3+14*x^2-61*x+84=0
- Express {x} in terms of y where:
- -18*x+13*y=-6
- -20*x+6*y=5
- -19*x-20*y=1
- 10*x-5*y=-4
-
Identical expressions
- three (x+ three)^ six - five =- seventeen
- 3(x plus 3) to the power of 6 minus 5 equally minus 17
- three (x plus three) to the power of six minus five equally minus seventeen
- 3(x+3)6-5=-17
- 3x+36-5=-17
- 3(x+3)⁶-5=-17
- 3x+3^6-5=-17
-
Similar expressions
- 3(x+3)^6-5=+17
- 3(x+3)^6+5=-17
- x^3*(2/5)^(x-1)*(7/10)^(x-1)*(9/10)^(x-1)=1
- 3(x-3)^6-5=-17
3(x+3)^6-5=-17 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$3 \left(x + 3\right)^{6} - 5 = -17$$
Because equation degree is equal to = 6 and the free term = -12 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$z = x + 3$$
then the equation will be the:
$$z^{6} = -4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -4$$
where
$$r = \sqrt[3]{2}$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$z_{1} = - \sqrt[3]{2} i$$
$$z_{2} = \sqrt[3]{2} i$$
$$z_{3} = - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$z_{4} = - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
$$z_{5} = \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$z_{6} = \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
do backward replacement
$$z = x + 3$$
$$x = z - 3$$
The final answer:
$$x_{1} = -3 - \sqrt[3]{2} i$$
$$x_{2} = -3 + \sqrt[3]{2} i$$
$$x_{3} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$x_{4} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
$$x_{5} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$x_{6} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
$$3 \left(x + 3\right)^{6} - 5 = -17$$
Because equation degree is equal to = 6 and the free term = -12 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$z = x + 3$$
then the equation will be the:
$$z^{6} = -4$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -4$$
where
$$r = \sqrt[3]{2}$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$z_{1} = - \sqrt[3]{2} i$$
$$z_{2} = \sqrt[3]{2} i$$
$$z_{3} = - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$z_{4} = - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
$$z_{5} = \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$z_{6} = \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
do backward replacement
$$z = x + 3$$
$$x = z - 3$$
The final answer:
$$x_{1} = -3 - \sqrt[3]{2} i$$
$$x_{2} = -3 + \sqrt[3]{2} i$$
$$x_{3} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$x_{4} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
$$x_{5} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
$$x_{6} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
Rapid solution
[src]
3 ___ x1 = -3 - I*\/ 2
$$x_{1} = -3 - \sqrt[3]{2} i$$
3 ___ x2 = -3 + I*\/ 2
$$x_{2} = -3 + \sqrt[3]{2} i$$
3 ___ 3 ___ ___
I*\/ 2 \/ 2 *\/ 3
x3 = -3 + ------- - -----------
2 2
$$x_{3} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
3 ___ ___ 3 ___
\/ 2 *\/ 3 I*\/ 2
x4 = -3 + ----------- - -------
2 2
$$x_{4} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
3 ___ 3 ___ ___
I*\/ 2 \/ 2 *\/ 3
x5 = -3 - ------- - -----------
2 2
$$x_{5} = -3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}$$
3 ___ 3 ___ ___
I*\/ 2 \/ 2 *\/ 3
x6 = -3 + ------- + -----------
2 2
$$x_{6} = -3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}$$
x6 = -3 + 2^(1/3)*sqrt(3)/2 + 2^(1/3)*i/2
Sum and product of roots
[src]
sum
3 ___ 3 ___ ___ 3 ___ ___ 3 ___ 3 ___ 3 ___ ___ 3 ___ 3 ___ ___
3 ___ 3 ___ I*\/ 2 \/ 2 *\/ 3 \/ 2 *\/ 3 I*\/ 2 I*\/ 2 \/ 2 *\/ 3 I*\/ 2 \/ 2 *\/ 3
-3 - I*\/ 2 + -3 + I*\/ 2 + -3 + ------- - ----------- + -3 + ----------- - ------- + -3 - ------- - ----------- + -3 + ------- + -----------
2 2 2 2 2 2 2 2
$$\left(\left(-3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}\right) + \left(\left(-3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}\right) + \left(\left(\left(-3 - \sqrt[3]{2} i\right) + \left(-3 + \sqrt[3]{2} i\right)\right) + \left(-3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}\right)\right)\right)\right) + \left(-3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}\right)$$
=
-18
$$-18$$
product
/ 3 ___ 3 ___ ___\ / 3 ___ ___ 3 ___\ / 3 ___ 3 ___ ___\ / 3 ___ 3 ___ ___\
/ 3 ___\ / 3 ___\ | I*\/ 2 \/ 2 *\/ 3 | | \/ 2 *\/ 3 I*\/ 2 | | I*\/ 2 \/ 2 *\/ 3 | | I*\/ 2 \/ 2 *\/ 3 |
\-3 - I*\/ 2 /*\-3 + I*\/ 2 /*|-3 + ------- - -----------|*|-3 + ----------- - -------|*|-3 - ------- - -----------|*|-3 + ------- + -----------|
\ 2 2 / \ 2 2 / \ 2 2 / \ 2 2 /
$$\left(-3 - \sqrt[3]{2} i\right) \left(-3 + \sqrt[3]{2} i\right) \left(-3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}\right) \left(-3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}\right) \left(-3 - \frac{\sqrt[3]{2} \sqrt{3}}{2} - \frac{\sqrt[3]{2} i}{2}\right) \left(-3 + \frac{\sqrt[3]{2} \sqrt{3}}{2} + \frac{\sqrt[3]{2} i}{2}\right)$$
=
733
$$733$$
733
Numerical answer
[src]
x1 = -4.09112363597172 + 0.629960524947437*i
x2 = -3.0 + 1.25992104989487*i
x3 = -3.0 - 1.25992104989487*i
x4 = -1.90887636402828 + 0.629960524947437*i
x5 = -4.09112363597172 - 0.629960524947437*i
x6 = -1.90887636402828 - 0.629960524947437*i
x6 = -1.90887636402828 - 0.629960524947437*i