Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^4+16=0
-
Equation (7-2x)(9-2x)-35=0
-
Equation sin^2((П/4)-x)=sin^2((П/4)+x)
- Equation cos^2((2*pi/3)-x)=cos^2((2*pi/3)+x)
- Express {x} in terms of y where:
- 5*x-5*y=20
- 8*x-7*y=-13
- 9*x+18*y=6
- 15*x-15*y=16
- Factor polynomial:
- x^4+16
- Graphing y =:
- x^4+16
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Identical expressions
- x^ four + sixteen = zero
- x to the power of 4 plus 16 equally 0
- x to the power of four plus sixteen equally zero
- x4+16=0
- x⁴+16=0
- x^4+16=O
-
Similar expressions
- x^4-16=0
x^4+16=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{4} + 16 = 0$$
Because equation degree is equal to = 4 and the free term = -16 < 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:
$$z^{4} = -16$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -16$$
where
$$r = 2$$
- 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:
$$z_{1} = - \sqrt{2} - \sqrt{2} i$$
$$z_{2} = - \sqrt{2} + \sqrt{2} i$$
$$z_{3} = \sqrt{2} - \sqrt{2} i$$
$$z_{4} = \sqrt{2} + \sqrt{2} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \sqrt{2} - \sqrt{2} i$$
$$x_{2} = - \sqrt{2} + \sqrt{2} i$$
$$x_{3} = \sqrt{2} - \sqrt{2} i$$
$$x_{4} = \sqrt{2} + \sqrt{2} i$$
$$x^{4} + 16 = 0$$
Because equation degree is equal to = 4 and the free term = -16 < 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:
$$z^{4} = -16$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -16$$
where
$$r = 2$$
- 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:
$$z_{1} = - \sqrt{2} - \sqrt{2} i$$
$$z_{2} = - \sqrt{2} + \sqrt{2} i$$
$$z_{3} = \sqrt{2} - \sqrt{2} i$$
$$z_{4} = \sqrt{2} + \sqrt{2} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \sqrt{2} - \sqrt{2} i$$
$$x_{2} = - \sqrt{2} + \sqrt{2} i$$
$$x_{3} = \sqrt{2} - \sqrt{2} i$$
$$x_{4} = \sqrt{2} + \sqrt{2} i$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___ - \/ 2 - I*\/ 2 + - \/ 2 + I*\/ 2 + \/ 2 - I*\/ 2 + \/ 2 + I*\/ 2
$$\left(- \sqrt{2} - \sqrt{2} i\right) + \left(- \sqrt{2} + \sqrt{2} i\right) + \left(\sqrt{2} - \sqrt{2} i\right) + \left(\sqrt{2} + \sqrt{2} i\right)$$
=
0
$$0$$
product
___ ___ ___ ___ ___ ___ ___ ___ - \/ 2 - I*\/ 2 * - \/ 2 + I*\/ 2 * \/ 2 - I*\/ 2 * \/ 2 + I*\/ 2
$$\left(- \sqrt{2} - \sqrt{2} i\right) * \left(- \sqrt{2} + \sqrt{2} i\right) * \left(\sqrt{2} - \sqrt{2} i\right) * \left(\sqrt{2} + \sqrt{2} i\right)$$
=
16
$$16$$
Rapid solution
[src]
___ ___ x_1 = - \/ 2 - I*\/ 2
$$x_{1} = - \sqrt{2} - \sqrt{2} i$$
___ ___ x_2 = - \/ 2 + I*\/ 2
$$x_{2} = - \sqrt{2} + \sqrt{2} i$$
___ ___ x_3 = \/ 2 - I*\/ 2
$$x_{3} = \sqrt{2} - \sqrt{2} i$$
___ ___ x_4 = \/ 2 + I*\/ 2
$$x_{4} = \sqrt{2} + \sqrt{2} i$$
Numerical answer
[src]
x1 = -1.4142135623731 + 1.4142135623731*i
x2 = 1.4142135623731 - 1.4142135623731*i
x3 = -1.4142135623731 - 1.4142135623731*i
x4 = 1.4142135623731 + 1.4142135623731*i
x4 = 1.4142135623731 + 1.4142135623731*i
The graph