What are the solutions to the equation (z-3)^4+(z-5)^4=-8 ?

The solutions to the equation (z-3)^4+(z-5)^4=-8 are No Solution for z\in\mathbb{R}

Find the zeros of (z-3)^4+(z-5)^4=-8

The zeros of (z-3)^4+(z-5)^4=-8 are No Solution for z\in\mathbb{R}

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(z - 3)^{4} + (z - 5)^{4} = - 8

No Solution for z \in R

Solution steps

(z - 3)^{4} + (z - 5)^{4} = - 8

Expand

(z - 3)^{4} + (z - 5)^{4} : 2 z^{4} - 32 z^{3} + 204 z^{2} - 608 z + 706

2 z^{4} - 32 z^{3} + 204 z^{2} - 608 z + 706 = - 8

Move

8

to the left side

2 z^{4} - 32 z^{3} + 204 z^{2} - 608 z + 714 = 0

Find one solution for

2 z^{4} - 32 z^{3} + 204 z^{2} - 608 z + 714 = 0

using Newton-Raphson:

No Solution for

z \in R

The solution is

No Solution for z \in R

x, - 5 x^{3} + 4 y^{2} = 16

x^{3} - x^{2} + 2 x + 5 = 0

A^{4} - 5 A^{3} + 2 A^{2} + 8 A = 0

3 - 3 x^{5} = 0

70 x^{4} - 181 x^{3} + 116 x^{2} + x = 6

What are the solutions to the equation (z-3)^4+(z-5)^4=-8 ?
The solutions to the equation (z-3)^4+(z-5)^4=-8 are No Solution for z\in\mathbb{R}
Find the zeros of (z-3)^4+(z-5)^4=-8
The zeros of (z-3)^4+(z-5)^4=-8 are No Solution for z\in\mathbb{R}