What are the solutions to the equation 2x(4-2)=4+x^8 ?

The solutions to the equation 2x(4-2)=4+x^8 are No Solution for x\in\mathbb{R}

Find the zeros of 2x(4-2)=4+x^8

The zeros of 2x(4-2)=4+x^8 are No Solution for x\in\mathbb{R}

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2 x (4 - 2) = 4 + x^{8}

No Solution for x \in R

Solution steps

2 x (4 - 2) = 4 + x^{8}

Expand

2 x (4 - 2) : 4 x

4 x = 4 + x^{8}

Switch sides

4 + x^{8} = 4 x

Move

4 x

to the left side

4 + x^{8} - 4 x = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{8} - 4 x + 4 = 0

Find one solution for

x^{8} - 4 x + 4 = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

No Solution for x \in R

4 x^{3} - 4 x^{2} = 0

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

13 x = x^{3}

x^{4} - 32 x^{2} + 175 = 0

0 = 6 - 6 x^{5}

What are the solutions to the equation 2x(4-2)=4+x^8 ?
The solutions to the equation 2x(4-2)=4+x^8 are No Solution for x\in\mathbb{R}
Find the zeros of 2x(4-2)=4+x^8
The zeros of 2x(4-2)=4+x^8 are No Solution for x\in\mathbb{R}