What are the solutions to the equation f^4(0.5f-0.25)=6+f ?

The solutions to the equation f^4(0.5f-0.25)=6+f are f\approx 1.84741…

Find the zeros of f^4(0.5f-0.25)=6+f

The zeros of f^4(0.5f-0.25)=6+f are f\approx 1.84741…

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f^{4} (0.5 f - 0.25) = 6 + f

f \approx 1.84741 \dots

Solution steps

f^{4} (0.5 f - 0.25) = 6 + f

Expand

f^{4} (0.5 f - 0.25) : 0.5 f^{5} - 0.25 f^{4}

0.5 f^{5} - 0.25 f^{4} = 6 + f

Move

f

to the left side

0.5 f^{5} - 0.25 f^{4} - f = 6

Move

6

to the left side

0.5 f^{5} - 0.25 f^{4} - f - 6 = 0

Find one solution for

0.5 f^{5} - 0.25 f^{4} - f - 6 = 0

using Newton-Raphson:

f \approx 1.84741 \dots

Apply long division:

\frac{0.5 f ^{5} - 0.25 f ^{4} - f - 6}{f - 1.84741 \dots} = 0.5 f^{4} + 0.67370 \dots f^{3} + 1.24461 \dots f^{2} + 2.29931 \dots f + 3.24778 \dots

0.5 f^{4} + 0.67370 \dots f^{3} + 1.24461 \dots f^{2} + 2.29931 \dots f + 3.24778 \dots \approx 0

Find one solution for

0.5 f^{4} + 0.67370 \dots f^{3} + 1.24461 \dots f^{2} + 2.29931 \dots f + 3.24778 \dots = 0

using Newton-Raphson:

No Solution for

f \in R

The solution is

f \approx 1.84741 \dots

(x + 4) (x + 2) (x - 1) = 0

3 x^{3} - 12 x^{2} + 9 x - 4 = 0

x, x^{3} + y^{3} = a^{3}

- x^{3} - 1 = 0

(u + 10)^{3} = - (u + 10)^{2}

What are the solutions to the equation f^4(0.5f-0.25)=6+f ?
The solutions to the equation f^4(0.5f-0.25)=6+f are f\approx 1.84741…
Find the zeros of f^4(0.5f-0.25)=6+f
The zeros of f^4(0.5f-0.25)=6+f are f\approx 1.84741…