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

The solutions to the equation 4x+4=(5-4x)^3 are x\approx 0.76995…

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

The zeros of 4x+4=(5-4x)^3 are x\approx 0.76995…

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4 x + 4 = (5 - 4 x)^{3}

x \approx 0.76995 \dots

Solution steps

4 x + 4 = (5 - 4 x)^{3}

Expand

(5 - 4 x)^{3} : 125 - 300 x + 240 x^{2} - 64 x^{3}

4 x + 4 = 125 - 300 x + 240 x^{2} - 64 x^{3}

Switch sides

125 - 300 x + 240 x^{2} - 64 x^{3} = 4 x + 4

Move

4

to the left side

- 64 x^{3} + 240 x^{2} - 300 x + 121 = 4 x

Move

4 x

to the left side

- 64 x^{3} + 240 x^{2} - 304 x + 121 = 0

Find one solution for

- 64 x^{3} + 240 x^{2} - 304 x + 121 = 0

using Newton-Raphson:

x \approx 0.76995 \dots

Apply long division:

\frac{- 64 x ^{3} + 240 x ^{2} - 304 x + 121}{x - 0.76995 \dots} = - 64 x^{2} + 190.72280 \dots x - 157.15179 \dots

- 64 x^{2} + 190.72280 \dots x - 157.15179 \dots \approx 0

Find one solution for

- 64 x^{2} + 190.72280 \dots x - 157.15179 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 0.76995 \dots

x^{3} - x^{2} - 5 x + 21 = 0

x^{3} + 3 x^{2} - 6 x = 0

x (6 - 2 x)^{2} = 0

y, x = 1 - ((1 - y)^{12})

(x^{2} + 6 x) (x + 3)^{2} = 400

What are the solutions to the equation 4x+4=(5-4x)^3 ?
The solutions to the equation 4x+4=(5-4x)^3 are x\approx 0.76995…
Find the zeros of 4x+4=(5-4x)^3
The zeros of 4x+4=(5-4x)^3 are x\approx 0.76995…