What are the solutions to the equation (1-x)(1-2x+x^2)=x ?

The solutions to the equation (1-x)(1-2x+x^2)=x are x\approx 0.31767…

Find the zeros of (1-x)(1-2x+x^2)=x

The zeros of (1-x)(1-2x+x^2)=x are x\approx 0.31767…

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(1 - x) (1 - 2 x + x^{2}) = x

x \approx 0.31767 \dots

Solution steps

(1 - x) (1 - 2 x + x^{2}) = x

Expand

(1 - x) (1 - 2 x + x^{2}) : - x^{3} + 3 x^{2} - 3 x + 1

- x^{3} + 3 x^{2} - 3 x + 1 = x

Move

x

to the left side

- x^{3} + 3 x^{2} - 4 x + 1 = 0

Find one solution for

- x^{3} + 3 x^{2} - 4 x + 1 = 0

using Newton-Raphson:

x \approx 0.31767 \dots

Apply long division:

\frac{- x ^{3} + 3 x ^{2} - 4 x + 1}{x - 0.31767 \dots} = - x^{2} + 2.68232 \dots x - 3.14789 \dots

- x^{2} + 2.68232 \dots x - 3.14789 \dots \approx 0

Find one solution for

- x^{2} + 2.68232 \dots x - 3.14789 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 0.31767 \dots

(x + 2)^{3} + (x - 1)^{2} = 0

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

x^{4} - 16 x = 0

(2 x + 1)^{3} = 125

x^{4} - 9 = 0

What are the solutions to the equation (1-x)(1-2x+x^2)=x ?
The solutions to the equation (1-x)(1-2x+x^2)=x are x\approx 0.31767…
Find the zeros of (1-x)(1-2x+x^2)=x
The zeros of (1-x)(1-2x+x^2)=x are x\approx 0.31767…