What are the solutions to the equation x^4-4x^3+5x^2-4x+1=0 ?

The solutions to the equation x^4-4x^3+5x^2-4x+1=0 are x=(-sqrt(5)+3)/2 ,x=(sqrt(5)+3)/2

Find the zeros of x^4-4x^3+5x^2-4x+1=0

The zeros of x^4-4x^3+5x^2-4x+1=0 are x=(-sqrt(5)+3)/2 ,x=(sqrt(5)+3)/2

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x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 1 = 0

x = \frac{- 5 + 3}{2}, x = \frac{5 + 3}{2}

Decimal Notation

x = 0.38196 \dots, x = 2.61803 \dots

Solution steps

x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 1 = 0

Find one solution for

x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 1 = 0

using Newton-Raphson:

x \approx 0.38196 \dots

Apply long division:

\frac{x ^{4} - 4 x ^{3} + 5 x ^{2} - 4 x + 1}{x - 0.38196 \dots} = x^{3} - 3.61803 \dots x^{2} + 3.61803 \dots x - 2.61803 \dots

x^{3} - 3.61803 \dots x^{2} + 3.61803 \dots x - 2.61803 \dots \approx 0

Find one solution for

x^{3} - 3.61803 \dots x^{2} + 3.61803 \dots x - 2.61803 \dots = 0

using Newton-Raphson:

x \approx 2.61803 \dots

Apply long division:

\frac{x ^{3} - 3.61803 \dots x ^{2} + 3.61803 \dots x - 2.61803 \dots}{x - 2.61803 \dots} = x^{2} - x + 1

x^{2} - x + 1 \approx 0

Find one solution for

x^{2} - x + 1 = 0

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx 0.38196 \dots : x = \frac{- 5 + 3}{2}

x \approx 2.61803 \dots : x = \frac{5 + 3}{2}

x = \frac{- 5 + 3}{2}, x = \frac{5 + 3}{2}

The solutions are

x = \frac{- 5 + 3}{2}, x = \frac{5 + 3}{2}

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

s^{3} + 9 s^{2} + 32 s + 32 = 0

(x + 7)^{3} = 0

(s^{2} - 2)^{2} = s^{2}

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

What are the solutions to the equation x^4-4x^3+5x^2-4x+1=0 ?
The solutions to the equation x^4-4x^3+5x^2-4x+1=0 are x=(-sqrt(5)+3)/2 ,x=(sqrt(5)+3)/2
Find the zeros of x^4-4x^3+5x^2-4x+1=0
The zeros of x^4-4x^3+5x^2-4x+1=0 are x=(-sqrt(5)+3)/2 ,x=(sqrt(5)+3)/2