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

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

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

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

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

x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

Decimal Notation

x = 0.61803 \dots, x = - 1.61803 \dots

Solution steps

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

Find one solution for

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

using Newton-Raphson:

x \approx 0.61803 \dots

Apply long division:

\frac{x ^{4} + x ^{3} + x ^{2} + 2 x - 2}{x - 0.61803 \dots} = x^{3} + 1.61803 \dots x^{2} + 2 x + 3.23606 \dots

x^{3} + 1.61803 \dots x^{2} + 2 x + 3.23606 \dots \approx 0

Find one solution for

x^{3} + 1.61803 \dots x^{2} + 2 x + 3.23606 \dots = 0

using Newton-Raphson:

x \approx - 1.61803 \dots

Apply long division:

\frac{x ^{3} + 1.61803 \dots x ^{2} + 2 x + 3.23606 \dots}{x + 1.61803 \dots} = x^{2} + 2

x^{2} + 2 \approx 0

Find one solution for

x^{2} + 2 = 0

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx 0.61803 \dots : x = \frac{5 - 1}{2}

x \approx - 1.61803 \dots : x = - \frac{5 + 1}{2}

x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

The solutions are

x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

x^{5} + x^{3} + x + 1 = 0

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

x^{5} - 7 x^{4} + 14 x^{3} - 2 x^{2} - 15 x + 9 = 0

x^{4} - 4 x^{3} - x^{2} + 14 x + 10

x^{4} + x^{2} + 1

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