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

The solutions to the equation x^4-3x^2+4x+1=0 are x\approx-0.21565…,x\approx-2.15435…

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

The zeros of x^4-3x^2+4x+1=0 are x\approx-0.21565…,x\approx-2.15435…

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

x \approx - 0.21565 \dots, x \approx - 2.15435 \dots

Solution steps

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

Find one solution for

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

using Newton-Raphson:

x \approx - 0.21565 \dots

Apply long division:

\frac{x ^{4} - 3 x ^{2} + 4 x + 1}{x + 0.21565 \dots} = x^{3} - 0.21565 \dots x^{2} - 2.95349 \dots x + 4.63694 \dots

x^{3} - 0.21565 \dots x^{2} - 2.95349 \dots x + 4.63694 \dots \approx 0

Find one solution for

x^{3} - 0.21565 \dots x^{2} - 2.95349 \dots x + 4.63694 \dots = 0

using Newton-Raphson:

x \approx - 2.15435 \dots

Apply long division:

\frac{x ^{3} - 0.21565 \dots x ^{2} - 2.95349 \dots x + 4.63694 \dots}{x + 2.15435 \dots} = x^{2} - 2.37001 \dots x + 2.15236 \dots

x^{2} - 2.37001 \dots x + 2.15236 \dots \approx 0

Find one solution for

x^{2} - 2.37001 \dots x + 2.15236 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx - 0.21565 \dots, x \approx - 2.15435 \dots

(z - 3)^{4} + (z - 5)^{4} = - 8

x, - 5 x^{3} + 4 y^{2} = 16

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

A^{4} - 5 A^{3} + 2 A^{2} + 8 A = 0

3 - 3 x^{5} = 0

What are the solutions to the equation x^4-3x^2+4x+1=0 ?
The solutions to the equation x^4-3x^2+4x+1=0 are x\approx-0.21565…,x\approx-2.15435…
Find the zeros of x^4-3x^2+4x+1=0
The zeros of x^4-3x^2+4x+1=0 are x\approx-0.21565…,x\approx-2.15435…