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

The solutions to the equation x^3+x^2-2x=1 are x\approx-0.44504…,x\approx 1.24697…,x\approx-1.80193…

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

The zeros of x^3+x^2-2x=1 are x\approx-0.44504…,x\approx 1.24697…,x\approx-1.80193…

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

x \approx - 0.44504 \dots, x \approx 1.24697 \dots, x \approx - 1.80193 \dots

Solution steps

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

Move

1

to the left side

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

Find one solution for

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

using Newton-Raphson:

x \approx - 0.44504 \dots

Apply long division:

\frac{x ^{3} + x ^{2} - 2 x - 1}{x + 0.44504 \dots} = x^{2} + 0.55495 \dots x - 2.24697 \dots

x^{2} + 0.55495 \dots x - 2.24697 \dots \approx 0

Find one solution for

x^{2} + 0.55495 \dots x - 2.24697 \dots = 0

using Newton-Raphson:

x \approx 1.24697 \dots

Apply long division:

\frac{x ^{2} + 0.55495 \dots x - 2.24697 \dots}{x - 1.24697 \dots} = x + 1.80193 \dots

x + 1.80193 \dots \approx 0

x \approx - 1.80193 \dots

The solutions are

x \approx - 0.44504 \dots, x \approx 1.24697 \dots, x \approx - 1.80193 \dots

t^{4} - 256 = 0

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

4 x^{3} + 3 x^{2} - 60 x - 45 = 0

12 x^{4} - 9 x^{2} - 3 = 0

5 = c^{3}

What are the solutions to the equation x^3+x^2-2x=1 ?
The solutions to the equation x^3+x^2-2x=1 are x\approx-0.44504…,x\approx 1.24697…,x\approx-1.80193…
Find the zeros of x^3+x^2-2x=1
The zeros of x^3+x^2-2x=1 are x\approx-0.44504…,x\approx 1.24697…,x\approx-1.80193…