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Popular >

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

Solution

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

Solution

x \approx 0.61803 \dots, x \approx 0.61803 \dots, x \approx - 1.61803 \dots, x = - \frac{1597}{987}

+1

Decimal Notation

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

Solution steps

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

Expand

x (x + 2) (x^{2} - 1) : x^{4} - x^{2} + 2 x^{3} - 2 x

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

Move

1

to the left side

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Find one solution for

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

using Newton-Raphson:

x \approx 0.61803 \dots

Apply long division:

\frac{x ^{4} + 2 x ^{3} - x ^{2} - 2 x + 1}{x - 0.61803 \dots} = x^{3} + 2.61803 \dots x^{2} + 0.61803 \dots x - 1.61803 \dots

x^{3} + 2.61803 \dots x^{2} + 0.61803 \dots x - 1.61803 \dots \approx 0

Find one solution for

x^{3} + 2.61803 \dots x^{2} + 0.61803 \dots x - 1.61803 \dots = 0

using Newton-Raphson:

x \approx 0.61803 \dots

Apply long division:

\frac{x ^{3} + 2.61803 \dots x ^{2} + 0.61803 \dots x - 1.61803 \dots}{x - 0.61803 \dots} = x^{2} + 3.23606 \dots x + 2.61803 \dots

x^{2} + 3.23606 \dots x + 2.61803 \dots \approx 0

Find one solution for

x^{2} + 3.23606 \dots x + 2.61803 \dots = 0

using Newton-Raphson:

x \approx - 1.61803 \dots

Apply long division:

\frac{x ^{2} + 3.23606 \dots x + 2.61803 \dots}{x + 1.61803 \dots} = x + 1.61803 \dots

x + 1.61803 \dots \approx 0

x \approx - 1.61803 \dots

Approximate to algebraic form

x = - \frac{1597}{987}

The solutions are

x \approx 0.61803 \dots, x \approx 0.61803 \dots, x \approx - 1.61803 \dots, x = - \frac{1597}{987}

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