What are the solutions to the equation (x+1)*x^3=x+11 ?

The solutions to the equation (x+1)*x^3=x+11 are x\approx-2.04578…,x\approx 1.67895…

Find the zeros of (x+1)*x^3=x+11

The zeros of (x+1)*x^3=x+11 are x\approx-2.04578…,x\approx 1.67895…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1) \cdot x^{3} = x + 11

x \approx - 2.04578 \dots, x \approx 1.67895 \dots

Solution steps

(x + 1) x^{3} = x + 11

Expand

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

x^{4} + x^{3} = x + 11

Move

11

to the left side

x^{4} + x^{3} - 11 = x

Move

x

to the left side

x^{4} + x^{3} - 11 - x = 0

Write in the standard form

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

x^{4} + x^{3} - x - 11 = 0

Find one solution for

x^{4} + x^{3} - x - 11 = 0

using Newton-Raphson:

x \approx - 2.04578 \dots

Apply long division:

\frac{x ^{4} + x ^{3} - x - 11}{x + 2.04578 \dots} = x^{3} - 1.04578 \dots x^{2} + 2.13946 \dots x - 5.37689 \dots

x^{3} - 1.04578 \dots x^{2} + 2.13946 \dots x - 5.37689 \dots \approx 0

Find one solution for

x^{3} - 1.04578 \dots x^{2} + 2.13946 \dots x - 5.37689 \dots = 0

using Newton-Raphson:

x \approx 1.67895 \dots

Apply long division:

\frac{x ^{3} - 1.04578 \dots x ^{2} + 2.13946 \dots x - 5.37689 \dots}{x - 1.67895 \dots} = x^{2} + 0.63316 \dots x + 3.20252 \dots

x^{2} + 0.63316 \dots x + 3.20252 \dots \approx 0

Find one solution for

x^{2} + 0.63316 \dots x + 3.20252 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx - 2.04578 \dots, x \approx 1.67895 \dots

(x + 4) (x + 2) (x - 3) = 0

x^{10} = 10

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

x^{4} + 2 x^{3} - 43 x^{2} - 8 x + 336 = 0

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

What are the solutions to the equation (x+1)*x^3=x+11 ?
The solutions to the equation (x+1)*x^3=x+11 are x\approx-2.04578…,x\approx 1.67895…
Find the zeros of (x+1)*x^3=x+11
The zeros of (x+1)*x^3=x+11 are x\approx-2.04578…,x\approx 1.67895…