The real root of x³ + x² + 3x + 4 = 0 correct to four decimal places, obtained using Newton Raphson method is

Concept:

Newton-Raphson Method:

The iteration formula is given by

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Where x0 is the initial value/root of the equation f(x) = 0

Given,

f(x) = x³ + x² + 3x + 4 = 0

f'(x) = 3x² + 2x + 3

∴ f(-1) = 1 > 0 and f(-2) = -6 < 0

∴ f(-1).f(-2) < 0

⇒ ∃ a root lies in [-1, -2]

Let, x₀ = -1

By Newton Raphson method

First approximation

\(x_1=x_0-\frac{f(x_0)}{f'(x_0)}\)

\(=-1-\frac{1}{4}=-1-0.25\)

x₁ = -1.25

\(x_2=x_1-\frac{f(x_1)}{f'(x_1)}\)

\(=-1.25-\frac{(-1.25)^3+(-1.25)^2+3(-1.25)+4}{3(-1.25)^2+2(-1.25)+3}=-1.2229\)