If 2.5 is the initial root of the equation x³ - x - 10 = 0 then by method of Newton - Raphson, the next approx root will be equal to-

Concept:

Newton-Raphson Method:

If x₁ is an initial root of the equation f(x) = 0 then the next approx root can be found by: \(x_{n + 1} = x_n - \frac{f{(x_n)}}{f'{(x_n)}}\)

Calculation:

Given: 2.5 is the initial root of the equation x3 - x - 10 = 0.

Let f(x) =  x3 - x - 10 and x₁ = 2.5

Here, we have to find the next approx root of f(x) = 0

⇒ f'(x) = 3x² - 1

⇒ f(2.5) = 3.125 and f'(2.5) = 17.75

As we know that, the next approx root is given by: \(x_{n + 1} = x_n - \frac{f{(x_n)}}{f'{(x_n)}}\)

⇒ \(x_{2} = x_1 - \frac{f{(x_1)}}{f'{(x_1)}}\)

⇒ \(x_{2} = 2.5 - \frac{3.125}{17.75} = 2.3239\)

So, the most approx option is A.