**method of Newton Raphson**, find the real roots of the equation x

^{3}– 3x – 5 = 0 correct to four places of decimals. This question comes from Numerical Analysis, Engineering Book.

**roots of the equation x**Follow the steps to solve the questions.

^{3}– 3x – 5 up to 5 decimal places using the Newton Raphson Method.Given equation f(x) = x^{3} – 3x – 5

Differentiate with respect to x, we get f ‘(x) = 3x^{2} – 3

Now, first find the range, where the real roots lie in i.e. f(2) = –3 and f(3) = 13

Since f(2) is a negative value and f(3) is a positive value. Therefore, our one real root of the equation lies between x = 2 and 3.

Now, by **Newton-Raphson formula**

Now, substitute the value of f(x) and f'(x) in this formula

Choose the initial approximate value of the root x_{0} = 2 and putting n = 0, we have the first approximate value

Now, for the second approximation, replacing n = 1 in equation (1) we get

Now, for the third approximation, replacing n = 2 in equation (1) we get

Now, for the Fourth approximation, replacing n = 3 in equation (1) we get

Here x_{3} and x_{4} obtained the same value; therefore one **root of the given equation = x ^{3} – 3x – 5 using the Newton Raphson Method **is 2.2790.

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