Question

The Newton-Raphson method is used to find the root of the equation f(x) \(\equiv\) x\(^2\) \(-\) x \(-\) 1 \(=\) 0. If the initial guess for the root is 1, then the estimate of the root after two iteration is

• A. 2
• B. 1.80
• C. 1.67
• D. 1.82

Hint:

Newton-Raphson tips:
- Double check your derivative: \( \frac{d}{dx}(x^2 - x - 1) = 2x - 1 \).
- Write out intermediate steps clearly: \( x_1 = 2 \rightarrow x_2 = 2 - 1/3 = 1.67 \).
This prevents algebraic or arithmetic errors during multi-step numerical calculations.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks to calculate the approximate root of a quadratic algebraic equation after performing exactly two iterations of the Newton-Raphson numerical method, starting with an initial guess of 1.

Step 2: Key Formula or Approach:
The Newton-Raphson iterative formula is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
For the given function:
\[ f(x) = x^2 - x - 1 \]
Its first derivative is:
\[ f'(x) = 2x - 1 \]

Step 3: Detailed Explanation:

• First Iteration (\( n = 0 \)) starting from \( x_0 = 1 \):
- Evaluate function and derivative at \( x_0 = 1 \):
\[ f(x_0) = f(1) = 1^2 - 1 - 1 = -1 \]
\[ f'(x_0) = f'(1) = 2(1) - 1 = 1 \]
- Calculate \( x_1 \):
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-1}{1} = 1 + 1 = 2 \]

• Second Iteration (\( n = 1 \)) starting from \( x_1 = 2 \key \):
- Evaluate function and derivative at \( x_1 = 2 \):
\[ f(x_1) = f(2) = 2^2 - 2 - 1 = 4 - 2 - 1 = 1 \]
\[ f'(x_1) = f'(2) = 2(2) - 1 = 3 \]
- Calculate \( x_2 \):
\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{1}{3} = \frac{5}{3} \approx 1.6667 \]
- Rounding to two decimal places yields 1.67.


Step 4: Final Answer:
The estimate of the root after two iterations is approximately 1.67.
Therefore, the correct choice is option (C).