Question

Newton-Raphson method is used to find the root of \(x^2 - 3 = 0\). If the initial solution is taken as \(x = -2\), then the iterations tend to

• A. \(-\infty\)
• B. \(-\sqrt{3}\)
• C. \(\sqrt{3}\)
• D. zero

Hint:

Newton-Raphson iterations preserve the sign of the root for symmetric functions like parabolic equations.
Since \(x_0 = -2 < 0\), it must converge to the negative root \(-\sqrt{3}\).

Answer 1

The Correct Option is B

Concept:
• The Newton-Raphson iteration formula is given by: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\).
• The roots of \(x^2 - 3 = 0\) are \(+\sqrt{3}\) and \(-\sqrt{3}\).
• The method generally converges to the root closest to the initial guess \(x_0\), provided the derivative does not vanish.

Step 1: Set up the iteration function
Given \(f(x) = x^2 - 3 \implies f'(x) = 2x\).
Substitute into the formula: \[ x_{n+1} = x_n - \frac{x_n^2 - 3}{2x_n} = \frac{2x_n^2 - x_n^2 + 3}{2x_n} = \frac{x_n^2 + 3}{2x_n} \]

Step 2: Compute the first iteration with \(x_0 = -2\)
\[ x_1 = \frac{(-2)^2 + 3}{2(-2)} = \frac{4 + 3}{-4} = -1.75 \]

Step 3: Analyze the direction of convergence
Since \(x_0 = -2\) is negative, every subsequent iteration remains negative.
The values approach the negative square root of 3: \[ \lim_{n \to \infty} x_n = -\sqrt{3} \approx -1.732 \]