Question

Approximate positive root of the equation \[ x^2-7x+9=0 \] using Newton-Raphson method with initial guess \(x_0=2\).

• A. \(\frac{56}{33}\)
• B. \(\frac{12}{5}\)
• C. \(23\)
• D. \(\frac{1}{2}\)

Hint:

In Newton-Raphson problems, one or two iterations are usually sufficient for objective questions. Carefully compute \(f(x_n)\) and \(f'(x_n)\) before substituting into \[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. \]

Answer 1

The Correct Option is A

Concept: Newton-Raphson method is an iterative technique used to approximate the roots of nonlinear equations. For an equation \[ f(x)=0, \] the iterative formula is \[ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. \] Starting from an initial approximation \(x_0\), successive approximations converge to the required root.

Step 1: Define the function and its derivative.
Given \[ f(x)=x^2-7x+9. \] Differentiating, \[ f'(x)=2x-7. \]

Step 2: Use the initial approximation \(x_0=2\).
Evaluate \[ f(2)=4-14+9=-1. \] Also, \[ f'(2)=4-7=-3. \] Applying Newton-Raphson formula, \[ x_1 = 2-\frac{-1}{-3}. \] \[ = 2-\frac13. \] \[ = \frac53. \]

Step 3: Compute the next approximation.
For \[ x_1=\frac53, \] \[ f\!\left(\frac53\right) = \frac{25}{9}-\frac{35}{3}+9 = \frac{1}{9}. \] Also, \[ f'\!\left(\frac53\right) = \frac{10}{3}-7 = -\frac{11}{3}. \] Hence, \[ x_2 = \frac53 -\frac{\frac19}{-\frac{11}{3}}. \] \[ = \frac53+\frac1{33}. \] \[ = \frac{55+1}{33}. \] \[ = \frac{56}{33}. \]

Step 4: Identify the approximate root.
Therefore, after the second iteration, \[ \boxed{x\approx\frac{56}{33}}. \] Hence the required answer is \[ \boxed{\frac{56}{33}}. \]