Question:
By Newton-Raphson method, the positive root of \(x^4 - x - 10 = 0\) is:
By Newton-Raphson method, the positive root of \(x^4 - x - 10 = 0\) is:
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Newton-Raphson converges quadratically; choose initial guess close to the root.
Updated On: Apr 16, 2026
- 1.871
- 1.868
- 1.856
- None of these
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The Correct Option is A
Solution and Explanation
Concept:
Newton-Raphson iteration: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), where \(f(x) = x^4 - x - 10\) and \(f'(x) = 4x^3 - 1\).
Step 1: Choose initial guess. \(x_0 = 2\) (since \(2^4 - 2 - 10 = 16 - 12 = 4>0\), and \(1.8^4 \approx 10.5\)).
Step 2: First iteration. \[ x_1 = 2 - \frac{16 - 2 - 10}{32 - 1} = 2 - \frac{4}{31} \approx 2 - 0.129 = 1.871 \]
Step 3: Second iteration (for verification). \[ f(1.871) \approx (1.871)^4 - 1.871 - 10 \approx 12.25 - 11.871 \approx 0.379 \] \[ f'(1.871) = 4(1.871)^3 - 1 \approx 4(6.55) - 1 \approx 26.2 - 1 = 25.2 \] \[ x_2 = 1.871 - \frac{0.379}{25.2} \approx 1.871 - 0.015 = 1.856 \]
Step 1: Choose initial guess. \(x_0 = 2\) (since \(2^4 - 2 - 10 = 16 - 12 = 4>0\), and \(1.8^4 \approx 10.5\)).
Step 2: First iteration. \[ x_1 = 2 - \frac{16 - 2 - 10}{32 - 1} = 2 - \frac{4}{31} \approx 2 - 0.129 = 1.871 \]
Step 3: Second iteration (for verification). \[ f(1.871) \approx (1.871)^4 - 1.871 - 10 \approx 12.25 - 11.871 \approx 0.379 \] \[ f'(1.871) = 4(1.871)^3 - 1 \approx 4(6.55) - 1 \approx 26.2 - 1 = 25.2 \] \[ x_2 = 1.871 - \frac{0.379}{25.2} \approx 1.871 - 0.015 = 1.856 \]
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