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Question: For finding real roots of the equation \(x^{2} - x = 2\) by Newton-Raphson method, choose \(x_{0} = ...

For finding real roots of the equation x2−x=2x^{2} - x = 2 by Newton-Raphson method, choose x0=1x_{0} = 1, then the value of x2x_{2} is

A

–1

B

3

C

115\frac{11}{5}

D

None of these

Answer

115\frac{11}{5}

Explanation

Solution

Let f(x)=x2−x−2f(x) = x^{2} - x - 2. Given x0=1x_{0} = 1.

By Newton-Raphson method,x1=x0−f(x0)f′(x0)x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}

f′(x)=2x−1=2(1)−1=1f'(x) = 2x - 1 = 2(1) - 1 = 1 andf(1)=−2f(1) = - 2. Now x1=1−−21=3x_{1} = 1 - \frac{- 2}{1} = 3

f(x1)=f(3)=9−3−2=4f(x_{1}) = f(3) = 9 - 3 - 2 = 4

f′(x1)=f′(3)=2(3)−1=5f'(x_{1}) = f'(3) = 2(3) - 1 = 5; x2=3−45=115x_{2} = 3 - \frac{4}{5} = \frac{11}{5}