Question

Use Newton Raphson method to find the root of the equation x^3-2x-5=0 correct to 3 decimal point?​

Answer 1

Given : x³ - 2x - 5=0

 

To Find : one root by  by Newton-Raphson Method

correct 3 decimal  

 

Solution:

f(x) =   x³ - 2x - 5

f'(x) = 3x² - 2

x₀  = 2  

xₙ₊₁  = xₙ ​−f(xₙ​)​./ f′(xₙ​)

x₁  = 2 -  (2³ - 2*2- 5)/  (3*2² - 2)

x₁  =  2 -  (-1)/10

x₁  = 2.1

x₂ = 2.1 -  (2.1³ - 2*2.1- 5)/  (3*2.1² - 2)

=> x₂ = 2.09457

x₃ =  2.09455  

x₄ = 2.09455

hence x = 2.095  upto 3 decimal  places

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