Find the positive root of x^4-x-10=0 by using Iteration method.
Answers
Answer:
What is the root of the equation x4−x−10=0 correct to 4 decimal places using the Newton-Raphson method?
x4−x−10=0⇒x4=x+10.
⇒x4≈10⇒x≈±10−−√4≈±1.75.
The equation has only two real roots.
Using the Newton-Raphson method, we can solve this equation as under:
Let f(x)=x4−x−10⇒f′(x)=4x3−1.
We want to determine the value of x for which f(x)=0.
Let the first estimate of x be x1.
Then, the second and better estimate would be x2=x1−f(x1)f′(x1).
The third and better estimate would be x3=x2−f(x2)f′(x2).
We continue in this manner until the difference between two successive estimates is lesser than the tolerable error.
For this particular case, taking the first estimate x1=−1.75, the details of the iterations are as under:
For the second root, taking the first estimate x1=1.75, the details of the iterations are as under:
So, we get the solutions, correct to 4 decimal places, as x=−1.6975 and x=1.8556.
Answer:
The positive root of x⁴ -x -10 = 0 is 1.856
Step-by-step explanation:
Given: x⁴ - x -10 = 0
Let the equation be f(x) = x⁴ - x - 10 ----eq. (1)
then the derivative of f(x), f'(x) = 4x³ - 1
now for x = 0,1,2,..., we get
f(0) =o⁴ - 0- 10 = -10 (negative)
f(1) = 1⁴ -1 -10 = -10 (negative)
f(2) = 2⁴ - 2 -10 =16-12=4 (positive)
∴ a root of f(x)=0 lies between the values x=1 and x=2
Let x₀ = 2
Then By Newton Raphson Method ,
xₙ₊₁ = xₙ - f(xₙ)/f'(x)
=xₙ -(xₙ⁴ -xₙ - 10)/4xₙ³-1
=(4xₙ⁴-xₙ-xₙ⁴+xₙ+10)/4xₙ³-1
xₙ₊₁ = (3xₙ⁴+10)/4xₙ³-1
To find approximations
Now putting n=0 ,
x₀₊₁ = 3x₀⁴+10/4x₀³-1 = (3(2)⁴+10)/4(2)³-1
x₁ =(3×16+10)/32-1
x₁ =48+10/31 = 58/31
x₁ = 1.871
n = 1, x₁₊₁ = 3x₁⁴+10/4x₁³-1
x₂ ={3(1.871)⁴+10}/4(1.871)³-1 =1.85578
x₂ =1.856
n = 2, x₃ = 3x₂⁴+10/4x₂³-1
=3(1.856)⁴+10/4(1.856)³-1
x₃ =1.856 we stop here because of getting same value.
∵ x₂ = x₃ = 1.856
Hence the positive root of x⁴-x-10=0 is 1.856
#spj 2