2x^4-3x^2-7x+1=0

Estimate what the real solution will be using Newton's method. Answers must be 4 decimal places. Let the first guess be x=1.75

1 answer

let y = 2x^4 - 3x^2 - 7x + 1
y ' = 8x^3 - 6x - 7

newx = x - y/y'
= x - (2x^4 - 3x^2 - 7x + 1)/(8x^3 - 6x - 7)
=(6x^4- 3x^2 - 1)/(8x^3 - 6x - 7)

x = 1.75
newx = 1.8162 ---> x
newx = 1.8111 ---> x
newx = 1.811102

I would call 1.8111 a good estimate