# Use Newton's method to find all roots of the equation correct to six decimal places? Square root (x+1)=x^2-x

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Let us define a function

`f(x)=x^2-x-sqrt(x+1)`

and find its derivative with respsct to x

`f'(x)=2x-1-(1/2)1/sqrt(x+1)` ,

from above expression we conclude that if x=-1 ,then f'(x) is not defined.

The Newton method is

`x_(k+1)=x_k-f(x_k)/(f'(x_k))` , prvided `f'(x_k)!=0` for any `x_k.`

Now draw the graph of f(x) to estimate the initial approximation

From graph we observe that roots of the given equation lies in (-1,0) and (1,2).

1. let approximation of first root be -.5 i.e `x_0=-.5`

`f(-.5)=(-.5)^2-(-.5)-sqrt(-.5+1)=.0428932`

`f'(-.5)=2(-.5)-1-.5/sqrt(.5)=-2.7071068`

`x_1=-.5-.0428932/(-2.7071068)=-.4841553`

`f(-.4841553)=(-.4841553)^2+.4841553-sqrt(-.4841553+1)`

`=.00033836`

`` `f'(-.4841553)=2(-.4841553)-1-.5/sqrt(-.4841553+1)`

`=``-2.6644729`

`x_2=-.4841553+.00033836/2.6644729=-.4840283`

Simlarly second root can be calculated which is near t the point x=2.