Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 18

sqrt(x + 1) = x^2-x
Set the left side equal to zero.
0=x^2-x-sqrt(x - 1)
To solve using Newton's method, apply the formula:
x_(n+1)=x_n-(f(x_n))/(f'(x_n))
Let the function f(x) be:
f(x)=x^2-x-sqrt(x+1)
Then, take its derivative.
f'(x)=2x-1-1/(2sqrt(x+1))
Plug-in f(x) and f'(x) to the formula of Newton's method.
x_(n+1)=x_n - (x_n^2-x_n-sqrt(x_n+1))/(2x_n-1-1/(2sqrt(x_n+1)))
And that simplifies to:
x_(n+1) = x_n - (2x_n^2sqrt(x_n+1)-2x_nsqrt(x_n+1)-2(x_n+1))/(4x_nsqrt(x_n+1) - 2sqrt(x_n+1)-1)
For the initial value of x, let's refer to the graph of f(x). (See attached figure.)
Notice that there are two values of x in which f(x)=0. The roots of the f(x) are near x=-0.5 and x=2.
To approximate the value of the first root to six decimal places, plug-in the initial value x_1=-0.5 to x_(x+1).
x_0=-0.5
x_2= x_1 - (2x_1^2sqrt(x_1+1)-2x_1sqrt(x_1+1)-2(x_1+1))/(4x_1sqrt(x_1+1) - 2sqrt(x_1+1)-1)=-0.4841553280
x_3= x_2 - (2x_2^2sqrt(x_2+1)-2x_2sqrt(x_2+1)-2(x_2+1))/(4x_2sqrt(x_2+1) - 2sqrt(x_2+1)-1)=-0.4840283103
x_4= x_3 - (2x_3^2sqrt(x_3+1)-2x_3sqrt(x_3+1)-2(x_3+1))/(4x_3sqrt(x_3+1) - 2sqrt(x_3+1)-1)=-0.4840283022
Notice that two approximation have now the same six decimal places. Thus, one of the approximate solution of the equation is x=-0.484028 .
Next, to approximate the second root to six decimal places, plug-in the initial value x_1=2 to x_(n+1).
x_1=2
x_2= x_1 - (2x_1^2sqrt(x_1+1)-2x_1sqrt(x_1+1)-2(x_1+1))/(4x_1sqrt(x_1+1) - 2sqrt(x_1+1)-1)=1.901174073
x_3= x_2 - (2x_2^2sqrt(x_2+1)-2x_2sqrt(x_2+1)-2(x_2+1))/(4x_2sqrt(x_2+1) - 2sqrt(x_2+1)-1)=1.897185922
x_4= x_3 - (2x_3^2sqrt(x_3+1)-2x_3sqrt(x_3+1)-2(x_3+1))/(4x_3sqrt(x_3+1) - 2sqrt(x_3+1)-1)=1.897179401
x_5= x_4 - (2x_4^2sqrt(x_4+1)-2x_4sqrt(x_4+1)-2(x_4+1))/(4x_4sqrt(x_4+1) - 2sqrt(x_4+1)-1)=1.897179401
Notice that the two approximation have the same six decimal places. Thus, the other approximate solution is x=1.897179 .

Therefore, the approximate solution (to six decimal places) of sqrt(x+1)=x^2-x are x={-0.484028,1.897179} .