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

`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...

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`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}` .

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