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

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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.

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