For the function $f(x)=x^2-a, a>0,$ the Newton iterative steps are given by?
$$ x_{n+1} = \frac 12 [x_n + \frac {a}{x_n}]$$ $$ x_{n+1} = 2[x_n + \frac {a}{x_n}]$$ $$ x_{n+1} = \frac 12 [x_n -ax_n]$$ $$ x_{n+1} = \frac 12 [x_n + ax_n]$$
Clueless regarding how to proceed. Here is my attempt for the sake of trying:
$$f(x) = x^2 - a$$ $$f'(x) = 2x$$ $$ x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)}$$
This is an iterative method where, starting from an initial approximation $x_0$, we form a sequence with each element, $x_{n+1}$, being computed from the previous, $x_n$, through the formula $$ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. $$
In this case, we have that $f(x) = x^2-a$ and therefore $$f(x_n)=x_n^2-a, \quad f'(x_n)= 2 x_n.$$
If you substitute this back into the formula and make some elementary simplifications you will get to one of the options.