If we have: $$x_n=\left\lbrace \begin{matrix} b\in\mathbb{R}\setminus \{0\} & ,n=1 \\ \dfrac{a+x_{n-1}^2}{2x_{n-1}} & , n>1\end{matrix} \right.$$
Then, is easy to prove that $(x_n)\to +\sqrt{a}$ if $b>0$, and $(x_n)\to -\sqrt{a}$ if $b<0$.
My question is: For $a\in\mathbb{R}^+$ there exists a sequence $(x_n^p)$ wich converges to $\sqrt[p]{a}$ for every $p\in\mathbb{N}$?
This is a special case of Newton–Rapheson: you look at $$ x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}, $$ with $f(x) = x^p-a$. Then the iteration simplifies to $$ x_n = x_{n-1} - \frac{x_{n-1}^p-a}{px_{n-1}^{p-1}} = \frac{a+(p-1)x_{n-1}^p}{px_{n-1}^{p-1}} $$
Certainly if it does converge, you would have $x_n , x_{n-1} \to x $, so $$ x = \frac{a+(p-1)x^p}{px^{p-1}} \\ px^{p} = a + (p-1)x^p, $$ so $x^p$ is a $p$th root of $a$. Further, it is easy to see that choosing $x_0>0$ makes every successive $x_n>0$, so it is the positive $p$th root.
Checking it converges is explained in the linked article, but it's likely that it works in exactly the same way as the $p=2$ case. Equally, you could look at $$ x_n = \frac{a+(k-1)x_{n-1}^p}{kx_{n-1}^{p-1}} $$ for any $k \neq 0$, although you will probably find that $k=p$ is the best option.