solve sum of square roots using a single functiom using Newton's Method I have to estimate $n= \sqrt{3} + \sqrt{7}$ using Newton's Method of approximation but I have to determine a single function which can be used to estimate $n$. 
P.S: The function should not involve radical expressions (nth root of constants or variables)
So, I just need to know what the single function is. I can solve the rest myself
Thanks!

Considering that radical roots appear in conjugate pairs, you could construction a rational function that has the root $n=\sqrt{7}+\sqrt{3}$ as follows,
$$[x-(\sqrt{7}+\sqrt{3})][x-(\sqrt{7}-\sqrt{3})] = x^2 +4 -2\sqrt{7}x$$
$$ (x^2 +4 -2\sqrt{7}x)(x^2 +4 + 2\sqrt{7}x) = (x^2+4)^2-28x^2 $$
Thus, the function
$$f(x)=(x^2+4)^2-28x^2$$ contains the roots $n=\sqrt{7}\pm \sqrt{3}$, as well as another pair.
