solve sum of square roots using a single functiom using Newton's Method

Question

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!

Answer 1

HINT Newton's method is a root finder, meaning given a function $f(x)$ it computes the root of $f(x)$, i.e. the special argument $\bar{x}$ such that $f(\bar{x}) = 0$.

You need for Newton's Method to find $\sqrt{3} + \sqrt{7}$. Can you find a function $f(x)$, which will have a root at this magic place?

Answer 2

HINT

Say $x=\sqrt{3}+\sqrt{7}$. Then $x^2=10+2\sqrt{21}$. There is still a radical left. You need to remove it to get to the function $f(x)$.

Answer 3

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.