Is there a rule for $\sqrt{a+b}$? You learn in algebra that 
$$\sqrt{ab}=\sqrt{a} \sqrt{b}$$
and that 
$$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$
You also learn to never make the fatal mistake of thinking
$$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$
However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.
$$\sqrt{a+b}=\text{?}$$

There is no finite rule but there is the binomial series:
$$
(1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k
$$
where
$$
{\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}
$$
This series converges for $|x|<1$.
Taking $\alpha=1/2$, we get
$$
(1+x)^{1/2} =
1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+\cdots
$$
We can apply this to $\sqrt{a+b}$ as follows. Assume wlog that $a>b$. Then
$$
\sqrt{a+b}=(a+b)^{1/2} = \sqrt{a}(1+x)^{1/2}
$$
with $x=b/a<1$.
We then have
$$
\sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right)
$$
