Is there a rule for $\sqrt{a+b}$?

Solution 1:

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 $$ \sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right) $$

Solution 2:

Provided that $a>0$ and $a+b>0$, we have $$\sqrt{a+b}=\sqrt{a}\cdot\sqrt{1+\frac{b}{a}}$$