Binomial expansion question. easy! I'm trying to do the binomial expansion of 
$(x-2)^{1/2}$.

How do you do it? As far as I'm aware the expansion only works for $(1+x)^n$. How could I get it in that form?

Thanks.
One way to do this is to rewrite the expression as $(1+(x-3))^{1/2}$, and then substitute $u=x-3$ and expand $(1+u)^{1/2}$ using the Binomial Series
$\displaystyle(1+u)^r=\sum_{n=0}^{\infty}\binom{r}{n}u^n$.
If what you meant to ask about was a binomial expansion, then the approach is as follows: 

First, note that the binomial theorem states that 
\begin{align*}
(a+b)^n = \sum_{k=0}^{\infty}\binom{n}{k}a^kb^{n-k}
\end{align*}
Thus, by substituting for the above values, we have
\begin{align*}
(x-2)^\frac{1}{2} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k(-2)^{\frac{1}{2}-k}
\end{align*}