<s>[INST] Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let 
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of 
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$
This is what I should get:
$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$
Which I can simplify to:
$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$
However, this answer can't be found in my multiple choice question here:

Any hints on what I'm doing wrong?
 [/INST] The answer is b, which is equivalent to your answer after simplifying.
This is because $ 9 - \sqrt{45} = 9 - \sqrt{9 * 5} = 9 - 3\sqrt{5}$. Then,$ \frac{9 - 3\sqrt{5}}{18} = \frac{3 - \sqrt{5}}{6}. $ 
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