Given $c\neq7$ and $st(c)=7$, simplify $\frac{3-\sqrt{c+2}}{\sqrt{c-7}}$
My inclination, based on one of the examples in the book, is to multiply by $3+\sqrt{c+2}$, yielding:
$\frac{(3-\sqrt{c+2})\cdot(3+\sqrt{c+2})}{\sqrt{c-7}\cdot(3+\sqrt{c+2})}$ $=\frac{9-(c+2)}{3\sqrt{c-7}+\sqrt{c-7}\sqrt{c+2}}$
Substituting in $c=7+\epsilon$ leads to:
$=\frac{9-(9+\epsilon)}{3\sqrt{\epsilon}+\sqrt{\epsilon}\sqrt{9+\epsilon}}$
To get the answer in the back of the book, $-\frac{1}{6}$, I think one must somehow convert this into:
$=-\frac{\epsilon}{6\epsilon}$
I don't see a road from the simplification after the substitution to this, which makes me think I've made more fundamental errors. I specifically don't see a way to get from $\sqrt{\epsilon}\sqrt{9+\epsilon}$ to $3\sqrt{\epsilon}$, which seems necessary if I haven't made errors earlier.
Thanks for your help!
Edit: I have transcribed the problem wrong (I originally wrote $\sqrt{7-c}$ instead of $\sqrt{c-7}$) which explains the spurious absolute value in Andre's answer, but I think the question still stands.
Edit 2: I have either transcribed the problem wrong twice (or more), or the 1st edition copy of the book I have at home is different from the 2nd edition copy online. Either way, the question is flawed. I have asked the corrected question again here.