-$\frac{2\sqrt{2}-6}{7}$ = $\frac{6-2\sqrt{2}}{7}$ correct?

When asked to rationalize the denominator for $\frac{2}{\sqrt{2}+3}$, I came up with $\frac{6-2\sqrt{2}}{7}$ but my algebra book gives -$\frac{2\sqrt{2}-6}{7}$ as the answer. I think we're both correct or am I missing something here? I'm 99% sure I'm right, but I'm not feeling confident... :P

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forget the $7$ in the denominator. Just note that $-(a-b) = b - a$. So, yes, they're the same. – JavaMan Apr 30 '11 at 4:36
Your answer is the better one, since your numerator is positive. The book's numerator is negative, but that is taken care of by the minus sign in front. I thought almost no one liked negative numbers, they are so --- negative. – André Nicolas Apr 30 '11 at 4:51
Ah, I see what the book did. The instructions were to "Multiply the irrational denominator by its conjugate binomial (one having the same terms but opposite middle sign) in order to eliminate the middle term of the resulting trinomial." I was a weirdo and multiplied by (-$\sqrt{2}+3$) instead of ($\sqrt{2}-3$) as instructed. :) – DaveG Apr 30 '11 at 5:08
Stranger and stranger! Most mathematicians would consider $-\sqrt{2}+3$ to be the conjugate of $\sqrt{2}+3$. It is also the conjugate of $3+\sqrt{2}$. The conjugate of a number does not depend on the order of the addition. The book's authors either do not know about the way the word "conjugate" is actually used in mathematics, or decided to alter the meaning. – André Nicolas Apr 30 '11 at 5:24
@user6312 Interesting. I'll have to keep that in mind as I go through the rest of the book. Thanks for the insight! It just seemed the most straightforward way to do it to me at the time. – DaveG Apr 30 '11 at 5:44

Suppose you went $6$ units to the right and $2\sqrt{2}$ units to the left, reaching a point $P$. Now suppose you went $2\sqrt{2}$ units to the right and $6$ units to the left, reaching a point $Q$. What's the relation between $P$ and $Q$?