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
-$\frac{2\sqrt{2}-6}{7}$ = $\frac{6-2\sqrt{2}}{7}$ correct?
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$\begingroup$
algebra-precalculus
irrational-numbers
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0@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. – 2011-04-30
2 Answers
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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 $Q$?
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0Thanks, I always appreciate a different way of looking at it. – 2011-04-30
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yes, both you and the book are correct.