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I'm trying to help my little bro, a bit rusty here... Wolfram Alpha is telling me that:

$$ x\sqrt{1+{\frac{x^2}{16-x^2}}} $$

simplifies to:

$$ 4x\sqrt{\frac1{16-x^2}} $$

I can't for the life of me figure out why. I'm thinking there's a simple rule I'm forgetting about..

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up vote 5 down vote accepted

Find the common denominator within the radical sign. $$x\sqrt {1 + \frac {x^2}{16 - x^2}} = x\sqrt {\frac {(16 - x^2) + x^2}{16 - x^2}} = x\sqrt{\frac{16}{16- x^2}}=4x \sqrt{\frac{1}{16 - x^2}}$$

In the last step, we simplify $\sqrt{16} = 4$.

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there it is... i tried 100 things. can you make inline equations in comments? i didn't see the $-x^2$ and $x^2$ cancel out. i was blind, and now i see, ty – Mike Lewis Oct 14 '13 at 23:56
You're welcome! We all overlook simple things, sometimes! – amWhy Oct 14 '13 at 23:57


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ok, amWhy beat me. – Krishna Oct 14 '13 at 23:58
yea, but i gave you (+1) 10 points anyway. thanks! – Mike Lewis Oct 14 '13 at 23:59

Hint: Think back to adding fractions. Also, remember how to move factors in and out of a square root.

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