# Simplifying the sum of a fraction and an integer under a radical sign

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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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
$x\sqrt{1+\frac{x^{2}}{16-x^{2}}}=x\sqrt{\frac{16-x^{2}}{16-x^{2}}+\frac{x^{2}}{16-x^{2}}}=x\sqrt{\frac{16-x^2+x^2}{16-x^2}}=x\sqrt{\frac{16}{16-x^2}}=4x\sqrt{\frac{1}{16-x^2}}$