Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question

3 Answers 3

up vote 6 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 $\sqrt{16} = 4$.

share|improve this answer
    
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
1  
You're welcome! We all overlook simple things, sometimes! –  amWhy Oct 14 '13 at 23:57

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

share|improve this answer
    
ok, amWhy beat me. –  MisterSpock 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.