1. ## Simplification II

Would this simplification be correct?

$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{1-2B^2+B^4+4B^2C^2}} = \sqrt{\frac{A^2(1-2B^2+B^4+4B^2C^2)}{1-2B^2+B^4+4B^2C^2}} = \sqrt{{A^2}} = A$

2. Hello,
Originally Posted by Air
Would this simplification be correct?

$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{1-2B^2+B^4+4B^2C^2}} = \sqrt{\frac{A^2(1-2B^2+B^4+4B^2C^2)}{1-2B^2+B^4+4B^2C^2}} = \sqrt{{A^2}} = {\color{red}|}A{\color{red}|}$
This would be better

3. ... And what if it was:

$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{(1-2B^2+B^4+4B^2C^2)^2}}$

How would I simplify that?

4. Originally Posted by Air
... And what if it was:

$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{(1-2B^2+B^4+4B^2C^2)^2}}$

How would I simplify that?
Hmmm well, you'd have :

$\displaystyle \sqrt{\frac{A^2}{|1-2B^2+B^4+4B^2C^2|}}$ after simplifying (what's in a square root must be positive)

$\displaystyle =\frac{|A|}{\sqrt{|1-2B^2+B^4+4B^2C^2|}}$

now if you know that one or another is positive, you can remove the absolute values