Would this simplification be correct?
$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{1-2B^2+B^4+4B^2C^2}} = \sqrt{\frac{A^2(1-2B^2+B^4+4B^2C^2)}{1-2B^2+B^4+4B^2C^2}} = \sqrt{{A^2}} = A$
Printable View
Would this simplification be correct?
$\displaystyle \sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{1-2B^2+B^4+4B^2C^2}} = \sqrt{\frac{A^2(1-2B^2+B^4+4B^2C^2)}{1-2B^2+B^4+4B^2C^2}} = \sqrt{{A^2}} = A$
... 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