Simplification II

Would this simplification be correct?

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

Hello,

Quote:

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Originally Posted by Air

Would this simplification be correct?

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

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This would be better (Tongueout)

... 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?

Quote:

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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?

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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