Simplification II

**Simplicity** · November 4th 2008, 09:55 AM

Would this simplification be correct?

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

**Moo** · November 4th 2008, 10:20 AM

Hello,

Originally Posted by Air

Would this simplification be correct?

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

This would be better

**Simplicity** · November 4th 2008, 10:59 AM

... And what if it was:

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

How would I simplify that?

**Moo** · November 4th 2008, 11:02 AM

Originally Posted by Air

... And what if it was:

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

How would I simplify that?

Hmmm well, you'd have :

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

(what's in a square root must be positive)

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

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

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