Simplification II

Printable View

Nov 4th 2008, 09:55 AM
Simplicity

Simplification II

Would this simplification be correct?

$\sqrt{\frac{A^2 - 2A^2B^2+A^2B^4+4A^2B^2C^2}{1-2B^2+B^4+4B^2C^2}} = \sqrt{\frac{A^2(1-2B^2+B^4+4B^2C^2)}{1-2B^2+B^4+4B^2C^2}} = \sqrt{{A^2}} = A$
Nov 4th 2008, 10:20 AM
Moo

Hello,

Quote:

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 (Tongueout)
Nov 4th 2008, 10:59 AM
Simplicity

... 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?
Nov 4th 2008, 11:02 AM
Moo

Quote:

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

All times are GMT -8. The time now is 02:51 AM.

Copyright © 2005-2017 Math Help Forum. All rights reserved.