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Math Help - Simplification II

  1. #1
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    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
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  2. #2
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    Hello,
    Quote Originally Posted by Air View Post
    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
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  3. #3
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    ... 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?
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  4. #4
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    Quote Originally Posted by Air View Post
    ... 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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