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 Author Comment/Response yehuda 01/31/13 08:58am Sure you can You need to do the following steps 1. Expand the expression inside the square root 2. Factor it you do this using pattern matching + relevant functions expr = 1.01792*10^-40 \[Sqrt](2.6336*10^72 + 4.80538*10^77 kx^2 + 8.76809*10^82 kx^4 + 4.80538*10^77 ky^2 + 3.65593*10^83 kx^2 ky^2 + 8.76809*10^82 ky^4 - 9.61076*10^77 kz^2 + 3.65593*10^83 kx^2 kz^2 + 3.65593*10^83 ky^2 kz^2 + 8.76809*10^82 kz^4) expr /. Sqrt[s_] :> Sqrt[Factor[Expand[s]]] Notice that kz has the highest degree (4) so its coefficient will be normalized to 1, so the result is 30.1416 \[Sqrt](3.00362*10^-11 + 5.48053*10^-6 kx^2 + 1. kx^4 + 5.48053*10^-6 ky^2 + 4.16959 kx^2 ky^2 + 1. ky^4 - 0.0000109611 kz^2 + 4.16959 kx^2 kz^2 + 4.16959 ky^2 kz^2 + 1. kz^4) yehuda URL: ,

 Subject (listing for 'factor inside root') Author Date Posted factor inside root Simone 01/30/13 09:16am Re: factor inside root Bill Simpson 01/30/13 3:46pm Re: factor inside root yehuda 01/31/13 08:58am
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