# How do you interpret this Solve output ?

Posted 9 years ago
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 I do not understand what the parameter # represents, since the variables in the problem are all represented in the solution. I want to be able to write an equation of the form v=something resembling real written math that I can actually use. Any feedback would be greatly appreciated.
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Posted 9 years ago
 If you have a quadratic equation then the quadratic formula is fairly simple. A cubic has a far more complicated formula. A quartic far more than that. Mathematica introduced a shorthand, an abbreviated way of describing these without all the details. That shorthand is Root. What this gives is Root[polynomial, n] where n is the n'th root and they have a consistent way of numbering the roots.Now they make it a little more complicated, instead of just using v or x, they use #1 and put an & at the end of that. There are reasons for that, but they might not matter much to you at the moment. You can just mentally replace #1 with v and that is supposed to represent the first root, because of that &, 1 at the very end.If you can assign values to your rho, k and nu before or after the Solve then you can do things like N[Root[...stuff...]] and see a numerical approximation of the root.On the other hand, if you don't have numeric constants to assign to rho, k and nu then often for degree 5 polynomials and beyond there simply isn't a degree 5 version of the quadratic equation that will use powers and roots of constants and variables and say that is "the root."It looks like you are expecting rho to not be zero. You might simplify the solution a little, but not much, by replacing Sqrt[k]/Sqrt[rho] with u inside your Solve. That doesn't get rid of Root, that doesn't give you something like what the quadratic equation would, but it might make it a little easier to read.See if that works for you
Posted 9 years ago
Posted 9 years ago
 Your code is hard to read and full of syntax errors. For example Sqrt[[Rho]] can't be correct. As far as the final answer you show in Out[1] then you simply use N to convert it to numerical values.  N[%] {{x -> -0.486389}, {x -> -0.701874 - 0.879697 I}, {x -> -0.701874 + 0.879697 I}, {x -> 0.945068 - 0.854518 I}, {x -> 0.945068 + 0.854518 I}} These are the solutions. If you do not want to see the x there then write  x /. % {-0.486389, -0.701874 - 0.879697 I, -0.701874 + 0.879697 I, 0.945068 - 0.854518 I, 0.945068 + 0.854518 I}