Mathematica How to convert the root of equation in Mathematica

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To convert the root of an equation from symbolic form to ordinary form in Mathematica, the N command may not suffice, as it often retains the symbolic representation. The ToRadicals[] function can sometimes transform Root[] objects into nested radicals, but this may not yield a fully numeric result due to the presence of variables. A suggested approach involves substituting variables and simplifying the equation, particularly by dividing exponents by two to convert to a cubic form, which can help extract roots without using Root[] objects. However, this method still results in expressions containing variables, making it challenging to achieve a purely numeric solution. Overall, while there are techniques to manipulate the equation, fully converting it to an ordinary form may remain difficult.
kaizen.moto
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Hi all,
How to convert the root of an equation from symbolic form into an ordinary from? I have tried to use the N command, but still displaying in the symbolic form.

Thanks
 

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Root[] objects are sometimes returned for complicated roots of higher degree polynomials.

Sometimes, but no always, ToRadicals[] can convert Root[] objects into usually much more complicated nested radicals and get closer to what I think you would consider "ordinary form", but you still have variables in those expressions and that is going to make it more difficult or impossible to get an answer like 3.64.

Unfortunately, after typing your data back in it looks like ToRadicals cannot help you with this one.

All your lambda do appear to even powers. Sometimes you can make a little more progress by substituting z=lambda^2 and working on z instead. Then after you are done you can look at the two square roots of z. Unfortunately this doesn't seem to help in this case either.
 
Last edited:
I gave this one more try.

All your variables, m, n, a and lambda appear as even powers. Divide all exponents on your variables by 2. If I haven't made any mistakes typing this back in, that gives you a cubic in lambda and reduces the other variable exponents enough that Solve[eqn==0,lambda] is able to recognize it can give you the three roots of the cubic and not use Root[] objects. Then multiply the exponents on m, n and a by two, find the two square roots of each of those three roots of the cubic and you should have the six roots of your original polynomial in lambda.

But that is still going to leave you with all the m, n and a variables in those six roots.
 
thanks for your comments...i really appreciate it...
 

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