# How to get the equation inside that of root(function,z) equation ,i.e the function only

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Apurva Suman on 2 Dec 2021
Commented: Jon on 2 Dec 2021
root(z^4 - (190266216202962831605969987684001433013625*z^3)/142934430860719287430314339936263093557948 + (1101852793399817336417575000084289382730783*z^2)/2286950893771508598885029438980209496927168 - (23080530109547214937644974638002755076096*z)/893340192879495546439464624601644334737175 + 8413353246836900979045156398614448177152/22333504821987388660986615615041108368429375, z)
I'm getting this as a solution , how can i get that equation alone ,i want to extract that equation and use fzero to solve for the root in the interval i want?

Star Strider on 2 Dec 2021
The vpa funciton usually works —
syms z
Zfcn = root(z^4 - (190266216202962831605969987684001433013625*z^3)/142934430860719287430314339936263093557948 + (1101852793399817336417575000084289382730783*z^2)/2286950893771508598885029438980209496927168 - (23080530109547214937644974638002755076096*z)/893340192879495546439464624601644334737175 + 8413353246836900979045156398614448177152/22333504821987388660986615615041108368429375, z)
Zfcn = Zvpa = vpa(Zfcn)
Zvpa = .

Jon on 2 Dec 2021
This is a polynomial, you should be able to obtain all of the roots (real and complex) using:
r = roots([1.0 ...
-(190266216202962831605969987684001433013625)/142934430860719287430314339936263093557948 ...
(1101852793399817336417575000084289382730783)/2286950893771508598885029438980209496927168 ...
-(23080530109547214937644974638002755076096)/893340192879495546439464624601644334737175 ...
8413353246836900979045156398614448177152/22333504821987388660986615615041108368429375])
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Jon on 2 Dec 2021
I see another answer was already posted by the time I answered. Note I don't have the symbolic toolbox shown in @Star Strider's, answer. I'm not sure if you do. In anycase you can do it this way with the core MATLAB functionality