solve, command reveals solution for each part of my polynomial

Hi I am using this http://uk.mathworks.com/help/symbolic/solve.html

however the solution presented is for each part of the polynomial:

>> solve((1211*x^6)/2500 + (9953*x^5)/10000 + (193*x^4)/125 - (681*x^3)/200 - (9549*x^2)/10000 + (3117*x)/1000 + 304/25 == 0,x) 

ans =

   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 1)
   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 2)
   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 3)
   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 4)
   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 5)
   root(z^6 + (9953*z^5)/4844 + (3860*z^4)/1211 - (17025*z^3)/2422 - (9549*z^2)/4844 + (15585*z)/2422 + 30400/1211, z, 6)
 whereas I want a single X value  for y  = 0

Will someone tell me the correct command?

g.

Answer 1

[John D'Errico]

There are no real roots.

You can ask for a "single" X value that works, but no matter what you do, you cannot make magic, unless of course, your name is Harry Potter. It might be, I guess. He did leave Hogwarts.

I suppose that possibly your question is really how to turn that mess into actual umbers. If so, just use vpa.

syms x
r = solve((1211*x^6)/2500 + (9953*x^5)/10000 + (193*x^4)/125 - (681*x^3)/200 - (9549*x^2)/10000 + (3117*x)/1000 + 304/25 == 0,x);

vpa(r)
ans =
  - 1.3989312138161239546724799986628 + 1.9811978680634290116114610605855i
  - 1.3989312138161239546724799986628 - 1.9811978680634290116114610605855i
 - 0.92035026602535783679025120533384 - 1.0426356256188857001302004733061i
 - 0.92035026602535783679025120533384 + 1.0426356256188857001302004733061i
   1.2919280529215808831225165054004 - 0.73309990384079756483990991478414i
   1.2919280529215808831225165054004 + 0.73309990384079756483990991478414i

Feel free to pick any of those complex roots.

Mathematics can be cruel. It refuses to do what you want, instead, doing what it wishes, blithely ignoring the needs of others. :)

[Gavin Seddon]

Finding a root X value seems far too complicated and there in the realms of Magic. I will manually choose the equilibrium X value. I fear a single value will be of no help.

Thanks for these explanations. g.

Answer 2

[Torsten]

Use "roots" instead of "solve".

Best wishes

Torsten.

Answer 3

[Gavin Seddon]

Thank you both. Reading further I think vpasolve may condense these roots to a single value. Are you able to assist in using this. Further I am aware of what the value should be from the initial distance time plot. Would it be advisable to use this Value? g.

[Walter Roberson]

In all newer releases, when vpasolve() sees that you have a polynomial to solve, it is going to return all of the roots. In order releases, it was potentially random as to which of the roots it returned.

If you know the approximate root, you could determine which of the 6 roots is closest to the one you are looking for. However, it seems improbable to me that you would be seeing an imaginary root on a distance/time plot, which suggests that you have given the wrong equations.