how can i solve this equation i keep get error

clc,clear;

syms x

eqn=-15730900+(-1473100)+(501831/(1+x))+(501831/(1+x)^2)+(501831/(1+x)^3)+...

+(501831/(1+x)^4)+(501831/(1+x)^5)+(501831/(1+x)^6)+(501831/(1+x)^7)+...

+(501831/(1+x)^8)+(501831/(1+x)^9)+(501831/(1+x)^10)==0;

solx=solve(eqn,x)

this is my equation and i cant take the value for x im wonder why

[Walter Roberson]

What error do you get?

Those continuation lines would have to be immediately after the previous lines; you show a blank line between the lines of the eqn and that is not permitted syntax.

If you are using an earlier version of the Symbolic Toolbox, before about R2011b (I think) you might need to remove the "==0"

[John D'Errico]

Note that there will surely be no analytical solution.

Answer 1

[John D'Errico]

As I pointed out, there will surely be no analytical solution. You should expect that with near certainty, as this is the equivalent of a high order polynomial.

I see no error however. Only the indication (from solve, in the form of the results it yields) that it was unable to compute analytical roots.

solve(eqn)
ans =
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 1) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 2) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 3) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 4) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 5) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 6) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 7) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 8) - 1
  1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 9) - 1
 1/root(z^10 + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z - 1564000/45621, z, 10) - 1

Again, that was completely expected. However, nothing stops you from getting approximate roots.

vpa(solve(eqn))
ans =
                                         -1.6661435339699159760690274098319
                                        -0.17803633410337122738678393539001
 - 0.82944078945054912679553055169416 + 0.67487342116657470095168018942751i
  - 1.2392357796067300876301190456248 - 0.63453525124017787853359333001222i
  - 1.5503593102427657212185969586246 - 0.38032263725340143579333271938087i
  - 1.5503593102427657212185969586246 + 0.38032263725340143579333271938087i
  - 1.2392357796067300876301190456248 + 0.63453525124017787853359333001222i
 - 0.82944078945054912679553055169416 - 0.67487342116657470095168018942751i
 - 0.44428946799323473628513677400307 - 0.47475279125570655866734145566345i
 - 0.44428946799323473628513677400307 + 0.47475279125570655866734145566345i