Matlab unable to find solution to cubic polynomial

2 views (last 30 days)
The following cubic equation has three roots.
syms a
solve((1225*a)/2 - 6125 == ((2*a - 35)^2*(60*a + 4200))/840, a)
Matlab's output is:
root(z^3 + 35*z^2 - (8575*z)/2 + 42875, z, 1)
root(z^3 + 35*z^2 - (8575*z)/2 + 42875, z, 2)
root(z^3 + 35*z^2 - (8575*z)/2 + 42875, z, 3)
Can cubic equations like this be solved analytically in Matlab?

Accepted Answer

James Tursa
James Tursa on 1 Feb 2021
Edited: James Tursa on 1 Feb 2021
Tell the solve( ) function that the max degree of the polynomial is 3 to force explicit solutions for the result:
syms a
p = (1225*a)/2 - 6125 - ((2*a - 35)^2*(60*a + 4200))/840
solve(p,a,'MaxDegree',3)
which gives
ans =
28175/(18*(- 5187875/108 + (432^(1/2)*659937359375^(1/2)*1i)/432)^(1/3)) + ((432^(1/2)*659937359375^(1/2)*1i)/432 - 5187875/108)^(1/3) - 35/3
- 28175/(36*(- 5187875/108 + (432^(1/2)*659937359375^(1/2)*1i)/432)^(1/3)) - ((432^(1/2)*659937359375^(1/2)*1i)/432 - 5187875/108)^(1/3)/2 - (3^(1/2)*(28175/(18*(- 5187875/108 + (432^(1/2)*659937359375^(1/2)*1i)/432)^(1/3)) - ((432^(1/2)*659937359375^(1/2)*1i)/432 - 5187875/108)^(1/3))*1i)/2 - 35/3
- 28175/(36*(- 5187875/108 + (432^(1/2)*659937359375^(1/2)*1i)/432)^(1/3)) - ((432^(1/2)*659937359375^(1/2)*1i)/432 - 5187875/108)^(1/3)/2 + (3^(1/2)*(28175/(18*(- 5187875/108 + (432^(1/2)*659937359375^(1/2)*1i)/432)^(1/3)) - ((432^(1/2)*659937359375^(1/2)*1i)/432 - 5187875/108)^(1/3))*1i)/2 - 35/3
Then you can also note
>> simplify(ans)
ans =
(35*2^(1/3)*(- 121 - 1077^(1/2)*3i)^(1/3))/6 + (35*2^(1/3)*(- 121 + 1077^(1/2)*3i)^(1/3))/6 - 35/3
- (35*2^(1/3)*(- 121 + 1077^(1/2)*3i)^(1/3))/12 - (35*2^(1/3)*(1 + 3^(1/2)*1i)*(- 121 - 1077^(1/2)*3i)^(1/3))/12 + (2^(1/3)*3^(1/2)*(- 121 + 1077^(1/2)*3i)^(1/3)*35i)/12 - 35/3
- (35*2^(1/3)*(- 121 + 1077^(1/2)*3i)^(1/3))/12 + (35*2^(1/3)*(- 1 + 3^(1/2)*1i)*(- 121 - 1077^(1/2)*3i)^(1/3))/12 - (2^(1/3)*3^(1/2)*(- 121 + 1077^(1/2)*3i)^(1/3)*35i)/12 - 35/3
>> imag(ans)
ans =
0
0
0
So you can pick off the real part for the answer.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!