number format using solve

Why when I use (solve) for quadratic equations I get This format numbers, Example:

solve(X^3+X-2==0)

ans =

1

- (7^(1/2)*1i)/2 - 1/2

(7^(1/2)*1i)/2 - 1/2

or sometimes the number will be so long

- (288230376151711744*889388876854208557621367400315380756942440049^(1/2))/1299318122813130077998372749837905 - 66276228678941179381385795875815555072/1299318122813130077998372749837905

(288230376151711744*889388876854208557621367400315380756942440049^(1/2))/1299318122813130077998372749837905 - 66276228678941179381385795875815555072/1299318122813130077998372749837905

Answer 1

[KSSV]

REad about double and vpasolve

Answer 2

[Walter Roberson]

Why not? Those are the answers to the question you are asking of MATLAB. If you put them back through the equations, you will find that they are correct.

The goal of solve is to find exact closed form solutions whenever possible, preferably algebraic numbers. The results you are getting back are algebraic numbers.

Perhaps you were expecting results closer to:

>> roots([1 0 1 -2])

ans =

                       -0.5 +       1.3228756555323i
                       -0.5 -       1.3228756555323i
                          1 +                     0i

However, those are not exact solutions, only approximations:

>> ans.^3 + ans - 2

ans =

      -1.55431223447522e-15 -  2.22044604925031e-16i
      -1.55431223447522e-15 +  2.22044604925031e-16i
       8.88178419700125e-16 +                     0i
       

Notice that back substituting does not give exact zeros -- because the numeric approximations are not exact solutions.

If you are looking for numeric solutions then use roots() or use vpasolve()

>> vpasolve(X^3+X-2==0)

ans =

1.0

- 0.5 - 1.3228756555322952952508078768196i

- 0.5 + 1.3228756555322952952508078768196i