1 | initial version |

The output of `solve`

is a list of solutions in the form of equations `x == ...`

:

sage: sol = solve(x^3 - x == 3*x^2 - 1, x)

Get the value of each solution as the right-hand side ("rhs"):

sage: a, b, c = [s.rhs() for s in sol]

Or use `solution_dict=True`

to get solutions in dictionary form:

```
sage: sol = solve(x^3 - x == 3*x^2 - 1, x, solution_dict=True)
```

Then each solution `s`

is a dictionary of the form `{x: ...}`

and
one can get its value as `s[x]`

:

```
sage: a, b, c = [s[x] for s in sol]
```

In each case, we get:

```
sage: a
-1/2*(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3)*(I*sqrt(3) + 1) - 2/3*(-I*sqrt(3) + 1)/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
sage: b
-1/2*(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3)*(-I*sqrt(3) + 1) - 2/3*(I*sqrt(3) + 1)/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
sage: c
(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 4/3/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
```

so it looks as though the solutions are non-real, since they involve `I`

.

This is because there are well-known formulas to express solutions of cubic equations using radicals, sometimes involving square roots of negative numbers even when the final result is real.

Indeed, the results are real here, as we can check with the `numerical_approx`

method
(`n`

for short`):

```
sage: a.n()
0.460811127189111
sage: b.n()
-0.675130870566646 - 1.11022302462516e-16*I
sage: c.n()
3.21431974337754 - 5.55111512312578e-17*I
```

The numerical approximation is imperfect, and leaves a tiny imaginary component which however is only there due to rounding errors.

To get the results directly as exact algebraic numbers, and see directly
whether or not they are real, and what their decimal expansion looks like,
work with polynomials and their `roots`

method rather than with the symbolic ring
and the `solve`

function.

```
sage: x = polygen(ZZ)
sage: p = x^3 - x
sage: q = p - p.derivative()
sage: q
x^3 - 3*x^2 - x + 1
```

The method `roots`

will show `(root, multiplicity)`

pairs:

```
sage: q.roots(AA)
[(-0.6751308705666461?, 1), (0.4608111271891109?, 1), (3.214319743377535?, 1)]
```

unless one specifies `multiplicities=False`

:

```
sage: q.roots(AA, multiplicities=False)
[-0.6751308705666461?, 0.4608111271891109?, 3.214319743377535?]
```

2 | No.2 Revision |

The output of `solve`

is a list of solutions in the form of equations `x == ...`

:

`sage: sol = solve(x^3 - x == 3*x^2 - 1, `~~x)~~x)

Get the value of each solution as the right-hand side ("rhs"):

`sage: a, b, c = [s.rhs() for s in `~~sol]~~sol]

Or use `solution_dict=True`

to get solutions in dictionary form:

```
sage: sol = solve(x^3 - x == 3*x^2 - 1, x, solution_dict=True)
```

Then each solution `s`

is a dictionary of the form `{x: ...}`

and
one can get its value as `s[x]`

:

```
sage: a, b, c = [s[x] for s in sol]
```

In each case, we get:

```
sage: a
-1/2*(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3)*(I*sqrt(3) + 1) - 2/3*(-I*sqrt(3) + 1)/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
sage: b
-1/2*(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3)*(-I*sqrt(3) + 1) - 2/3*(I*sqrt(3) + 1)/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
sage: c
(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 4/3/(1/9*I*sqrt(37)*sqrt(3) + 1)^(1/3) + 1
```

so it looks as though the solutions are non-real, since they involve `I`

.

This is because there are well-known formulas to express solutions of cubic equations using radicals, sometimes involving square roots of negative numbers even when the final result is real.

Indeed, the results are real here, as we can check with the `numerical_approx`

method
(`n`

for short`):

```
sage: a.n()
0.460811127189111
sage: b.n()
-0.675130870566646 - 1.11022302462516e-16*I
sage: c.n()
3.21431974337754 - 5.55111512312578e-17*I
```

The numerical approximation is imperfect, and leaves a tiny imaginary component
(`e-16`

means `*10^-16`

) which however is only there due to rounding errors.

To get the results directly as exact algebraic numbers, and see directly
whether or not they are real, and what their decimal expansion looks like,
work with polynomials and their `roots`

method rather than with the symbolic ring
and the `solve`

function.

```
sage: x = polygen(ZZ)
sage: p = x^3 - x
sage: q = p - p.derivative()
sage: q
x^3 - 3*x^2 - x + 1
```

The method `roots`

will show `(root, multiplicity)`

pairs:

```
sage: q.roots(AA)
[(-0.6751308705666461?, 1), (0.4608111271891109?, 1), (3.214319743377535?, 1)]
```

unless one specifies `multiplicities=False`

:

```
sage: q.roots(AA, multiplicities=False)
[-0.6751308705666461?, 0.4608111271891109?, 3.214319743377535?]
```

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