Question

How do you find all the real and complex roots of `f(x)=x^3+x-3`?

Answer 1

Use Cardano's method to find the Real zero:

`x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))`

and related Complex zeros.

Explanation:

The discriminant `Delta` of a general cubic in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

So in our example, with `a=1`, `b=0`, `c=1` and `d=-3`, we find:

`Delta = 0-4-0-243+0 = -247 < 0`

Since `Delta < 0` this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

It can be solved using Cardano's method.

Let `x = u+v`

Then:

`u^3+v^3+(3uv+1)(u+v)-3 = 0`

Add the constraint `v = -1/(3u)` to eliminate the term in `(u+v)` and get:

`u^3-1/(27u^3)-3 = 0`

Multiply through by `27u^3` and rearrange a little to get:

`27(u^3)^2-81(u^3)-1 = 0`

Then use the quadratic formula to find:

`u^3 = (81+-sqrt(81^2-(4*27*-1)))/(2*27)`

`=(81+-sqrt(6561+108))/54`

`=(81+-sqrt(6669))/54`

`=(81+-3sqrt(741))/54`

Since this is Real and the derivation was symmetric in `u` and `v`m we can use one of these roots for `u^3` and the other for `v^3` to find the Real zero:

`x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))`

and Complex zeros:

`x_2 = 1/3(omega root(3)((81+3sqrt(741))/2)+omega^2 root(3)((81-3sqrt(741))/2))`

`x_3 = 1/3(omega^2 root(3)((81+3sqrt(741))/2)+omega root(3)((81-3sqrt(741))/2))`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.