Question

How do you solve `0 = 4x^3 + 2x-1`?

Answer 1

Check there are no rational roots, then use Cardano's method to find the Real and two Complex roots.

Explanation:

Let `f(x) = 4x^3+2x-1`

By the rational root theorem, any roots of `4x^3+2x-1 = 0` will be expressible in the form `p/q` for integers `p` and `q` where `p` is a divisor of the constant term `-1` and `q` a divisor of the coefficient `4` of the leading term.

So the only possible rational roots are:

`+-1/4`, `+-1/2`, `+-1`

We find `f(x) != 0` for all of these values, so `f(x) = 0` has no rational roots.

Use Cardano's method, substituting `x = u + v` to get:

`4u^3 + 4v^3 + 2(6uv+1)(u+v) - 1 = 0`

Add the constraint `v = -1/(6u)` to eliminate the term in `(u+v)` to get:

`0 = 4u^3 - 4/(6u)^3 - 1 = 4u^3 - 1/(54u^3) - 1`

Multiply through by `54u^3` to get:

`216(u^3)^2-54(u^3)-1 = 0`

Use the quadratic formula to find:

`u^3 = (54+-sqrt(54^2+4*216))/(2*216)`

`=(54+-sqrt(3780))/432`

`=(54+-6sqrt(105))/432`

`=(27+-3sqrt(105))/216`

Due to the symmetry of the derivation in `u` and `v`, we can take one of these roots as `u^3` and the other as `v^3` to find the Real root:

`x_1 = 1/6(root(3)(27+3sqrt(105))+root(3)(27-3sqrt(105)))`

and Complex roots:

`x_2 = 1/6(omega root(3)(27+3sqrt(105))+omega^2 root(3)(27-3sqrt(105)))`

`x_3 = 1/6(omega^2 root(3)(27+3sqrt(105))+omega root(3)(27-3sqrt(105)))`

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