Question

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

Answer 1

Use a trigonometric method to find roots:

`x_k = (4sqrt(3))/3 cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3) " "` for `k = 0, 1, 2.`

Explanation:

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`f(x) = x^3-4x-2`

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Descriminant

The discriminant `Delta` of a cubic polynomial 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`

In our example, `a=1`, `b=0`, `c=-4` and `d=-2`, so we find:

`Delta = 0+256+0-108+0 = 148`

Since `Delta > 0` this cubic has `3` Real zeros.

Trigonometric solution

Use a substitution of the form `x = k cos theta`, choosing `k` so that the resulting expression contains `4cos^3 theta - 3 cos theta = cos 3 theta`.

Let `k = (4sqrt(3))/3`

Then:

`0 = x^3-4x-2`

`= k^3 cos^3 theta - 4k cos theta - 2`

`= (64 sqrt(3))/9 cos^3 theta - (16 sqrt(3))/3 cos theta - 2`

`= (16 sqrt(3))/9 (4 cos^3 theta - 3 cos theta) - 2`

`= (16 sqrt(3))/9 cos 3 theta - 2`

Hence:

`cos 3 theta = 2* 9/(16 sqrt(3)) = (3 sqrt(3))/8`

So:

`3 theta = +-cos^(-1)((3 sqrt(3))/8) + 2kpi`

`theta = +-1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3`

`cos theta = cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3)`

Hence distinct roots:

`x_k = (4sqrt(3))/3 cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3) " "` for `k = 0, 1, 2.`