Question

What are the roots of `x^3+5x^2-29x-129 = 0` ?

Answer 1

This has roots:

`x_n = 1/3(-5+8sqrt(7)cos(1/3cos^(-1)((241sqrt(7))/784)+(2npi)/3))" "n = 0, 1, 2`

Explanation:

`f(x) = x^3+5x^2-29x-129`

Discriminant

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=5`, `c=-29` and `d=-129`, so we find:

`Delta = 21025+97556+64500-449307+336690 = 70464`

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

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

`0=27f(x)=27(x^3+5x^2-29x-129)`

`=27x^3+(27*5)x^2-(27*29)x-(27*129)`

`=27x^3+135x^2-783x-3483`

`=(27x^3+135x^2+225x+125)-(1008x+1680)-1928`

`=(3x+5)^3-336(3x+5)-1928`

`=t^3-336t-1928`

where `t=(3x+5)`

Trigonometric substitution

Since this cubic has `3` real zeros, Cardano's method will encounter its casus irreducibilis, resulting in expressions in terms of irreducible cube roots of complex numbers.

So I prefer to express in terms of real trigonometric functions.

Let:

`t = k cos theta`

Then:

`t^3-336t-1928 = (kcos theta)^3-336(kcos theta)-1928`

Putting `k = sqrt(4/3*336) = sqrt(448) = 8sqrt(7)`

this becomes:

`t^3-336t-1928 = 896sqrt(7)(4cos^3 theta-3cos theta)-1928`

`color(white)(t^3-336t-1928) = 896sqrt(7)cos 3theta-1928`

which is `0` when:

`cos 3 theta = 1928/(896sqrt(7)) = 241/(112sqrt(7)) = (241sqrt(7))/784`

So:

`3 theta = +-cos^(-1)((241sqrt(7))/784)+2npi`

So:

`theta = +-1/3cos^(-1)((241sqrt(7))/784)+(2npi)/3`

Hence distinct roots:

`t_n = 8sqrt(7)cos(1/3cos^(-1)((241sqrt(7))/784)+(2npi)/3)" "n = 0, 1, 2`

So distinct roots of our original cubic:

`x_n = 1/3(-5+8sqrt(7)cos(1/3cos^(-1)((241sqrt(7))/784)+(2npi)/3))" "n = 0, 1, 2`