Question

What are the roots of `x^3+52x^2+1060x-4624 = 0`?

Answer 1

The real root is:

`x = 2/3(-26+root(3)(21232+3sqrt(50275887))+root(3)(21232-3sqrt(50275887)))`

There are two related complex roots.

Explanation:

`f(x) = x^3+52x^2+1060x-4624`

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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=52`, `c=1060` and `d=-4624`, so we find:

`Delta = 3038214400-4764064000+2600685568-577297152-4587747840 = -4290209024`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

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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)=27x^3+1404x^2+28620x-124848`

`=(3x+52)^3+1428(3x+52)-339712`

`=t^3+1428t-339712`

where `t=(3x+52)`

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Cardano's method

We want to solve:

`t^3+1428t-339712=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv+476)(u+v)-339712=0`

Add the constraint `v=-476/u` to eliminate the `(u+v)` term and get:

`u^3-107850176/u^3-339712=0`

Multiply through by `u^3` and rearrange slightly to get:

`(u^3)^2-339712(u^3)-107850176=0`

Use the quadratic formula to find:

`u^3=(339712+-sqrt((-339712)^2-4(1)(-107850176)))/(2*1)`

`=(339712+-sqrt(115404242944+431400704))/2`

`=(339712+-sqrt(115835643648))/2`

`=(339712+-48sqrt(115835643648))/2`

`=169856+-24sqrt(115835643648)`

`=8(21232+-3sqrt(50275887))`

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

`t_1=2root(3)(21232+3sqrt(50275887))+2root(3)(21232-3sqrt(50275887))`

and related Complex roots:

`t_2=2 omega root(3)(21232+3sqrt(50275887))+2 omega^2 root(3)(21232-3sqrt(50275887))`

`t_3=2 omega^2 root(3)(21232+3sqrt(50275887))+2 omega root(3)(21232-3sqrt(50275887))`

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

Now `x=1/3(-52+t)`. So the roots of our original cubic are:

`x_1 = 2/3(-26+root(3)(21232+3sqrt(50275887))+root(3)(21232-3sqrt(50275887)))`

`x_2 = 2/3(-26+omega root(3)(21232+3sqrt(50275887))+omega^2 root(3)(21232-3sqrt(50275887)))`

`x_3 = 2/3(-26+omega^2 root(3)(21232+3sqrt(50275887))+omega root(3)(21232-3sqrt(50275887)))`