Question

How do you solve `x^3+2x^2 = 2` ?

Answer 1

Use Cardano's method to find real root:

`x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))`

and related Complex roots.

Explanation:

First subtract `2` from both sides to get this equation into standard form:

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

`color(white)()`
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=2`, `c=0` and `d=-2`, so we find:

`Delta = 0+0+64-108+0 = -44`

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

`color(white)()`
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=27(x^3+2x^2-2)`

`color(white)(0)=27x^3+54x^2-54`

`color(white)(0)=(3x+2)^3-12(3x+2)-38`

`color(white)(0)=t^3-12t-38`

where `t=(3x+2)`

`color(white)()`
Cardano's method

We want to solve:

`t^3-12t-38=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-4)(u+v)-38=0`

Add the constraint `v=4/u` to eliminate the `(u+v)` term and get:

`u^3+64/u^3-38=0`

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

`(u^3)^2-38(u^3)+64=0`

Use the quadratic formula to find:

`u^3=(38+-sqrt((-38)^2-4(1)(64)))/(2*1)`

`=(38+-sqrt(1444-256))/2`

`=(38+-sqrt(1188))/2`

`=(38+-6sqrt(33))/2`

`=19+-3sqrt(33)`

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=root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33))`

and related Complex roots:

`t_2=omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33))`

`t_3=omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33))`

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

Now `x=1/3(-2+t)`. So the roots of our original cubic are:

`x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))`

`x_2 = 1/3(-2+omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33)))`

`x_3 = 1/3(-2+omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33)))`