Question

`alpha, beta` are real numbers such that `alpha^3-3alpha^2+5 alpha - 17=0` and `beta^3-3beta^2+5beta+11 = 0`. What is the value of `alpha+beta` ?

Answer 1

`alpha+beta=2`

Explanation:

First use Tschirnhaus transformations to simplify the cubics:

`0 = alpha^3-3alpha^2+5alpha-17 = gamma^3+2gamma-14`

`0 = beta^3-3beta^2+5beta+11 = delta^3+2delta+14`

where `gamma = alpha-1` and `delta = beta-1`

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`

We can check the discriminant of these cubics, finding that they are both negative, so they have `1` Real root and `2` Complex roots each.

By Descartes' rule of signs, the Real roots result in `gamma > 0` and `delta < 0`. We use this later.

Adding these two equations, we get:

`0 = gamma^3+delta^3+2gamma+2delta`

`color(white)(0) = (gamma+delta)(gamma^2-gamma delta+delta^2)+2(gamma+delta)`

`color(white)(0) = (gamma+delta)(gamma^2-gamma delta+delta^2+2)`

So either `gamma+delta = 0` or `gamma^2-gamma delta+delta^2+2 = 0`

Note however that `gamma*delta < 0`, hence `gamma^2-gamma delta+delta^2 + 2 > 0` for any Real values of `gamma` and `delta`.

So the only Real solution gives us `gamma+delta = 0` and hence `alpha+beta = 2`