Question

What are the real roots of `x^8+x^3=1` ?

Answer 1

Use a numerical method to find real zeros:

`x_1 ~~ -1.1148`

`x_2 ~~ 0.872495`

Explanation:

Given:

`x^8+x^3=1`

We can write this in standard form as:

`x^8+x^3-1 = 0`

Let:

`f(x) = x^8+x^3-1`

Note that the pattern of signs of the coefficients is `+ + -`. With one change of sign, Descartes' Rule of Signs tells us that this octic polynomial has exactly one positive real zero.

The pattern of signs of coefficients of `f(-x)` is `+ - -`. With one change of sign, we can deduce that `f(x)` has one negative real zero.

So `f(x)` has two real zeros and must have `6` non-real complex zeros (counting multiplicity).

Note that:

`{ (f(-2) = 128-8-1 = 119), (f(-1) = 1-1-1 = -1), (f(0) = 0+0-1 = -1), (f(1) = 1 + 1 - 1 = 1) :}`

So the negative zero is somewhere in `(-2, -1)` (probably quite close to `-1`) and the positive zero is somewhere in `(0, 1)`.

They are both irrational numbers, inexpressible in terms of `n`th roots or standard functions.

About the best we can do is find numerical approximations using Durand-Kerner to find all `8` zeros at once, or Newton-Raphson to find them one at a time.

If `a_i` is an approximation to a zero of `f(x)`, then a better approximation is:

`a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i-(a_i^8+a_i^3-1)/(8a_i^7+3a_i^2)`

We can put this formula into a spreadsheet with `a_0 = -1` to find the negative zero, or `a_0 = 1/2` to find the positive zero.

Hence we find approximate zeros:

`x_1 ~~ -1.1148`

`x_2 ~~ 0.872495`