Question

How do you find the solutions to the equation `x^4-6x^3+5x^2+30x-50=0`?

Answer 1

This quartic has roots: `3+-i` and `+-sqrt(5)`

Explanation:

`f(x) = x^4-6x^3+5x^2+30x-50`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-50` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-5, +-10, +-25, +-50`

None of these work. So `f(x)` has no rational zeros.

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Tschirnhaus transformation

First make a linear substitution called a Tschirnhaus transformation in order to simplify the quartic:

`16f(x) = 16x^4-96x^3+80x^2+480x-800`

`=(2x-3)^4-34(2x-3)^2+144(2x-3)-143`

`=t^4-34t^2+144t-143`

where `t = (2x-3)`

Since this quartic in `t` is monic and has no `t^3` term, it can be expressed as the product of two monic quadratics with opposite middle terms:

`t^4-34t^2+144t-143`

`=(t^2-ax+b)(t^2+ax+c)`

`=t^4+(b+c-a^2)t^2+(b-c)at+bc`

Equating coefficients and rearranging a little, we get:

`{ (b+c = a^2-34), (b-c = 144/a), (bc = -143) :}`

So we find:

`(a^2-34)^2 = (b+c)^2 = (b-c)^2+4bc = 20736/a^2-572`

Expanding, this becomes:

`(a^2)^2-68(a^2)+1156 = 20736/((a^2))-572`

Multiply through by `(a^2)` and rearrange a little to get:

`(a^2)^3-68(a^2)^2+1728(a^2)-20736=0`

By the rational roots theorem, the possible rational roots of this cubic are any factors of `20736 = 2^8*3^4`

That's rather a lot to try, but there are various methods to speed up the search and eventually find `a^2=36` is a zero.

Let `a=6`

Then:

`b = 1/2(a^2-34+144/a) = 1/2(36-34+24) = 13`

`c = 1/2(a^2-34-144/a) = 1/2(36-34-24) = -11`

Hmmm. We could have spotted that one fairly easily from the simultaneous equations by looking at possible factorings of `bc =-143`.

Well however we get there we now have two quadratics to solve:

`0 = t^2-6t+13 = (t-3)^2 + 2^2 = (t-3-2i)(t-3+2i)`

`0 = t^2+6t-11 = (t+3)^2-(2sqrt(5))^2 = (t+3-2sqrt(5))(t+3+2sqrt(5))`

Then `x=1/2(t+3)`

So we get solutions to our original quartic:

`x=1/2((3+-2i)+3)=3+-i`

`x=1/2((-3+-2sqrt(5))+3) = +-sqrt(5)`