Question

How to solve this inequation ?

`x^4-x^3-x>0`

Answer 1

`x<0` and `x>~1.4656`

Explanation:

We need to find out when the polynomial is positive. To do this, we need to find the zeroes of the polynomial, since that is where it could potentially cross the `x`-axis and go from positive to negative or vice-versa.

We can quite quickly find that `x=0` is a zero by factoring out an `x`:

`x(x^3-x^2-1)=0`

`x_1=0`

The remaining cubic doesn't have any rational zeroes, so I'm just going to use an approximation:

`x_2~~1.4656`

This means that we need to look at the intervals `(-oo,0)`, `(0,1.4656)` and `(1.4656,oo)` and determine if the function is possible or negative in those intervals. To do this, we can just pick a point anywhere in those intervals:

`(-oo,0) -> x=-1 -> (-1)^4-(-1)^3-(-1)=3->` Positive

`(0,1.4656) -> x=1 -> 1^4-1^3-1=-1->` Negative

`(1.4656,oo) -> x=2 -> 2^4-2^3-2=6->` Positive

This means that the inequality holds when:
`x<0` and `x>~1.4656`

If we were interested in the exact answer, we could also use one of the methods for solving cubic equations to get the exact value for the `1.4656` root:
`x<0` and `x>root(3)(29/54+sqrt93/18)+root(3)(29/54-sqrt93/18)+1/3`