Question

Show that the equation `x^6+x^2-1=0` has exactly one positive root. Justify your response. Name the theorems on which your response depends and the properties of `f(x)` that you must use?

Answer 1

Here are a couple of methods...

Explanation:

Here are a couple of methods:

Descartes' Rule of Signs

Given:

`f(x) = x^6+x^2-1`

The coefficients of this sextic polynomial have signs in the pattern `+ + -`. Since there is one change of signs, Descartes' Rule of Signs tells us that this equation has exactly one positive zero.

We also find:

`f(-x) = f(x) = x^6+x^2-1`

which has the same pattern of signs `+ + -`. Hence `f(x)` has exactly one negative zero too.

Turning points

Given:

`f(x) = x^6+x^2-1`

Note that:

`f'(x) = 6x^5+2x = 2x(3x^4+1)`

which has exactly one real zero, of multiplicity `1`, namely at `x=0`

Since the leading term of `f(x)` has positive coefficient, that means that `f(x)` has a minimum at `x=0` and no other turning points.

We find `f(0) = -1`. So `f(x)` has exactly two zeros, either side of the minimum.