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?

1 Answer
Nov 24, 2017

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.