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.