# 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

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

We also find:

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

which has the same pattern of signs

**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

Since the leading term of

We find