# 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?

Nov 24, 2017

Here are a couple of methods...

#### Explanation:

Here are a couple of methods:

Descartes' Rule of Signs

Given:

$f \left(x\right) = {x}^{6} + {x}^{2} - 1$

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:

$f \left(- x\right) = f \left(x\right) = {x}^{6} + {x}^{2} - 1$

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

Turning points

Given:

$f \left(x\right) = {x}^{6} + {x}^{2} - 1$

Note that:

$f ' \left(x\right) = 6 {x}^{5} + 2 x = 2 x \left(3 {x}^{4} + 1\right)$

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

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

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