How to prove that the quartic equation $x^4-3x+1=0$ has exactly 2 real roots? So I have this quartic equation here:
$x^4-3x+1=0$
I'm supposed to prove this equation has exactly 2 roots.
I defined $f(x)=x^4-3x+1=0$
Then I used the Intermediate value theorem at $f(0)$ and $f(1)$ to show it goes from $(0,1)$ to $(1,-1)$ so it must have crossed the x-axis since this polynomial is continuous. So there is at least 1 root.
The derivative is $f'(x)=4x^3-3$ and the second derivative is $f''(x)=12x^2$. 
There is a critical point at $x=(3/4)^{\frac{1}{3}}$ where the function changes from increasing from $(-\infty,(3/4)^{\frac{1}{3}}]$ and decreasing from $[(3/4)^{\frac{1}{3}},\infty)$.
There is a point of inflection at $x=0$ where the concavity changes from concave up to concave down.
I'm not really sure how to use this information to determine how to show there is a second root though...
No I can't use Descartes's rule of signs. It would be easy otherwise.

The Sturm Chain for $x^4-3x+1$ is
$$
\left\{x^4-3x+1,\,\,4x^3-3,\,\,\tfrac94x-1,\,\,\tfrac{1931}{729}\right\}
$$
There are $2$ sign changes at $-\infty$: $\{+,-,-,+\}$.
There are $0$ sign changes at $+\infty$: $\{+,+,+,+\}$.
This means there are $2$ real roots.

In fact, there are $2$ sign changes at $0$: $\{+,-,-,+\}$.
This means that there are no negative roots and $2$ positive roots.
There is $1$ sign change at $1$: $\{-,+,+,+\}$.
There are $0$ sign changes at $2$: $\{+,+,+,+\}$.
Thus, one of the roots is between $0$ and $1$, and one is between $1$ and $2$.
