Does $f{(x)} = a x^4 - x^3 + ax - a$ have exactly $1$ positive root given $a>0$? Let $$f{(x)} = a x^4 - x^3 + ax - a$$ where $a>0$.
We know that it contains $4$ roots. Also, by the Descartes' rule of signs, we know that


*

*There are either $1$ or $3$ positive roots.

*There is exactly $1$ negative root.


I observed from plotting the function that it may contain exactly one positive root.
Therefore, can we prove/disprove that the function always contain a pair of conjugate complex roots?
Equivalently, can we prove/disprove that it has exactly $1$ positive root?

Let $f(x)=0$.
Thus, $$a=\frac{x^3}{x^4+x-1}.$$
Now, by the  Descartes' rule of signs we see that $x^4+x-1$ has an unique positive root $x_0$.
Thus, for $x>x_0$ we have $\frac{x^3}{x^4+x-1}>0$ and for $0<x<x_0$ we have $\frac{x^3}{x^4+x-1}<0$.
Now, $$\left(\frac{x^3}{x^4+x-1}\right)'=-\frac{x^2(x^4-2x+3)}{(x^4+x-1)^2}<0$$ for any $x>x_0$,
$$\lim_{x\rightarrow x_0^+}\frac{x^3}{x^4+x-1}=+\infty,$$
$$\lim_{x\rightarrow+\infty}\frac{x^3}{x^4+x-1}=0,$$ and $\frac{x^3}{x^4+x-1}$ is a continuous on $(x_0,+\infty),$ which says that the equation 
$$a=\frac{x^3}{x^4+x-1}$$ or $$f(x)=0$$ has an unique positive root.
By the same way we see that the equation $f(x)=0$ has an unique negative root.
