First show $k=2$, otherwise $u^k(x)-a$ would have a complex root. Next, for $k=2$, the problem reduces to $p(x) - q(x) = 2\sqrt{a}$ where $p(x) \cdot q(x) = x(x+1)(x+2)(x+4)f(x)$. Without loss assume $x+4$ divides $p(x)$. If $x+2$ also divides $p(x)$, then we have $q(-4) = q(-2)$. But obviously the former has larger absolute value (because roots of $f$ are positive), contradiction. Thus $x+2$ divides $q(x)$. If $x+1$ divides $p(x)$, then $q(-4) = q(-1)$, which is not possible because they have different signs. Thus $x+1$ divides $q(x)$. If $x$ divides $p(x)$, then $q(-4) = q(0)$, again impossible because the former has larger absolute value.
We are left with the case where $x+4$ divides $p(x)$ while $x(x+1)(x+2)$ divides $q(x)$. Hence $p(0) = p(-1) = p(-2)$. But by AM-GM, $p(-1)^2 > p(-2)p(0)$, contradiction.