Consider
$$p(x) = x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a$$
First, we note that if $x< 0$, each term is negative, hence there are no negative roots. Also, $p(0) = -a < 0$. Further,
$$p'(x) = 5x^4-4(3-a)x^3+3(3-2a)x^2-2ax+2a$$
So it is sufficient to show that $p'(x) > 0$ for $x > 0, \; a \in (0, \frac12)$.
For this, note that by AM-GM inequality, $\frac12 ax^2+2a \ge 2ax$, so it is sufficient to show that:
$5x^4+\frac12(18-13a)x^2 > 4(3-a)x^3$. By AM-GM we again have:
$$5x^4+\tfrac12(18-13a)x^2 \ge 2\sqrt{\frac{5(18-13a)}2}x^3$$
So it is enough to show $5(18-13a) > 8(3-a)^2 \iff 8a^2+17a < 18$, which is true for $a \in (0, \frac12)$.
To answer this and make it answered, here is a statement.
Fact
A polynomial $p(x)=\sum_{0=1}^ma_ix^i$ has a nonzero root, i.e. there exists a nonzero complex root $y\neq0:p(y)=0$, iff there exist $i,j\in\{0,\dotsc,m\}$ with $i\neq j$ such that $a_i$ and $a_j$ are both nonzero.
Proof.
Suppose the condition holds. Suppose wlog $i<j$, and that $i$ is the least integer for which $a_i\neq0$. Then:
$$p(x)=x^i\cdot\sum_{k=i}^ma_kx^{k-i}.$$
For $k=i$, the term in the sum is $a_i\neq0$, so that polinomial doesn't have 0 as a root. Yet by the Fundamental Theorem of Algebra $\mathbb{C}$ is algebraically closed, and since the polinomial in that sum is nonconstant (as $a_j\neq0$ and $j\neq i$) it must have a complex root, in fact, it must have $m-i+1$ complex roots counted with multiplicity. Hence, $p$ has at least a nonzero root.
Suppose now that the condition does not hold, so for any two $i\neq j$ you have either $a_i=0$ or $a_j=0$. This means the polynomial is of the form:
$$p(x)=ax^k,$$
for some $a\in\mathbb{C},k\in\{0,\dotsc,m\}$. If $k=0$, then $p$ is constant and has no roots. Otherwise, the only root is 0.
$$\tag*{$\square$}$$
Best Answer
This is (a particular case of) Descartes' rule of signs
Since $x^n-f(x)$ has exactly one sign change, the number of positive real roots is either 1 or an odd number less than 1. This means it has exactly 1 positive root.
You can find a proof for example here, or many other places