Approximation of a root of a polynomial of degree k I know that solving every polynomial of degree higher than 5 is not possible via a closed formula. However, how is it possible to find an approximation for $x_{k, a}$ the biggest positive root of
$$P(x) = x (x-1)^k - (x-a) x^k$$
where $a, k \in \mathbb{N}$  and $1 \leq a \leq k$ ?
For example, when $k=15$ and $a=11$, one has to find a solution of 
P:=-4*x^15+105*x^14-455*x^13+1365*x^12-3003*x^11+5005*x^10-6435*x^9+6435*x^8-5005*x^7+3003*x^6-1365*x^5+455*x^4-105*x^3+15*x^2-x

for which the real roots are $0$, $0.4475$ and $x_{15,11} = 21.6566$.
Is it possible to find an approximation for $x_{k,a}$ (which can involve a $O(1/k)$ term or such asymptotic error terms)?

This is linked to the problem here.
 A: I'll find approximations (without rigorously analyzing the error) in three large-$k$ regimes: $a$ is much smaller than $k$ ($2 \leq a \ll k$), $a$ is in the midrange between $1$ and $k$ ($a \propto k$), and $a$ is very close to $k$ ($a = k-b$ with $1 \leq b \ll k$).

The regime where $a$ is much smaller than $k$.
Here we'll assume that $2 \leq a \ll k$.
Numerical experiments suggest that if $a$ is fixed and $k \to \infty$, then $x \to a^+$. Let's then suppose that $x = a + \epsilon$ and $0 < \epsilon \ll a$.
Substituting this into the equation $x(x-1)^k - (x-a)x^k = 0$ gives
$$
(a+\epsilon)(a-1+\epsilon)^k - \epsilon (a+\epsilon)^k = 0.
$$
Now we apply the method of dominant balance. The first balance is achieved by replacing $a + \epsilon$ in the first term with $a$, and so we get
$$
a(a-1+\epsilon)^k - \epsilon (a+\epsilon)^k \approx 0,
$$
or, after rearranging,
$$
(a-1+\epsilon)^k \approx \frac{\epsilon}{a} (a+\epsilon)^k.
$$
Now take logarithms to get
$$
k\log(a-1+\epsilon) \approx \log \frac{\epsilon}{a} + k\log(a+\epsilon). \tag{1}
$$
Approximating the logarithms by
$$
\log(a-1+\epsilon) \approx \log(a-1) + \frac{\epsilon}{a-1}
$$
and
$$
\log(a+\epsilon) \approx \log a + \frac{\epsilon}{a},
$$
$(1)$ becomes
$$
k \log \frac{a-1}{a} \approx \log \frac{\epsilon}{a} + \frac{k\epsilon}{a(1-a)}. \tag{2}
$$
The dominant balance is achieved by ignoring the last term on the right (no other balance satisfies $0 < \epsilon \ll a$), so that
$$
k \log \frac{a-1}{a} \approx \log \frac{\epsilon}{a},
$$
from which we get
$$
\epsilon \approx a\left(\frac{a-1}{a}\right)^k.
$$
In summary:

If $2 \leq a \ll k$ then we expect
  $$
x \approx a + a\left(\frac{a-1}{a}\right)^k.
$$


The regime where $a$ is in the midrange between $1$ and $k$.
Here we'll assume that $a \propto k$, i.e. that $\lim_{k \to \infty} a/k = c$, where $0 < c < 1$ is a constant.
With the assumption $a \approx ck$, our equation $x(x-1)^k - (x-a)x^k = 0$ becomes
$$
x(x-1)^k \approx (x-ck) x^k.
$$
Numerical experiments suggest $x \propto k$, so let's set $x = ky$ and solve for $y$ under the assumption $y = \Theta(1)$. This gives
$$
ky (ky-1)^k \approx (ky-ck) (ky)^k,
$$
and after a little rearranging
$$
y\left(1-\frac{1}{ky}\right)^k \approx y-c.
$$
But $(1-1/(ky))^k \approx e^{-1/y}$, so we really have
$$
ye^{-1/y} \approx y-c. \tag{3}
$$
Unfortunately this equation can only be solved numerically for $y$ given some $c$.
For your example $k=15$ and $a=11$ we have $c = a/k = 11/15$, and numerically we get $y \approx 1.52428$ from $(3)$. So, our approximation is
$$
x = ky \approx 22.86415.
$$
This is not far from the true answer of $x \doteq 21.65661$.
A little more analysis suggests the following improvement:

If $a \propto k$ and $k \gg 1$, then
  $$
x = ky + \frac{1}{2(1+y-y e^{1/y})} + O\!\left(k^{-1}\right) \qquad \text{as } k \to \infty,
$$
  where
  $$
ye^{-1/y} = y - \frac{a}{k}.
$$

When $k=15$ and $a=11$ this yields the improved approximation $x \approx 21.65424$.

The regime where $a$ is very close to $k$.
Here we'll assume that $a = k - b$ with $1 \leq b \ll k$.
Numerical experiments suggest that $x \gg k$ as $k \to \infty$, so we'll assume this for the analysis.
By substituting $a = k-b$ into $x(x-1)^k - (x-a)x^k = 0$ and rearranging, we get
$$
\left(1-\frac{1}{x}\right)^k = 1 - \frac{k}{x} + \frac{b}{x}. \tag{4}
$$
The left-hand side can be expanded as
$$
\left(1-\frac{1}{x}\right)^k = 1 - \frac{k}{x} - \frac{k}{2x^2} + \frac{k^2}{2x^2} + O\!\left((k/x)^3\right),
$$
and substituting this into $(4)$ yields
$$
-\frac{k}{2x^2} + \frac{k^2}{2x^2} \approx \frac{b}{x}.
$$
The second term on the left-hand side dominates the first, so the dominant balance is
$$
\frac{k^2}{2x^2} \approx \frac{b}{x},
$$
and so we get the estimate
$$
x \approx \frac{k^2}{2b}.
$$
A little more analysis suggests the following improvement:

If $a = k-b$ where $1 \leq b \ll k$, then
  $$
x \approx \frac{1}{2b}k^2 - \frac{2b+3}{6b} k + \frac{12-b}{18}.
$$
  If $b$ is fixed then the error of this approximation is $O(k^{-1})$.

Further terms in this series can be computed fairly easily using a CAS, and I would expect that the truncation error at any point in the tail of this series is $o(b)$ as $k \to \infty$.
For $k=15$ and $a = 11$ this yields the approximation $x \approx 21.69444$.
