FFedor Petrov showed that, for natural $n$, the equation \begin{equation*} x^{n+1}=(1-x)^n(n+x) \tag{10}\label{10} \end{equation*} has exactly one nonnegative root $x=x_n$, and $x_n\in(0,1)$.

For $x\in(0,1)$, rewriting \eqref{10} as \begin{equation*} \Big(\frac x{1-x}\Big)^n=1+\frac nx, \tag{20}\label{20} \end{equation*} we see that, for each real $h>0$, as $n\to\infty$,

(i) uniformly in $x\in(1/2+h,1)$, the left-hand side of \eqref{20} goes exponentially fast to $\infty$ whereas the right-hand side of \eqref{20} is $O(n)$.

(ii) uniformly in $x\in(0,1/2-h)$, the left-hand side of \eqref{20} goes exponentially fast to $0$ whereas the right-hand side of \eqref{20} is $>1$.

So, for the only nonnegative root $x=x_n$ of \eqref{10} we have \begin{equation*} x_n\to1/2 \end{equation*} (as $n\to\infty$).

Let us consider the possible negative roots $x$ of \eqref{10}. Here we have to distinguish two cases, whether $n$ is even or odd.

Consider first the case when $n$ is even. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} g_n(u):=n\ln\frac u{u+1}-\ln(1-n/u) \tag{30}\label{30} \end{equation*} for $u:=-x>n$. We have \begin{equation*} g'_n(u)=-\frac{n(n+1)}{u(u+1)(u-n)}<0 \end{equation*} and $g_n(u)\to0$ as $u\to\infty$. So, $g_n(u)>0$ for all $u>n$, and hence \eqref{10} has no negative roots $x$ if $n$ is even.

Consider finally the case when $n$ is odd. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} h_n(c):=(c-1)(1+c/n)^n=1\tag{40}\label{40} \end{equation*} for $c:=n/u>1$ and $u:=-x\in(0,n)$. Note that $h_n(c)$ is continuously and strictly increasing in $c>1$ from $0$ to $\infty$. So, equation \eqref{40} has a unique root $c=c_n>1$. Since $(1+c/n)^n\to e^c$, we see that \begin{equation} c_n\to c_\infty:=W(1/e)+1=1.278\ldots, \end{equation} where $c_\infty$ is the unique root $c>1$ of the equation $(c-1)e^c=1$ and $W$ is the Lambert function. So, when $n$ is odd, equation \eqref{10} has exactly one negative root $x=y_n$, and $y_n\sim -n/c_\infty$.

Fedor Petrov showed that, for natural $n$, the equation \begin{equation*} x^{n+1}=(1-x)^n(n+x) \tag{10}\label{10} \end{equation*} has exactly one nonnegative root $x=x_n$, and $x_n\in(0,1)$.

For $x\in(0,1)$, rewriting \eqref{10} as \begin{equation*} \Big(\frac x{1-x}\Big)^n=1+\frac nx, \tag{20}\label{20} \end{equation*} we see that, for each real $h>0$, as $n\to\infty$,

(i) uniformly in $x\in(1/2+h,1)$, the left-hand side of \eqref{20} goes exponentially fast to $\infty$ whereas the right-hand side of \eqref{20} is $O(n)$.

(ii) uniformly in $x\in(0,1/2-h)$, the left-hand side of \eqref{20} goes exponentially fast to $0$ whereas the right-hand side of \eqref{20} is $>1$.

So, for the only nonnegative root $x=x_n$ of \eqref{10} we have \begin{equation*} x_n\to1/2 \end{equation*} (as $n\to\infty$).

(As Emil Jeřábek noted, we also have $x_n>1/2$ for all $n$.)

Let us consider the possible negative roots $x$ of \eqref{10}. Here we have to distinguish two cases, whether $n$ is even or odd.

Consider first the case when $n$ is even. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} g_n(u):=n\ln\frac u{u+1}-\ln(1-n/u) \tag{30}\label{30} \end{equation*} for $u:=-x>n$. We have \begin{equation*} g'_n(u)=-\frac{n(n+1)}{u(u+1)(u-n)}<0 \end{equation*} and $g_n(u)\to0$ as $u\to\infty$. So, $g_n(u)>0$ for all $u>n$, and hence \eqref{10} has no negative roots $x$ if $n$ is even.

Consider finally the case when $n$ is odd. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} h_n(c):=(c-1)(1+c/n)^n=1\tag{40}\label{40} \end{equation*} for $c:=n/u>1$ and $u:=-x\in(0,n)$. Note that $h_n(c)$ is continuously and strictly increasing in $c>1$ from $0$ to $\infty$. So, equation \eqref{40} has a unique root $c=c_n>1$. Since $(1+c/n)^n\to e^c$, we see that \begin{equation} c_n\to c_\infty:=W(1/e)+1=1.278\ldots, \end{equation} where $c_\infty$ is the unique root $c>1$ of the equation $(c-1)e^c=1$ and $W$ is the Lambert function. So, when $n$ is odd, equation \eqref{10} has exactly one negative root $x=y_n$, and $y_n\sim -n/c_\infty$.