= Solution
Model the specified drag by a prescribed head loss, $(E+H)_x=-\gamma$. A modified conserved head is consequently
$$
\boxed{\mathcal E_\gamma=H+h+\frac{Q^2}{2gb^2h^2}+\gamma x,}
$$
with an arbitrary reference constant absorbed into $\mathcal E_\gamma$. More generally the last term is $\int_0^x\gamma(s)\,ds$. The effective bed elevation for the <shallow-water specific energy> calculation is $H+\gamma x$. The critical momentum equation becomes
$$
(1-F^2)h_x=F^2h\frac{b_x}b-H_x-\gamma.
$$
This is the <head-loss correction to hydraulic control>.
When $\beta=0$, $b=1$, and regularity at $F=1$ gives $-2\eta x_c+\gamma=0$. Therefore, for $\eta,\gamma>0$,
$$
\boxed{x_c=\frac{\gamma}{2\eta}.}
$$
The effective bed $-\eta x^2+\gamma x$ has a strict maximum there, with elevation increment $\Delta=\gamma^2/(4\eta)$ relative to $x=0$.
Let $d=h(0)$ and require $h_c=2d$. At the control $Q^2=g h_c^3=8gd^3$. At $x=0$ the <shallow-water specific energy> is $d+Q^2/(2gd^2)=5d$, while at the control it is $3h_c/2=3d$. Equality of the modified head gives $5d=3d+\Delta$. Hence
$$
\boxed{h(0)=\frac{\gamma^2}{8\eta},\qquad h_c=\frac{\gamma^2}{4\eta},\qquad
Q=\sqrt g\left(\frac{\gamma^2}{4\eta}\right)^{3/2}=\frac{\sqrt g\,\gamma^3}{8\eta^{3/2}}.}
$$
The inlet has $F(0)^2=8$ and is <supercritical flow>; the branch whose depth increases into the control makes the required smooth transition. At the control the increasing branch has slope $h_x=\sqrt{2\eta h_c/3}$, as follows by expanding the modified head to second order.
The numerical discharge uses a constant loss rate on the entire segment from $0$ to $x_c$. The wording specifies that rate only near the control, which is sufficient to locate $x_c$ but does not by itself determine the inlet-to-control head loss. If the loss elsewhere is unspecified, write $\Lambda_c=\int_0^{x_c}\gamma(s)\,ds$. The same calculation gives $h_c=H(x_c)-H(0)+\Lambda_c$ and $Q=\sqrt g\,h_c^{3/2}$; the boxed value is the constant-loss specialization.
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