Solving a constrained optimisation problem where the algebra doesn't make sense

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I have the following function to maximise with respect to $x$, $p$, and $\eta$:

$\frac{p\eta}{R(x)}[v(x)-\bar\theta R(x)]+(1-\frac{p\eta}{R(x)})p\eta-(1+\lambda)tpx-\lambda R(x) (1)$

subject to the following constraints:

$0\leq p \leq 1$

$\eta, x \geq 0 $

So my instinct with this problem is to just take first order conditions with respect to $p$, $x$, $\eta$. I want to focus on the conditions with respect to $p$ and $\eta$ for a second.

With respect to $p$, we get:

$\frac{\eta}{R(x)}[v(x)-\bar\theta R(x)]+\eta-\frac{2p\eta^2}{R(x)}-(1+\lambda)tx=0 (2)$

With respect to $\eta$, we get:

$\frac{p}{R(x)}[v(x)-\bar\theta R(x)]+p-\frac{2p^2\eta}{R(x)}=0 (3)$

So at this point, my instinct would be to isolate $p$ in Equation $(3)$ and substitute into Equation $(2)$. However, I get:

$p^* = \frac{v(x^*)+R(x^*)(1-\bar\theta )}{2\eta^*} (4)$

And plugging $(4)$ back into $(2)$ leads to:

$\eta[\frac{v(x^*)}{R(x^*)}+(1-\bar\theta )-(1-\bar\theta )-\frac{2v(x^*)}{2R(x^*)}]-(1+\lambda)tx=0 (5)$

And this is not ideal, I don't understand what to do here. I've made up some functions in R (e.g., let $v(x) = ax, R(x) = bx^2$) and run through the optim function with some test values just to see what $p$ and $\eta$ look like, and what I see is that $p$ is very close to zero and $\eta$ is as large as possible. And this makes some sense to me given the derivatives of the objective function. However, I want to show algebraically that $p$ would optimally be close to zero and am just unsure of how to do that as my mathematics background isn't strong enough.

EDIT:

The full Lagrangian here would be:

$\frac{p\eta}{R(x)}[v(x)-\bar\theta R(x)]+(1-\frac{p\eta}{R(x)})p\eta-(1+\lambda)tpx-\lambda R(x) + \delta_1x + \delta_2p + \delta_3(1-p) (6)$

Leading to conditions:

With respect to $p$:

$\frac{\eta}{R(x)}[v(x)-\bar\theta R(x)]+\eta-\frac{2p\eta^2}{R(x)}-(1+\lambda)tx + \delta_2 - \delta_3 =0 (7)$

And the condition with respect to $\eta$ remains unchanged:

$\frac{p}{R(x)}[v(x)-\bar\theta R(x)]+p-\frac{2p^2\eta}{R(x)}=0 (3)$

In addition $\bar\theta$ is a coefficient between zero and one. I don't see what I should be doing next here.