I am trying to solve the following equation for $d_2$ (from this paper here):
$$ \frac{c{\sqrt{(d_1+d_2)}}-\sqrt{d_1}z_1 -( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2}}{\sqrt{d_2}}= -z_{\beta}$$
I've tried various manipulations, moving $\sqrt{d_2}$ to the right hand side, then squaring, for example, but the hold up is always the $\sqrt{(d_1+d_2)}$ term.
My question is:
How do they solve for $d_2$?
Thank you!
Edit: they assume that rather than equality, there is an inequality, and they can make the denominator larger to satisfy it, then it becomes a quadratic of $d_2$.
$$ \frac{c{\sqrt{(d_1+d_2)}}-\sqrt{d_1}z_1 -( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2}}{\sqrt{d_2}} \leq -z_{\beta}$$
$$ \frac{c{\sqrt{(d_1+d_2)}}-\sqrt{d_1}z_1 -( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2}}{\sqrt{d1+d_2}} = -z_{\beta}$$
$$\begin{align*} \frac{c{\sqrt{(d_1+d_2)}}-\sqrt{d_1}z_1 -( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2}}{\sqrt{d_2}} &= -z_{\beta} \ \\ \\ c{\sqrt{(d_1+d_2)}}-\sqrt{d_1}z_1 -( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2} &= -z_{\beta}\sqrt{d_2} \ \\ \\ c{\sqrt{(d_1+d_2)}} &= -z_{\beta}\sqrt{d_2} +\sqrt{d_1}z_1 +( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2} \ \\ \\ c^2(d_1+d_2) &= \left[-z_{\beta}\sqrt{d_2} +\sqrt{d_1}z_1 +( \alpha)(d_2)\big(p_1(1-p_1)\big)^{1/2}\right]^2 \ \\ \end{align*}$$
Eliminates the $\sqrt{(d_1+d_2)}$ term.