Continuously replenished reaction Substrate ($m$) concentration in a reaction scheme is governed by the equation $$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}$$ where $c_1$ is the initial enzyme concentration and each $r_i$ is the reaction rate of the reactions present within this reaction scheme (which are positive). Now additionally suppose the substrate is being continuously replenished by a reaction $$\ce{A-&gt;[r_5]S,}$$ where $A$ has constant concentration $a_0$. Modify $dm/dt$ and show there are two distinct positive equilibria $0&lt;m_1&lt;m_2$ if $r_5&lt;r_5^C$ where $r_5^C$ is a critical value to be determined.



With this information we have 

$$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}+r_5a_0.$$

Equilibria when $dm/dt=0$. We rearrange this to produce the quadratic

$$r_2r_4r_5a_0 + r_1r_4(r_5a_0 - r_2c_1)m + r_1r_3r_5a_0m^2=0$$

So the equilibria occur at

$$m_{1,2} = \dfrac{r_1r_4(r_2c_1-r_5a_0)\pm\sqrt{r_1^2r_4^2(r_5a_0-r_2c_1)^2-4r_1r_2r_3r_4r_5^2a_0^2}}{2r_1r_3r_5a_0}$$

Now I need to justify these are positive and find the critical value for $r_5$ but I have no idea how to continue here.
The first step is to set the discriminant equal to zero, and thereby solve for the critical value of $r_5$. When I did this, I got $$r_5a_0=\frac{r_2c_1}{1+2\sqrt{\frac{r_2r_3}{r_1r_4}}}$$.  The next step is to substitute this into the equation for m, and find the two  corresponding equal values of m.