I apologize for my comments; I shouldn't be answering Chem.SE questions during my commute without writing stuff out...
I renamed some of your rate constants. In addition, notice that as written, the reverse of the first reaction does not lead directly to a component of $\frac{d\ce{[A]}}{dt}$ because two $\ce{A}$'s are being consumed.
$$\ce{A2 \underset{k_{-1}}{\overset{k_{1}}{<=>}} 2A}$$
$$\ce{A + B \overset{k_{2}}{->} P}$$
$$\mathrm{rate} = \frac{d\ce{[P]}}{dt} = k_{2}\ce{[A][B]}$$
Your steady state equation is almost correct:
$$\frac{d\ce{[A]}}{dt}=2k_{1}\ce{[A2]}−\frac{k_{-1}}{2}\ce{[A]}^{2}−k_{2}\ce{[A][B]}=0$$
I actually ended up solving the quadratic equation. It wasn't that bad:
$$\ce{[A]} = \frac{\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}}{k_{-1}}$$
so:
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$
Sanity check the two limits:
- $k_{2}\ce{[B]} \gg \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$
- $k_{2}\ce{[B]} \ll \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$
Case 1 means that the second reaction is very fast. Case 2 means that the second reaction is very slow.
Case 1:
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=\frac{k_{2}\left(k_{2}\ce{[B]}\sqrt{1 + \frac{4k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$
Taylor expand the root: $\sqrt{1+x^{2}} \approx 1+\frac{x^{2}}{2}$
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(k_{2}\ce{[B]}\left(1 + \frac{2k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}\right)-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=2k_{1}\ce{[A2]}$$
In other words, the intermediate is consumed in the forward direction at the rate that we can fragment $\ce{A2}$. This seems right.
Case 2:
$$$$
$\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}$ reduces to $\sqrt{4k_{1}k_{-1}\ce{[A2]}}$.
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\sqrt{4k_{1}k_{-1}\ce{[A2]}}\ce{[B]}}{k_{-1}}$$
The thing you want is somewhere between the complicated first rate I wrote and the limit in case 2. This seems reasonable in that it is first order in $\ce{B}$ as you might expect from a transition state that involves a single $\ce{B}$.