BIAMPING

Half an answer - Using your 125W example, assume a 5 ohm nominal speaker impedance for convenience. If the amp can supply the 125W to the 5 ohm load, it is supplying 5 amps. P=I*2 x R > 125 = I*2 x 5 > 25 = I*2 > I=5. To do this, the output must swing 25 volts. I = E/R > 5 = E/5 > E = 25. Now, bridging essentially puts the two amp outputs in series. Therefore there is a possible swing of 50 volts. E/R = I > 50/5 = 10 amps. P = I*2 x R > P = 10*2 x 5 > P= 100 x 5 > P = 500 watts. Apparently, though, the amps can't quite supply the 10 amps because of component limitations (transformer, transistors, etc), but can only supply about 9 amps, so 9*2 x 5 = 400 watts limit in that configuration. PS - a single one of those amps could also supply 9 amps, but that would be about a half ohm load, which it not where it's rated! :)

More answer - in the example, if you can't play your system loud enough (at least on peaks) then perhaps you are output voltage limited and need the higher voltage swing to drive your speakers (and ears) to their potential. Voltage limitation like current limitation current limitation can result in clipping and distortion or damage. That is the argument for bridging as long as the amps can supply the additional current. You must also know that the speakers can accomodate the extra current/power; that is, they were meant to play louder. If your system plays loud enough, even on peaks, you would just be running the level control lower with the bridged amps and not accomplishing much. Now comes the counter responses, I guess. :)

Sean's answer is excellent overall, but the fact that you double the voltage applied to the speaker and therefore double the current (if the amp has the ability) through the SAME load accounts for the 2*2 times power. (*2 mean to the power 2, or squared).

Time for fairness. A question on another thread about stability/load rating when bridging made me realize the point of Sean's perspective on "effective" load impedance. While the two amps being in series with the same physical load does account for (up to) twice the current in the circuit, indeed each amp is supplying twice the current that it alone would into the nominal load impedance of the speaker. Thus, each is operating as though it were seeing an impedance half that of the speaker. If I may simplify (dangerous), think of the other amp as a negative impedance because it is supplying current rather than impeding it. My apologies for being focused on my point.