How Close Can You Be to Speaker B for Destructive Interference?

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Destructive interference of sound waves from two speakers occurs when the path length difference is a half-integer multiple of the wavelength. Given a frequency of 172 Hz, the wavelength is calculated to be 2 meters. The distance from speaker A is 8 meters, which corresponds to 4 wavelengths. To achieve destructive interference, the distance to speaker B must create a path length difference of 1 meter, meaning speaker B must be 7 meters away from the listener. Thus, the closest distance to speaker B for destructive interference is 7 meters.
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[SOLVED] Interference of Sound Waves

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.
What is the closest you can be to speaker B and be at a point of destructive interference?
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I am having the hardest time trying to visualize the problem. I know that destructive interference occurs when the difference in path lengths traveled by sound waves is a half integer number of wavelengths. So I need to know the wavelength of the sound which is just 2m.

I also know that in general if d_a and d_b are paths traveled by two waves of equal frequency that are originally emitted in phase, the condition for destructive interference is d_a-d_b=n(wavelength)/2 where wavelength is what I calculated it to be (2m) and n=any nonzero odd integer. I think I need to know what the value of n is that corresponds to the shortest distance d_b to solve my prob. (is d_a=8m? then what is d_b?)

I'm going around in circles and getting nowhere. Please help!
 
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d_a=8m is given.

One has determined the wavelength of the sound, 2 m, so the distance 8 m is 4 wavelengths. The the distance to B must be out of phase by 180 degrees in order to destructively interfere.

A half wavelength is 1 m.
 
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