This should be absurd, because when m is determined as a value, each set is at most two solutions, so as long as the two solutions are different, their intersection is also an empty set.

Let's assume that there is a real number a, which makes the two equations hold, so there is a? +ma+ 1=0 and 8a? +a+m=0

So you can get m=-8a? -a, brought into the previous equation, simplified to -8a 3+ 1 = 0.

Get a= 1/2.

Then you get m=-5/2.

That is to say, only when m=-5/2, A∩B= 1/2, and at other times a ∩ b = φ, so as long as m ≦- 5/2 is enough.