Penseu en el conjunt
#A = x ^ 2 + (m-1) x-2 (m + 1) = 0
Ho sabem
# Delta_A = (m-1) ^ 2-4 (1) (- 2 (m + 1)) #
= m 2-2 m + 1 + 8 m + 8 #
= (m-3) ^ 2 #
# Delta_A = 0 => m = 3 => 1 # solució
# Delta_A gt 0 => m! = 3 => 2 # solucions
Ara volem
-
Un element de A, dos elements de B:
# => Delta_A = 0, Delta_B gt 0 # # => (m = 3) nn (m! = 2) => m = 3 -
Un element de B, dos elements de A
# => Delta_B = 0, Delta_A gt 0 # # => (m = 2) nn (m! = 3) => m = 2
Per tant, hi ha