Let a = {X & # 178; + (2a-3) x-3a = 0}, B = {X & # 178; + (A-3) x + A & # 178; - 3A = 0}, if a ≠ B, a ∩ B ≠ empty set, try to find a ∪ B

Let a = {X & # 178; + (2a-3) x-3a = 0}, B = {X & # 178; + (A-3) x + A & # 178; - 3A = 0}, if a ≠ B, a ∩ B ≠ empty set, try to find a ∪ B

A ∩ B ≠ empty set equation x & # 178; + (2a-3) x-3a = 0 and X & # 178; + (A-3) x + A & # 178; - 3A = 0 have at least one common solution. By subtracting the two equations, we get ax-a ^ 2 = 0 if a ≠ 0, x = a if a = 0, X ∈ R, a = {x │ x ^ 2-3x = 0}, B = {x │ x ^ 2-3x = 0}, a = B is not satisfactory, so if a ≠ 0, x = a is their common solution

Elden Ring Premiere

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