Question

Se (x-2) (x-3) = 0 e x ≠ 2, entao x = ?

Answer 1

Aplique distributiva
(x-2)(x-3)=x²-3x-2x+6 =x²-5x+6
x=(-b+ou-√ b²-4.a.c)/2a
x=(-(-5)+ou- √(-5)²-4.1.6)/2.1
x=(5 +ou-√25-24)/2
x=(5+ou- √1)/2
x1=(5+1)/2 = 6/2= 3
x2=(5-1)/2 = 4/2= 2