materi tentang persamaan kuadrat. soalnya : Px2-4x+3,memiliki D=0,tentukan nilai p?

px²-4x+3      a = p, b = -4, c = 3
D = b²-4ac
0 = (-4)² - 4(p)(3)
0 = 16 - 12p
12p = 16
p = 16/12 = [tex] \frac{4}{3} [/tex]

D=b^2-4ac
(-4)^2-4p.3=0
=16-12p
12p=16
p=4/3