persamaan(p+2)x²-10x+5=0 mempunyai akar akar kembar.tentukan nilai p

Diket :
a= p+2
b = -10
c = 5

ditanya : p -> akar kembar...?
jawab :

p -> akar kembar = D = 0

• D = 0
• B^2 -4(a)(c) = 0
• (-10)^2 -4(p+2)(5) = 0
• 100 -20(p+2) = 0
• 100 -20p-40 = 0
• 60 - 20p = 0
• 20p = 60
• p = 60/20
• p = 3

maaf klo slah
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