Tentukan nilai P agar persamaan berikut mempunyai akar kembar
a. x²-2px-p+2=0
a. x²-2px-p+2=0
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D=b²-4ac
0=(-2p)²-4(1.(-p+2))
0=4p²+4p-8
4p²+4p-8=0 *kedua ruas dibagi 4
p²+p-2=0
(p+2)(p-1)
jadi p= -2 atau p=1
*maaf klo salah