Respostas

2014-01-10T18:15:31-02:00
f(3) = x ²- (k + 2)x + 2k + 6 = 0
x²-(k+2)x+2k+6=0
(3)²-(k+2)3+2k+6=0
9-(3k+6)+2k+6=0
9-3k-6+2k+6=0
-3k+2k+9-6+6=0
-k+9=0
-k=-9
k=9

Tirando a prova:
9-(9+2)3+18+6=0
9-27-6+18+6=0
-33+33=0
0=0
(confere!)