Akar-akar persamaan x^3-7x^2+kx-12=0 adalah x1 , x2 dan x3 , jika x1x2=4 maka nilai k=

x³-7x²+kx-12 = 0

> x1+x2+x3 = 7
>> x1+x2 = 7-x3 ...(1)

> x1.x2.x3 = 12
>> (4).x3 = 12
>> x3 = 12/4 = 3 ...(2)

> x1.x2 + x1.x3 + x2.x3 = -k
>> (4) + x1.x3 + x2.x3 = -k
>> 4 + x3(x1 + x2) = -k
>> 4 + 3(7-x3) = -k
>> 4 + 3(7-3) = -k
>> 4 + 12 = -k
>> -16 = k

maka, nilai k adalah -16.