a-maths..........

a+a^2=12..(1)
a^3=k..(2)
solving (1)
a^2+a-12=0
(a+4)(a-3)=0
a=-4 or 3
Sub a=-4 or 3 into (2)
When a=-4
k=-64
When a=3
k=27

2007-08-12 21:28:16 補充：
Checking when k=-64the solutions are 16 or -4k=27The soltions are 3 or 9Both answers are correct!

α²-12α+k=0--------1
(α²)²-12α²+k=0--------2
α²+α=12
α² *α=k
2-1
(α²)²-13α²+12α=0
(α²)²+12α²+12α-25α²=0
(α²)²+12(α²+α)-25α²=0
(α²)²-25α²+144=0
α²=9或α²=16
α=+3或α=-4
k=27或-64

Sum of roots = 12/1 = a + a^2 ---------(1)
Product of roots = k/1 = a x a^2 ------------(2)

From (1),
a^2 + a -12 = 0
(a+4)(a-3) = 0
a = -4 or a = 3
From (2),
k = a^3
k = (-4)^3 = -64 or 3^3 = 27
Checking:
x^2 -12x-64 = 0
(x-16)(x+4) = 0
x=16 or -4 (correct)
x^2 -12x+27=0
(x-3)(x-9) =0
x=3 or 9 (correct)

Therefore, k is equal to -64 or 27