a-maths..........
回答 (3)
2007-08-13 5:25 am
✔ 最佳答案
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!
2007-08-13 5:42 am
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
(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
2007-08-13 5:29 am
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
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
參考: ME~
收錄日期: 2021-04-13 12:00:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070812000051KK04926