1.設方程式x平方-kx-27=0之一根為另一根之平方,則k值為? 2.方程式(x平方+5x)平方+10(x平方+5x)+24=0之四根為何?

1.
令兩根為 u , u^2

x^2 - kx - 27
= ( x - u )( x - u^2 )
= x^2 - ( u + u^2 )x + u^3

比較係數得
k = u + u^2 ..... (1式)
- 27 = u^3 ..... (2式)

由 (2式) 得 u = - 3 , 再代入(1式)得:
k = u + u^2 = - 3 + (-3)^2 = - 3 + 9 = 6
Ans: k = 6

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2.
令 u = x^2 + 5x , 則原式即為:
u^2 + 10u + 24 = 0
( u + 4 )( u + 6 ) = 0
u = - 4 , - 6

當 u = - 4 時:
u = x^2 + 5x = - 4
x^2 + 5x + 4 = 0
( x + 1 )( x + 4 ) = 0
x = - 1 , - 4

當 u = - 6 時:
u = x^2 + 5x = - 6
x^2 + 5x + 6 = 0
( x + 2 )( x + 3 ) = 0
x = - 2 , - 3

Ans: - 1 , - 2 , - 3 , - 4