# [挑戰]求有理根與公根的方程式

1)設aZ ,若x^2-3x+a=0 ,
2x^2+ax-4=0與ax^2+bx-3=0恰有一公根,則a=
□ , b=□

2)設x^2+px+q=0與x^2+qx+p=0僅有一公共根,則其他二非公共根之和=□

3)設f(x)=x^3+6x^2+11x+6
, g(x)=x^3+7x^2+14x+8 , 若實數h , 滿足f(h)*g(h)=0 ,

≠0,求h=

1)
x^2-3x+a=0…………..(1)
2x^2+ax-4=0…………(2)

f(x)=2x^2+ax-4=0, 可能有+-4, +-2, +-1, +-1/2之解

f(4)=32+4a-4=0,
a=-7, f(-1/2)=1/2-1/2a-4=0, a=-7,

f(-4)=32-4a-4=0,
a=7, f(1/2)=1/2+1/2a-4=0, a=7,

f(2)=8+2a-4=0,
a=-2, f(-1)=2-a-4=0, a=-2,

f(-2)=8-2a-4=0,
a=2, f(1)=2+a-4=0, a=2,

a=2

x=2,1
a=2

(2), 2x^2+2x-4=0, (x+2)(x-1)=0,
x=-2,1

a=2, x=1代入ax^2+bx-3=0,

: a=2,
b=1

2)將兩多項式相減得: (p-q)x-(p-q)=0, x-1=0,
x=1

x^2+px+q=0,

x^2+qx+p=0, 兩根和為-q, 故另一根為-q-1

-p-1-q-1=-p-q-2

:
-p-q-2

3)

f(x)=x^3+6x^2+11x+6=(x+1)(x+2)(x+3)
g(x)=x^3+7x^2+14x+8=(x+1)(x+2)(x+4)

h=-1,-2會使f(h)+g(h) =0,

h=-3,-4 使f(h)*g(h)=0 , 且f(h)+g(h) ≠0

答: h = -3,-4

mazilla解第一題

Z ,若x^2-3x+a=0 , 2x^2+ax-4=0與ax^2+bx-3=0恰有一公根,則a=□ ,
b=□

v^2 -3v +a=0…(1)
2v^2 +av -4=0…(2)
av^2 +bv
-3=0…(3)

2v^2 +(3v -v^2)v -4=0
v^3 -5v^2 +4
=0
(v-1)(v^2 -4v -4) =0…(4)

v=1，再代入(1)得
a=2

b=1