Question

关于x的一元二次方程x^2-x+p-1=0有两个实数根x1,x2,若(x1^2-x1-2)(x2^2-x2-2)=9,求p的值！

Answer 1

x^2-x+p-1=0 得x^2-x =1-p
x1,x2,代入x^2-x =1-p
x1^2-x1=1-p
x2^2-x2=1-p
(x1^2-x1-2)(x2^2-x2-2)=9即得（p+1）^2=9 得p=2或-4
若p=2 x^2-x+p-1=0 无解
p=-4时成立 所以p=-4

Answer 2

x^2-x+p-1=0
(-1)^2-4(p-1)>=0
p<=5/4

x1+x2=1,x1x2=p-1
(x1^2-x1-2)(x2^2-x2-2)=9,
[2+x1(1-x1)][2+x2(1-x2)]=9
4-2(x1+x2)+2(x1+x2)^2+x1x2[5-(x1+x2)+x1x2]=9
带入整理：
(p+4)(p-2)=0
p=-4,或p=2(舍，因为p<=5/4)
P的值p=-4

追问

哦哦，晓得了，谢谢

Answer 3

x^2-x+p-1=0
则x^2-x=1-p
(x1^2-x1-2)(x2^2-x2-2)=9
（1-p-2)^2=9
(p+1)^2=9
则
p1=2,p2=-4
△=1-4（p-1)>=0,4p<=5,p<5/4
∴p=-4