Question

对于任意实数x,等式（2x-1)^2-a(x+b)^2=px(a b p 均为常数）求p的值

Answer 1

（2x-1)^2-a(x+b)^2

=4x^2-4x+1-ax^2-2abx-ab^2

=(4-a)x^2-(4+2ab)x+(1-ab^2)

=px

对应项系数相等

所以4-a=0

-(4+2ab)=p

1-ab^2=0

a=4,1-4ab^2=0,b^2=1/4

b=1/2,b=-1/2

p=-(4+2ab)

所以

b=1/2,p=-(4+2ab)=-8

b=-1/2,p=-(4+2ab)=0