Question

一道1元2次函数题

Answer 1

x^2+2(k+1)+k^2+2k-5/4=0

解方程得,x1=k-1/2,x2=k+5/2

代入y^2-（x1-k-1/2)y+（x1-k)(x2-k)+1/4=0,得到

y^2+y-1=0,

解得a=(√5-1)/2或a=-(√5+1)/2

(1)当a=(√5-1)/2时,1/a=(√5+1)/2,

a/(a+1)=1/(1+1/a)=(3-√5)/2,(a+1)/4=(√5+1)/8,a^2-1=(1-√5)/2,

[1/a-a/(a+1)]/[4/(a+1)]*(a^2-1)=[(√5+1)/2-(3-√5)/2]*(√5+1)/8*(1-√5)/2=(1-√5)/4

(2)当a=-(√5+1)/2时,1/a=(1-√5)/2,

a/(a+1)=1/(1+1/a)=(3+√5)/2,(a+1)/4=(1-√5)/8,a^2-1=(√5+1)/2,

[1/a-a/(a+1)]/[4/(a+1)]*(a^2-1)=[(1-√5)/2-(3+√5)/2]*(1-√5)/8*(√5+1)/2=(√5+1)/4