Question

已知抛物线y=x^2+(k+1)x+(k-3)/4,设经X1、X2是此抛物线与x轴两交点的横坐标,且满足x1^2+x2^

Answer 1

y=x^2+(k+1)x+(k-3)/4

根据韦达定理得

x1+x2=-k-1

x1x2=(k-3)/4

已知：x1²+x2²=k²+5/2

(x1+x2)²-2x1x2=k²+5/2

将x1+x2=-k-1 x1x2=(k-3)/4代入得

(k+1)²-(k-3)/2=k²+5/2

等式两边同时乘以2得

2(k+1)²-(k-3)=2k²+5

2k²+4k+2-k+3=2k²+5

3k=0

k=0

将k=0代入y=x²+(k+1)x+(k-3)/4得：

y=x²+(0+1)x+(0-3)/4

y=x²+x-3/4

所以抛物线的解析式为：y=x²+x-3/4