Question

y=x+（k+1）x+k-3÷4 设x1,x2是此抛物线的于x轴两交点的横坐标,且满足x1+x2=k+5/2,求此抛物线

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：x1+x2=k+5/2 (x1+x2)-2x1x2=k+5/2 将x1+x2=-k-1 x1x2=(k-3)/4代入得 (k+1)-(k-3)/2=k+5/2 等式两边同时乘以2得 2(k+1)-(k-3)=2k+5 2k+4k+2-k+3=2k+5 3k=0 k=0 将k=0代入y=x+(k+1)x+(k-3)/4得：y=x+(0+1)x+(0-3)/4 y=x+x-3/4 所以抛物线的解析式为：y=x+x-3/4