Question

若关于x的方程x2-x+m=0和x2-x+n=0(m,n∈R,且m≠n)的四个根可组成首项为¼的等差数列求

Answer 1

设方程x²-x+m=0两根分别为x1,x4；方程x²-x+n=0两根分别为x2,x3.

由韦达定理得

x1+x4=1 x2+x3=1

x1+x4=x2+x3

x1、x2、x3、x4成等差数列,设公差为d.

又首项为1/4,x1=1/4

x4=1-x1=1-1/4=3/4

x4-x1=3d=3/4-1/4=1/2 d=1/6

x2=x1+d=1/4+1/6=5/12 x3=x1+2d=1/4+1/3=7/12

m=x1x4=(1/4)(3/4)=3/16 n=x2x3=(5/12)(7/12)=35/144

m+n=3/16+35/144=31/72