分类：教育/科学-学习帮助
问题：作业6若x的方程x^2
若x的方程x^2-x+a=0和x^2-x+b=0(a≠b)的四个根可组成首项为1/4的等差数列，则a+b的值是多少？
回答：x1+x2 =1 x1x2 =a (x1<x2) 
x3+x4 =1 x3x4 =b (x3<x4) 
等差数列,所以a1+a4 =a2+a3 
如果设 
x1 =1/4 ===>x2 =3/4 
则 
x3 =5/12 x4 =7/12 

反之,设X3 =1/4 , 
x3 =1/4 ==>x4 =3/4 
==>x1 =5/12 x2 =7/12 
显然,上述两个情况对a+b结果影响 

===>a+b =x1x2+x3x4 =3/16 +35/144 =31/72