关于x的方程(a+c)x^2+bx-(a+2c)=0的两根之和为1,两根的倒数和为-2.1 求两跟之积 2 求a:b:c.

1、
x1+x2=1
1/x1+1/x2=(x1+x2)/(x1x2)=1/x1x2=-2
两跟之积=x1x2=-1/2
2、
x1+x2=-b/(a+c)=1
a+c=-b
x1x2=-(a+2c)/(a+c)
所以1/x1+1/x2=(x1+x2)/(x1x2)=1/[-(a+2c)/(a+c)]=-(a+c)/(a+2c)=-2
a+c=2a+4c
a=-3c
a:c=-3:1
b=-(a+c)=-(-3c+c)=2c
b:c=2:1
所以a:b:c=-3:2:1