Question

』数学代数;:解一元二次方程关于X的方程（a+c）x2+bx-(a+2c)=0的两根之和为1,两根的倒数和为-2.(1)

Answer 1

设两根x1,x2,则x1+x2=1,1/x1+1/x2=(x1+x2)/(x1*x2)=-2,so x1*x2=-0.5.

x1+x2=1=-b/(a+c),x1*x2=-0.5=-(a+2c)/(a+c)=-(a+c+c)/(a+c)=-1-c/(a+c),

b/(a+c)=-1,c/(a+c)=-0.5,so b/c=2,代入b/(a+c)=-1,则a=-3c,so

a:b:c=-3:2:1