设x1、x2为方程ax²+bx+c=0(a≠0)的两根,则:
x1+x2=-b/a,x1x2=c/a
(1/x1+1)、(1/x2+1)为所求方程两根
(1/x1+1)+(1/x2+1)=(x1+x2)/x1x2+2=-b/c+2
(1/x1+1)(1/x2+1)=1+(x1+x2+1)/x1x2=(a+c-b)/c
∴所求方程为:
x²-(-b/c+2)x+(a+c-b)/c=0
整理后得:
cx²+(b-2c)x+(a+c-b)=0
设x1、x2为方程ax²+bx+c=0(a≠0)的两根,则:
x1+x2=-b/a,x1x2=c/a
(1/x1+1)、(1/x2+1)为所求方程两根
(1/x1+1)+(1/x2+1)=(x1+x2)/x1x2+2=-b/c+2
(1/x1+1)(1/x2+1)=1+(x1+x2+1)/x1x2=(a+c-b)/c
∴所求方程为:
x²-(-b/c+2)x+(a+c-b)/c=0
整理后得:
cx²+(b-2c)x+(a+c-b)=0