Question

已知方程x^2-10x+24=0,求作一个方程,使它们的根分别是原方程各根的倒数,

Answer 1

x1+x1=10

x1x2=24

则1/x1+1/x2

=(x1+x2)/x1x2

=10/24

=5/12

1/x*1/x2=1/(x1x2)=1/24

所以是x²-5x/12+1/24=0

即24x²-10x+1=0