- 方程三个实根成等差数列请证明:方程x^3+ax^2+bx+c=0
- 请证明:方程x^3+ax^2+bx+c=0三个实根成等差数列的充分条件是:2a^3-9ab+27c=0和a^2-3b≥0
- 为书写方便,设a=3p,则原题改为
方程x^3+3px^2+bx+c=0三个实根成等差数列的充分条件是:
2p^3-pb+c=0和3p^2-b≥0.
设f(x)=x^3+3px^2+bx+c
=(x^3+px^2)+(2px^2+2p^2x)+[(b-2p^2)x+p(b-2p^2)]+(c-pb+2p^3)
=(x+p)[x^2+2px+(b-2p^2)]+(c-pb+2p^3)
当2p^3-pb+c=0时,方程f(x)=0必有一实数根-p.
当3p^2-b≥0时,方程x^2+2px+(b-2p^2)=0中
△=4p^2-4(b-2p^2)=4(3p^2-b)≥0,有两个实数根,
且两实根之和为-2p
即方程f(x)=0有三个实数根,且成等差数列.
以p=a/3代入,2a^3-9ab+27c=0和a^2-3b≥0是方程x^3+ax^2+bx+c=0三个实根成等差数列的充分条件