Question

已知m、n是方程x^2-3x+1=0的实数根,求代数式2m^2+4n^2-6n+2006的值

Answer 1

根据一元二次方程根与系数的关系得到:m+n=3,m*n=1
则m=3-n代入m*n=1得(3-n)n=1即:3n-n^2=1
2m^2+4n^2-6n+2006
=2(3-n)^2+4n^2-6n+2006
=2(9-6n+n^2)+4n^2-6n+2006
=18-12n+2n^2+4n^2-6n+2006
=6n^2-18n+2024
=6(n^2-3n)+2024
把3n-n^2=1代入得
=-6+2024
=2018