Question

已知m.n是二元一次方程x^2-3x+1=0的两根,那么代数式2m^2+4n^2-6n+1999

Answer 1

m,n是方程的两根,那么

n^2-3n+1=0

m+n=3,mn=1

2m^2+4n^2-6n+1999

=2m^2+2n^2+2n^2-6n+1999

=2(m^2+n^2)+2(n^2-3n+1)+1997

=2[(m+n)^2-2mn]+1997

=2*(3^2-2)+1997

=14+1997

=2011