Question

若M,N是两个不相等的实数,且满足M^2 - 2m=1,n^2 - 2n=1,那么代数式2m^2+4n^2+1994=

Answer 1

m^2-2m-1=0

n^2-2n-1=0

那么m,n可以看成是方程x^2-2x-1=0的二个不相等的实根.

根据“韦达定理”得：

m+n=2

mn=-1

是不是这样的：

2m^2+2n^2+1994

=2(1+2m)+2(1+2n)+1994

=2+4m+2+4n+1994

=4(m+n)+1998

=4*2+1998

=2006