Question

一元二次题 若m，n是一元二次方程X^2-3X+1=0的两根，则代数式2m^2=4n^2-6n=1998的值为？？

若m，n是一元二次方程X^2-3X+1=0的两根，则代数式2m^2=4n^2-6n=1998的值为？？

Answer 1

若m，n是一元二次方程X^2-3X+1=0的两根
则m+n=3,mm=3m-1,nn=3n-1

2m^2+4n^2-6n+1998
=6m-2+12n-4-6n+1998
=6(m+n)+1992
=18+1992=2010

Answer 2

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+1998
=2m^2+2n^2+2n^2-6n+1998
=2(m^2+n^2)+2(n^2-3n+1)+1996
=2[(m+n)^2-2mn]+1996
=2*(3^2-2)+1996
=14+1996
=2010

Answer 3

解：∵m、n是一元二次方程x2-3x+1=0的两根，
∴m2-3m+1=0，n2-3n+1=0，m+n=3，
∴m2=3m-1，n2=3n-1，
∴2m2+4n2-6n+1999
=2（3m-1）+4（3n-1）-6n+1999
=6m-2+12n-4-6n+1999
=6（m+n）+1993
=6×3+1993
=2011．