已知m,n是方程X²-3X+1=0的两根,求代数式2m²+4n²-6n+1996的值如题

因为m,n是方程X²-3X+1=0的两根，所以有n^2-3n+1=0,而且由韦达定理知m+n=3,m*n=1
代数式2m²+4n²-6n+1996=2m^2+2n^2+2(n^2-3n+1)+1994=2(m+n)^2-4m*n+2(n^2-3n+1)+1994
将n^2-3n+1=0,m+n=3,m*n=1代入上式得
2m²+4n²-6n+1996=2(m+n)^2-4m*n+2(n^2-3n+1)+1994=18-4+0+1994=2008

一楼的回答挺好，就是最后答案错了，我更正如下：
∵m,n是方程X²-3X+1=0的两根,
∴m+m=3,mn=1,
2m²+4n²-6n+1996
=2m²+2n²+2n²-6n+1996
=2(m²+n²)+2(n²-3n)+1996
=2[(m+n)^2-2mn]+2*(-1)+1996
=2(3^2-2)-2+1996
=2*7+1994
=14+1994
=2008

∵m,n是方程X²-3X+1=0的两根,
∴m+m=3,mn=1,
2m²+4n²-6n+1996
=2m^2+2n^2+2n^2-6n+1996
=2(m^2+n^2)+2(n^2-3n)+1996
=2[(m+n)^2-2mn]+2*(-1)+1996
=2(3^2-2)-2+1996
=2*7+1994
=14+1994
=2018