用配方法证明代数式4x^2-6x+11的值恒大于0

4x^2-6x+11
=4[x^2-(3/2)x+11/4]
=4[x^2-(3/2)x+9/16-9/16+11/4]
=4[(x-3/4)^2-9/16+11/4]
=4[(x-3/4)^2+35/16]
=4(x-3/4)^2+35/4
因为:
4>0
(x-3/4)^2>0
35/4>0
所以4x^2-6x+11的值恒大于0

4x^2-6x+11
=(2x）^2-6x+9/4-9/4+11
=(2x-9/4)^2-9/4+11
=(2x-9/4)^2+35/4
(2x-9/4)^2+35/4>=0
所以恒大于0

=4(x^2-3/2x+9/16)+35/4=4(x-3/4)^2+35/4
4(x-3/4)^2>0 35/4>0
得证

4x^2-6x+11=3x^2+(x-3)^2+2>=2>0

4(x^2-3/2x)+11
=4(x-3/4)^2-9/4+11
=4(x-3/4)^2+35/4>0