1)1/4x^2-(m-2)x+m^2=0有两相等实根,说明判别式等于0即△=[-(m-2)]^2 - 4*1/4*m^2=0m^2-4m+4-m^2=04m-4=0m=1x= -[-(m-2)]/(2*1/4)= (1-2)/(1/2)= -22)由韦达定理,得x1+x2 = -[-(m-2)]/(1/4)= 4(m-2...
已知关于x的方程1/4x^2-(m-2)x+m^2=0
1)1/4x^2-(m-2)x+m^2=0有两相等实根,说明判别式等于0即△=[-(m-2)]^2 - 4*1/4*m^2=0m^2-4m+4-m^2=04m-4=0m=1x= -[-(m-2)]/(2*1/4)= (1-2)/(1/2)= -22)由韦达定理,得x1+x2 = -[-(m-2)]/(1/4)= 4(m-2...