设x1,x2是方程x^2-3x+1=0的两根,则√x1+√x2 =

设x1,x2是方程x^2-3x+1=0的两根,则√x1+√x2 =
设x1,x2是方程x^2-3x+1=0的两根,则√x1+√x2 =

设x1,x2是方程x^2-3x+1=0的两根,则√x1+√x2 =
根据韦达定理可知：x1+x2=3,x1x2=1
∴(√x1+√x2)²
=x1+x2+2√(x1x2)
=3+2√1
=5
∵√x1≥0,√x2≥0
∴ √x1+√x2 ≥0
即√x1+√x2 =√5