已知x1,x2是方程2x^2+3x-1=0的两个根,用韦达定理求x1-x2的值

由韦达定理有x1+x2=-3/2,x1x2=-1/2,
所以x1-x2
=±√(x1-x2)^2
=±√(x1^2+x2^2-2x1x2)
=±√[(x1^2+x2^2)-2(x1x2)]
=±√[(x1+x2)^2-2x1x2-2x1x2]
=±√[(x1+x2)^2-4x1x2]
=±√[(-3/2)^2-4*(-1/2)]
=±√[(9/4)+2]
=±√(17/4)
=±√17/2.