设方程x²+3[sqrt(3)]x+4=0的实数根为x1,x2,α=arctanx1,β=arctanx2,求α+β

设方程x²+3[sqrt(3)]x+4=0的实数根为x1,x2,α=arctanx1,β=arctanx2,求α+β
设方程x²+3[sqrt(3)]x+4=0的实数根为x1,x2,α=arctanx1,β=arctanx2,求α+β

设方程x²+3[sqrt(3)]x+4=0的实数根为x1,x2,α=arctanx1,β=arctanx2,求α+β
tan(alfh+beta)=(tan(alfh)+tan(beta))/1-tan(alfh).tan(beta)
But
tan(alth)=x1
tan(beta)=x2
So
tan(alfh+beta)=sum of roots/1-product of roots
tan(alfh+beta)=(-3.sqrt(3))/1-4
tan(alfh+beta)=sqrt(3)
tan(alfh+beta)=tan(60°)
alfh+beta=60°

计算机告诉我是-120

楼上的回答，很好。。。嘿嘿