如果方程lg²x-lgx²-2=0有两根a,b那么loga b+logb a的值等于

如果方程lg²x-lgx²-2=0有两根a,b那么loga b+logb a的值等于

a,b是方程（lgx）^2-lgx^2-2=0的两个根
即:
(lga)^2-2lga-2=0
(lgb)^2-2lgb-2=0
设m=lga,n=lgb
则:m^2-2m-2=0,n^2-2n-2=0
说明m,n是方程t^2-2t-2=0的二个根.
m+n=2,mn=-2
loga b+logb a=lgb/lga+lga/lgb=[(lgb)^2+(lga)^2]/(lga*lgb)
=[(m+n)^2-2mn]/(mn)
=[4-2*(-2)]/(-2)
=-4

(lgx)^2-2lgx-2=0
lga+lgb=2
lgalgb=-2
loga b+logb a=lgb/lga+lga/lgb
=[(lgb)^2+(lga)^2]/lgalgb
=[(lga+lgb)^2-2lgalgb]/lgalgb
=(4+4)/(-2)
=-4