解:(1)∵x,x是方程x-6x+k=0的两个根 ∴x+ x=6 x x=k······················1分 ∵xx—x—x=115 ∴k—6=115·············································2分 解得k=11,k=-11······································3分 当k=11时=36—4k=36—44<0 ,∴k=11不合题意·······4分 当k=-11时=36—4k=36+44>0∴k=-11符合题意·········5分 ∴k的值为—11············································6分 (2)x+x=6,xx=-11·····························7分 而x+x+8=(x+x)—2xx+8=36+2×11+8=66·····9分 (1)方程有两个实数根,必须满足△=b2-4ac≥0,从而求出实数的取值范围,再利用根与系数的关系,12-1-2=115.即12-(1+2)=115,即可得到关于的方程,求出的值. (2)根据(1)即可求得1+2与12的值,而12+22+8=(1+2)2-212+8即可求得式子的值 |