解:(1)∵x
,x
是方程x
-6x+k=0的两个根
∴x
+ x
=6 x
x
=k······················1分
∵x
x
—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,x
x
=-11·····························7分
而x
+x
+8=(x
+x
)
—2x
x
+8=36+2×11+8=66·····9分
(1)方程有两个实数根,必须满足△=b
2-4ac≥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)
2-2
12+8即可求得式子的值