题目内容
已知是关于的一元二次方程的两个实数根,且——=115
(1)求k的值;(2)求++8的值。
(1)求k的值;(2)求++8的值。
(1)-11(2)66
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
∴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
练习册系列答案
相关题目