方程4x平方+(m-2)+(m-5)=0问是否存在实数m使方程的两根都大于1?并证明(主要是证明哦)

方程应该是4x^2+(m-2)x+(m-5)=0吧
delta=m^2-4m+4-16m+80=m^2-20m+84=(m-6)(m-14)>0 ==> m<6 or m>14
x1+x2=(2-m)/4>2 ==> 2-m>8 ==> m<-6
(x1-1)(x2-1)=x1x2-(x1+x2)+1=(m-5)/4+(m-2)/4+1>0 ==> 2m-3>0 ==> m>1.5
取三者交集得 m不存在

4x^2+(m-2)x+(m-5)=0
delta=m^2-4m+4-16m+80=m^2-20m+84=(m-6)(m-14)>0 ==
> m<6 or m>14
x1+x2=(2-m)/4>2 ==> 2-m>8 ==> m<-6
(x1-1)(x2-1)=x1x2-(x1+x2)+1=(m-5)/4+(m-2)/4+1>0 ==> 2m-3>0 ==> m>1.解答完毕！~~希望采纳！~