關于x的方程(a+2)x*2-2ax+a=0有兩個不相等的實數根x1和x2,并且拋物線y=x*2-(2a+1)x+2a-5與x軸的兩個交點分別位于點(2,0)的兩旁.1;求實數a的取值范圍.2;當|x1|+|x2|=2根號2時,求a的值.

熱心網友

1. (a+2)x*2-2ax+a=0有兩個不相等的實數根== (2a)^2 - 4*(a+2)*a 0 === a (x3 -2)*(x4 -2) x3*x4 -2*(x3+x4)+4 (2a-5) -2*(2a+1) +4 a -3/2 ...(2)因此，a = (-3/2,0)2. |x1|+|x2|=2根號2 === (|x1|+|x2|)^2 = (2根號2)^2 = 8== (x1+x2)^2 -2*(x1*x2) +2*|x1*x2| =[2a/(a+2)]^2 -2*a/(a+2) +2*|a/(a+2)|= [2a/(a+2)]^2 -4*a/(a+2) = 8=== a = -1；-3（舍去）因此，a = -1