关于x的方程(a+2)x*2-2ax+a=0有两个不相等的实数根x1和x2,并且抛物线y=x*2-(2a+1)x+2a-5与x轴的两个交点分别位于点(2,0)的两旁.1;求实数a的取值范围.2;当|x1|+|x2|=2根号2时,求a的值.
热心网友
1. (a+2)x*2-2ax+a=0有两个不相等的实数根== (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根号2 === (|x1|+|x2|)^2 = (2根号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