It is known that the equation (a + b) x ^ 2-2ax + a = 0 about X has two unequal real roots x1, X2, and the parabola y = x ^ 2 - (2a + 1) x + 2a-5 is the same as X axis It is known that the equation (a + 2) x ^ 2-2ax + a = 0 about X has two unequal real roots x1, X2, and the two intersections of the parabola y = x ^ 2 - (2a + 1) x + 2a-5 and the X axis are located on both sides of the point (2,0). (1) find the value range of the real number a; (2) find the value of a when the absolute value of X1 plus the absolute value of X2 is equal to two root signs 2

It is known that the equation (a + b) x ^ 2-2ax + a = 0 about X has two unequal real roots x1, X2, and the parabola y = x ^ 2 - (2a + 1) x + 2a-5 is the same as X axis It is known that the equation (a + 2) x ^ 2-2ax + a = 0 about X has two unequal real roots x1, X2, and the two intersections of the parabola y = x ^ 2 - (2a + 1) x + 2a-5 and the X axis are located on both sides of the point (2,0). (1) find the value range of the real number a; (2) find the value of a when the absolute value of X1 plus the absolute value of X2 is equal to two root signs 2

one
4a^2-4a(a+2)＞0,
a(a-a-2)＞0,
a＜0;
Y = x ^ 2 - (2a + 1) x + 2a-5 is a parabola with the opening upward
So at x = 2, y < 0
y=2^2-(2a+1)2+2a-5
=-2a-3＜0
a＞-3/2
So - 3 / 2 < a < 0
two
x1+x2=2a/(a+2)
x1*x2=a/(a+2)
|x1|+|x2|=2√2
(|x1|+|x2|)^2=x1^2+x2^2+2|x1x2|
=x1^2+x2^2-2a/(a+2)
=x1^2+x2^2-2a/(a+2)
=(2ax1-a)/(a+2)+(2ax2-a)/(a+2)-2a/(a+2)
=(2ax1+2ax2)/(a+2)-4a/(a+2)
=2a(x1+x2)/(a+2)-4a/(a+2)
=4a^2/(a+2)^2-4a/(a+2)
=8
a^2/(a+2)^2-a/(a+2)-2=0
[a/(a+2)-2][a/(a+2)+1]=0
A / (a + 2) = 2 or a / (a + 2) = - 1
A = 2A + 4 or a + A + 2 = 0
A = - 4 or a = - 1
Because - 3 / 2 < a < 0
So a = - 1