已知抛物线y=-x2+2(m+1)x+m+3与X轴有两个交点AB,且A在X轴的正半轴,B在负半轴,设OA长为a,OB长为b

(1) A在X轴的正半轴,B在负半轴,OA长为a,OB长为b,A(a,0),B(-b,0),
-x²+2(m+1)x+m+3 = 0的二根为a和-b,其积为-ab = (m+3)/(-1) = -(m+3) = a(-b) = -ab < 0
m >-3
(2)
a：b=3:1,a = 3b
y = -x² + 2(m+1)x + m+3 = -(x - a)(x+ b) = -(x - 3b)(x +b) = -x² + 2bx + 3b²
比较系数:2b = 2(m+1)
3b² = m+3
3(m+1)² = m+ 3
m(3m+5) = 0
m = 0或m = -5/3 (此时b = m+1 = -2/3 < 0,舍去)
m = 0