函数y=-x^2+2(m-1)x+m+1的图像与x轴交于A,B两点,点A在负半轴上,点B在正半轴上,线段OA与OB的比为1:3,求m的值

A(-a,0),B(3a,0),a > 0
y = -(x + a)(x - 3a) = -x² + 2ax + 3a² = -x² +2(m - 1)x + m + 1
2a = 2(m - 1),a = m -1
3a² = 3(m - 1)² = m + 1
3m² - 7m + 2 = 0
(3m - 1)(m - 2) = 0
m = 1/3( a= -2/3 < 0,舍去)或m = 2
m = 2答案还有个三分之一、应舍去m= 1/3m = 1/3时, = -x² -4x/3 + 4/3= - (1/3)(3x^2 + 4x - 4)= (-1/3)(3x - 2)(x + 2)A(-2, 0), B(2/3, 0)OA : OB= 3 : 1与题意矛盾