Question

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

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

Answer 1

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

追问

答案还有个三分之一、

追答

应舍去m= 1/3
m = 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