Question

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）求m的取值范围
（2）若a、b满足a：b=3:1, 求m的值

Answer 1

(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

Answer 2

(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