一元二次方程的根的範圍

Let y = mx^2 + (2m+3)x + (1-m), the line of symmetry is x = -(2m+3)/(2m)
As the line of symmetry lies between -4 and 3, so
-4 < -(2m+3)/(2m) < 3
==> m < -3/8 or m > 1/2

Case when line of symmetry lies between -4 and -3, ie 1/2 < m < 3/4, then
2[(-3) + (2m+3)/(2m)] < |x1 - x2| < 2[-(2m+3)/(2m) - (-5)]
==> [(2m+3)/m - 6]^2 < [(2m+3)/m]^2 - 4(1-m)/m < [(10 - (2m+3)/m]^2
==> 36 - 12(2m+3)/m < -4(1-m)/m < 100 - 20(2m+3)/m
==> 1 < m < 4 ... which contradict 1/2 < m < 3/4

Case when it lies between -3 and -1/2, ie. m > 3/4, then
|x1 - x2| < 2[-(2m+3)/(2m) - (-5)]
==> -4(1-m)/m < 100 - 20(2m+3)/m
==> m < 0 or m > 1
As m > 3/4, so m > 1 .... (i)

Case when it lies between -1/2 and 2, ie m < -1/2, then
|x1 - x2| < 2[4 + (2m+3)/(2m)]
==> -4(1-m)/m < 64 + 16(2m+3)/m
==> m < -13/23 or m > 0
As m < -1/2, so m < -13/23 .... (ii)

Case when it lies between 2 and 4, ie -1/2 < m < -3/8, then
2[-(2m+3)/(2m) - 2] < |x1 - x2| < 2[4 + (2m+3)/(2m)]
==> [(2m+3)/m + 4]^2 < [(2m+3)/m]^2 - 4(1-m)/m < [(8 + (2m+3)/m]^2
==> 16 + 8(2m+3)/m < -4(1-m)/m < 64 + 16(2m+3)/m
==> -1 < m < -13/23 ... which contradict -1/2 < m < -3/8

So, from (i), (ii), the solution is m < -13/23 or m > 1

To 自由自在：
題中x1一定是較大根嗎？x2一定是較小根嗎？

http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-909.png

圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-909.png


2012-11-22 20:33:33 補充：
若其中一根大或等於2,該根為較大根
若其中一根小或等於-3,該根為較小根
若兩根都介乎-3及2之間,任何一根都有機會較大或較小
所以較大根a滿足4>a>=2或2>a>-3即4>a>-3
較小根b滿足-5<=-3或-3<2即2>b>-5