F.4 的 MATHS

∵the equation has two distinct real roots
∴b2 – 4ac > 0
(-a)2 – 4(1)(40) > 0
a2 –160 >0
a2 > 160
a > 4√10
the two possible values for a should be 13、14

2007-10-07 20:16:53 補充：
睇漏左個負號變成‘x^2 - ax＋40 = 0’應該是‘x^2 - ax - 40 = 0’(-a)^2 – 4(1)(-40) 0a^2 ＋160 0a^2 - 160a -4√10 ∴the two possible values for a should be -11、-10

2007-10-07 20:18:20 補充：
顯示唔到大於的符號(-a)^2 – 4(1)(-40) 0a^2 ＋160＞0a^2＞- 160a＞ -4√10 答案沒有影響

x^2-ax-40=0
∵the equation has 2 distinct real roots
∴△>0
(-a)^2-4(1)(-40)>0
a^2+160>0
a^2>-160
a>40 or a>-40

唔知係我計錯定條問有問題
suggest two possible values for a
我計到既係ｒａｎｇｅ範圍a>40 or a>-40

2007-10-07 20:08:42 補充：
ｓｏｒｒｙ係我計錯應該係a^2＞１６０a √160 or a -√160a 4√10 or a -4√10

Assume m and n are the 2 distinct real roots, then m*n = -40, m+ n = a

Let take the factor of -40, then we have when m = 1, n = -40 => a = -39
when m = -1, n = 40 => a = 39
when m = 2, n =-20 => a = -18
when m = -2, n = 20 => a = 18

etc.