(急!!)有關△ge 1題問題...f.5 [[15點]]

the roots of the quadratic equation are negative

話就話係有兩個root，實際上呢兩個root可能係相等...

可能唔相等 = △>0 , 相等△=0

所以係△≧0

x^2 + (1 - k)x + 625 = 0

(1 - k)^2 - 4(625) ≧ 0

1 - 2k + k^2 - 2500 ≧ 0

k^2 - 2k - 2499 ≧ 0

k = 2 +- sqrt(4 + 9996) / 2

= 2 +- 100 / 2

= 51 and -49

k^2 - 2k - 2499 ≧ 0

(k + 49)(k - 51) ≧ 0

k <= -49, k≧51




2011-06-13 21:48:55 補充：
多謝雨後陽光的提醒

兩根和 = k - 1 < 0

k ≧ 51 應捨去....

2011-06-13 21:49:32 補充：
多謝提醒，你不提我還真不記得呢....

All answers is -49 <= k < 1 by combining k < 1 and ( k <= -49 or k >= 51 ).
no ones' answer is correct.

Because the roots are negative, the sum of roots is negative.

-(1-k) <= 0
k - 1 <= 0
k <= 1

Besides, if there exists at least one real root in the equation, Delta >= 0
(1-k)^2 - 4(1)(625) >= 0
(k^2 - 2k + 1) - 2500 >= 0
k^2 - 2k - 2499 >= 0
(k - 51) (k + 49) >= 0
k >= 51 or k <= -49

But, as k <= 1, the possible range of k is k <= -49

2011-06-13 22:12:56 補充：
雨後陽光, 答案係 k <= - 49, 你漏左負號呀.

厲害 險些兒看漏了兩根之和是負數

x² + (1-k)x + 625 = 0

方程有兩負根,

所以 △ = (1-k)² - 4*625 ≥ 0 , ( 不是△>0 因為兩負根可以相同,當兩負根相同時△=0) ............(1)

兩根和 = k - 1 < 0 ........(2)

由 (1) :

k² - 2k - 2499 ≥ 0

(k - 51) (k + 49) ≥ 0

k ≥ 51(捨去因不合(2)式) 或 k ≤ 49

k ≤ 49