兩題中四數學功課........高手請進

1. x2 – 2(m + 1)x + (2m2 + 3) = 0
Discriminant
= [-2(m + 1)]2 – 4(1)(2m2 + 3)
= 4(m2 + 2m + 1) – 8m2 – 12
= -4m2 + 8m – 8
= -4(m2 – 2m + 1) + 4 – 8
= -4(m - 1)2 – 4 < 0 [Since -4(m - 1)2 is equal or less than 0]
So, the equation has no real roots for all real value of m.

2.a. 12x2 + 4x + (k - 1) = 0
Since the equal has real roots,
So, delta ≧ 0
(4)2 – 4(12)(k - 1) ≧ 0
16 – 48k + 48 ≧ 0
-48k + 64 ≧ 0
-48k ≧ -64
k ≦ 4/3

b.i. 12x2 + 4x – 1 = 0
From a, the value of k is 0 in this case.
Since 0 < 4/3
So this equation has real roots.

ii. 36x2 + 12x + 1 = 0
12x2 + 4x + 1/3 = 0
12x2 + 4x + (4/3) – 1 = 0
From a, the value of k is 4/3 in this case.
So, this equation has real roots.

(1) Delta = 4(m+1)^2-4(2m^2+3)
=4(m^2+2m+1-2m^2-3)
=-4(m^2-2m+2)
=-4(m^2-2m+1)-4
=-4(m-1)^2-4

任何實數m, -4(m-1)^2 最大係0, 還要-4, 所以Delta 必定是負, 所以任何實數m,x^2-2(m+1)x+(2m^2+3)=0 也沒有實根.

(2)(a) 你做左la.
(b)(i) 12x^2+4x-1=0 可寫成 12x^2+4x+0=1 ,即係當k=0, 乎合2(a)的條件, 所以 12x^2+4x-1=0 有實根

(ii) 36x^2+12x+1=0 ,兩邊除以3, 變成 12x^2+4x+1/3=0
可寫成 12x^2+4x+4/3=1 即m=4/3
亦乎合2(a)要求, 所以36x^2+12x+1=0 有實根