一元二之次方程怎樣計?-------10點**

1.)∵ (1–p)x2–2(2–p)x+(1–p) = 0 has no real roots
∴Δ ＜ 0
[–2(2–p)]2–4(1–p)(1–p) ＜ 0
[(2–p)]2–4(1–p)(1–p) ＜ 0
(4–4p+p2)–4(1–2p+p2) ＜ 0
4–4p+p2+8p–4p2＜ 0
–3p2+4p+4 ＜ 0
3p2–4p–4 ＜ 0
(3p+2)(p–2) ＜ 0
p＜-2/3 或 p＜2
∴p＜2

2a.)x[(2k–1)x+3](3k+1)+9 = 0
(3k + 1)(2k - 1)x2 + 3(3k + 1)x + 9 = 0
(6k2 – k - 1)x2 + (9k + 3)x + 9 = 0
∵the equation has two equal real roots
∴Δ = 0
(9k + 3)2 – 4(6k2 – k - 1)(9) =0
(81k2 + 54k + 9) – 216k2 + 36k + 36 =0
-135k2 + 90k + 45 =0
3k2 – 2k – 1 =0
(3k + 1)(k - 1) =0
k = -1/3 ( rejected ) or k = 1
∴k = 1

b)x [ ( 2k - 1 )x + 3 ] ( 3k + 1 ) + 9 = 0
x [ ( 2(1) - 1 )x + 3 ] [ 3(1) + 1 ] + 9 = 0
x [ x + 3 ] (4) + 9 = 0
4x2 +12x + 9 = 0
( 2x + 3 )2 = 0
x = -3/2


2007-09-28 15:36:34 補充：
第1題打漏左個2：[–2(2–p)]^2–4(1–p)(1–p) ＜ 0[2(2–p)]^2–4(1–p)(1–p) ＜ 02(4–4p p^2)–4(1–2p p^2) ＜ 08–8p 2p^2 8p–4p^2＜ 0–2p^2 8 ＜ 02p^2–8 ＜ 0(2p 4)(p–2) ＜ 0p＜-2 或 p＜2∴p＜2

2007-09-28 15:37:38 補充：
顯示唔到正號：[–2(2–p)]^2–4(1–p)(1–p) ＜ 0[2(2–p)]^2–4(1–p)(1–p) ＜ 02(4–4p＋p^2)–4(1–2p＋p^2) ＜ 08–8p＋2p^2＋8p–4p^2＜ 0–2p^2＋8 ＜ 02p^2–8 ＜ 0(2p＋4)(p–2) ＜ 0p＜-2 或 p＜2∴p＜2

2007-09-28 18:51:34 補充：
1.∴p＜–2

very detail & clear