數學 急救呀

1.
Delta = [2(a+b)]^2 - 4(a^2 + b^2)(2)
= 2a^2 + 2b^2 + 4ab - 8a^2 - 8b^2
= 4ab - 6a^2 - 6b^2
= -2(a^2 + 2ab + b^2) - 4a^2 - 4b^2
= -2(a - b)^2 - 4a^2 - 2b^2 < 0.
2.
(1) Delta > 0 for unequal real roots. That is
4^2 - 4p > 0
4 - p > 0
p < 4.
(2)
a + b = -4............(1) and
ab = p............(2)
a^2 + b^2 + a^b^2 + 3(a + b) -19
= (a + b)^2 - 2ab + p^2 + 3(a + b) - 19
= 16 - 2p + p^2 - 12 - 19
= -2p + p^2 -15 = 0
(p -5)(p+3) = 0
so p = 5 or -3.
3.
ar^4 - a = 15.............(1)
ar^3 - ar = 6..............(2)
(2)/(1) we get
(r^4 - 1)/(r^3 - r) = 15/6
6r^4 - 6 = 15r^3 - 15r
6r^4 - 15r^3 + 15r - 6 = 0.
By factor theorem, we can find that (r-1) and (r -2) are factors, that is
3(r - 1)(r -2)(2r^2 + r - 1) = 0
3(r-1)(r-2)(2r-1)(r+1) = 0
so r = 1, 2 , 1/2 and -1.
r = 1 rej. because r^3 - r = 0.
r = 2, a = 1.
r = 1/2, a = -16.
r = -2 rej. since it cannot match both (1) and (2).
so a3 = ar^2 = 4 or -4.

2008-11-28 22:49:36 補充：
Correction: There is mistake in Q1 although delta can still proved to be negative. Please take 001's answer for reference.

1)
△=4(a+b)²-4(a²+b²)(2)
=4a²+8ab+4b²-8a²-8b²
=-4a²+8ab-4b²
=-4(a-b)²<0 (a≠b)
∴(a²+b²)x²+2(a+b)x+2=0 沒有實數根

2)
因方程有2個不等根
∴△=4²-4(1)(p)>0
則p<4

之後果d都唔知你問咩@@

2008-11-28 22:45:15 補充：
一係問番好D