F.4數學3題~~~~~help me please~~~~

2010-08-06 8:36 pm

1)
Simplify (3√2+4)^5 (3√2-4)^6.2)
Let a(alpha) and b(beta) be two roots of the equation x^2+kx+9=0, where k is a constant.
(a) Express (a+2)(b+2) in terms of k.
(b) If a+2 and b+2 are two roots of the equation y^2+ay+b=0, where a and b are constants, express a and b in terms of k.
(c) If the equation y^2+ay+b=0 has equal real roots, find the values of k.3)
(a) Find the L.C.M. of x^2-3x-10 and x^3+10x^2+16x.
(b) Simplify 1/(x^2-3x-10) - 1/(x^3+10x^2+16x).

回答 (1)

知識就是力量

2010-08-06 9:24 pm

✔ 最佳答案

1)
Simplify
(3√2+4)^5 (3√2-4)^6
= (3√2+4)^5 (3√2-4)^5 (3√2-4)
= [(3√2+4)(3√2-4)]^5 (3√2-4)
= (18-16)^5 (3√2-4)
= 32(3√2-4)
2)(a)
α+β = -k
αβ = 9
(a+2)(β+2)
= αβ + 2α + 2β + 4
= 9 + 2(α+β) + 4
= 9 + 2(-k) + 4
= 13 - 2k
(b) Sum of root = α+2 + β+2 = α+β+4 = 4-k
Product of root = (a+2)(β+2) = 13-2k
For the equation y^2 + ay + b = 0,
Sum of root = -a
4-k = -a
a = k-4
Product of root = b
13-2k = b
b = 13-2k
(c) Δ = 0
a^2 - 4b = 0
(k-4)^2 - 4(13-2k) = 0
k^2 - 8k + 16 - 52 + 8k = 0
k^2 = 36
k = ±6

3)(a)
x^2 - 3x - 10 = (x-5)(x+2)
x^3 + 10x^2 + 16x = x(x^2+10x+16) = x(x+2)(x+8)
L.C.M. = x(x+2)(x-5)(x+8)
(b) 1/(x^2-3x-10) - 1/(x^3+10x^2+16x)
= 1/(x-5)(x+2) - 1/x(x+2)(x+8)
= [x(x+8) - (x-5)] / x(x+2)(x-5)(x+8)
= [x^2 + 8x - x + 5] / x(x+2)(x-5)(x+8)
= (x^2+7x+5) / x(x+2)(x-5)(x+8)

2010-08-06 13:29:51 補充：
Amendment:
2)(a)
α+β = -k
αβ = 9
(a+2)(β+2) , a should be α

參考: Knowledge is power.

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