F.4maths續多項式

1.
(√6 - √2) / 2√8
= (√6 - √2)(√2) / 2√8(√2)
= (√12 - √4) / 2√16
= (2√3 - 2) / 8
= (√3 - 1) / 4


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2.
3x - y = -9 ...... [1]
2x - 3y = -13 ...... [2]

[1] * 3 :
9x - 3y = -27 ...... [3]

[3] - [2] :
7x = -14
x = -2

將 x = -2 代入 [1] 中：
3(-2) - y = -9
y = 3


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3.
二次方程 (k - 2)x² - 4kx - 4 = 0 有一個二重實根。
則 判別式 ∆ = 0

(-4k)² - 4(k - 2)(-4) = 0
16k² + 16(k - 2) = 0
k² + k - 2 = 0
(k - 1)(k + 2) = 0
k = 1 或 k = -2


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4.
(a)
f(x) = x² + kx + 3

f(0) + f(1) = -3
(0)² + k(0) + 3 + (1)² + k(1) + 3 = -3
k + 7 = -3
k = -10

(b)
f(x) = x² - 10x + 3

f(a) = a² + 2
a² - 10a + 3 = a² + 2
10a = 1
a = 1/10


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5.
f(x) = 4x³ + px² + qx - 11

當 f(x) 除以 x + 3 時，餘式是 -170。
(-3) = -170
4(-3)³ + p(-3)² + q(-3) - 11 = -170
-108 + 9p - 3q - 11 = -170
9p - 3q = -51
3p - q = -17
6p - 2q = -34 ...... [1]

當 f(x) 除以 2x - 1 時，餘式是 -16。
4(1/2)³ + p(1/2)² + q(1/2) - 11 = -16
(1/2) + (1/4)p + (1/2)q - 11 = -16
2 + p + 2q - 44 = -64
p + 2q = -22 ...... [2]

[1] + [2] :
7p = -56
p = -8

將 p = -8 代入 [2] 中：
(-8) + 2q = -22
2q = -14
q = -7

2014-02-03 16:51:11 補充：
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