中四一元二次方程應用題急20分(要過程)

1. 設該兩數為a和b, a > b
a + b = 10 ... (1)
a2 + b2 = 52 ... (2)
(1) x (1) => a2 + 2ab + b2 = 100 ... (3)
(3) - (2) => 2ab = 48
So a2 - 2ab + b2 + 4ab = 100
(a - b)2 = 100 - 96 = 4
a - b = 2 ... (4)
(1) + (4) => 2a = 12 = > a = 6 = > b = 4
2. 設邊框的闊度為x cm
卡紙面積 = 24 * 20 = 480
相片長和闊分別為24 - 2x 和20 - 2x
相片面積 = (24 - 2x)(20 - 2x) = 480 * 2/3 = 320
(12 - x)(10 - x) = 80
120 - 22x + x2 = 80
x2 - 22x + 40 = 0
(x - 20)(x - 2) = 0
x = 2 或 x = 20(不合)
邊框的闊度為2 cm
3. 設邊長為a, a + 2 及a + 4
畢氏定理 a2 + (a + 2)2 = (a + 4)2
a2 + a2 + 4a + 4 = a2 + 8a + 16
a2 - 4 - 12 = 0
(a - 6)(a + 2) = 0
a = 6 或 a = -2(不合)
此三角形的三邊之長為6, 8 及10
4. 設收費共減少$100n, 可多吸引4n人參加
總人數 = 80 + 4n
總收費 = (80 + 4n)(4000 - 100n) = 350000
(80 + 4n)(40 - n) = 3500
3200 + 160n - 80n - 4n2 = 3500
4n2 - 80n + 300 = 0
n2 - 20n + 75 = 0
(n - 15)(n - 5) = 0
n = 5 或n = 15
該纖體計畫的收費應定為 4000 - 5(100) = $3500(比較合理)或4000 - 15(100) = $2500
5. 設三邊長度為a, b, c不妨假設c > a,b及c = b + 1
a + b + c = 30
a + b + b + 1 = 30
a = 29 - 2b ... (1)
a2 + b2 = c2
a2 + b2 = (b + 1)2
a2 = 2b + 1 ... (2)
代(1)入(2) => (29 - 2b)2 = 2b + 1
841 - 116b + 4b2 = 2b + 1
4b2 - 118b + 840 = 0
2b2 - 59b + 420 = 0
(2b - 35)(b - 12) = 0
b = 17.5 (不合,超過周界一半) 或 b = 12
b = 12 => c = 13 => a = 5
三邊的長度為5cm, 12cm, 13cm
6. 設其一正方形邊長 = x cm
該正方形周界 = 4x
另一正方形周界 = 32 - 4x, 其邊長為 8 - x
兩正方形的面積之和 = x2 + (8 - x)2 = 34
x2 + 64 - 16x + x2 = 34
2x2 - 16x + 30 = 0
x2 - 8x + 15 = 0
(x - 3)(x - 5) = 0
x = 3 或 x = 5
x = 3 => 8 - x = 5
x = 5 => 8 - x = 3
正方形邊長為 3cm 及5cm
7. (a) 18 = 12t - 2t2
t2 - 6t + 9 = 0
(t - 3)2 = 0
t = 3
3秒時離地面18米
(b) 0 = 12t - 2t2
2t(t - 6) = 0
t = 0 (不合) 或t = 6
6秒時跌回地上

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