4題數學題

1)
已知f(x)=2x和g(x)=x²
若g[f(x)]-f[g(x)]=16,求x的值
g[f(x)] - f[g(x)] = 16
g(2x) - f(x²) = 16
(2x)² - 2(x²) = 16
4x² - 2x² = 16
2x² = 16
x² = 8
x = ±√8
x = ±2√2
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2)
二次方程5x²-kx-36=0的其中一個根是k,
(a) 求k的值
設 f(x) = 5x²-kx-36
若 k 是 f(x) = 0 其中一個根，則 f(k) = 0
f(k) = 0
5(k)² - k(k) - 36 = 0
5k² - k² - 36 = 0
4k² = 36
k² = 9
k = ±√9
k = ±3
(b) 用(a)的結果,解方程5x²-kx-36=0
由(a), k = ±3
若 k = 3
f(x) = 0
5x²-3x-36 = 0
(x-3)(5x+12) = 0
x-3 = 0 或 5x+12 = 0
x = 3 或 x = -12/5
若 k = -3
f(x) = 0
5x²-(-3)x-36 = 0
5x²+3x-36 = 0
(x+3)(5x-12) = 0
x+3 = 0 或 5x-12 = 0
x = -3 或 x = 12/5
所以 x = ±3 或 x = ±12/5。
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3)
(a)若方程5kx²-8kx+3k+1=0有2個相等的實根,求k的值
設 f(x) = 5kx²-8kx+3k+1
首先 x² 常項 5k != 0
即 k != 0
判別式△
= (-8k)² - 4(5k)(3k+1)
= 64k² - 60k² - 20k
= 4k² - 20k
= 4k(k-5)
f(x) = 0 有兩個相等的根，即判別式 = 0
4k(k-5) = 0
k = 0 或 k-5 = 0
k = 0 (捨去) 或 k = 5
所以 k = 5
(b)用(a)的結果,解方程5kx²-8kx+3k+1=0
由 (a), k = 5
f(x) = 0
5(5)x² - 8(5)x + 3(5) + 1 = 0
25x² - 40x + 16 = 0
(5x)² - 2(5x)(4) + 4² = 0
(5x - 4)² = 0
5x - 4 = 0
x = 4/5
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4)
(a)二次方程x²-2(k+1)x+(k²+5)=0有二重根
i) 求k的值
設 f(x) = x²-2(k+1)x+(k²+5) ... (*)
判別式△
= [2(k+1)]² - 4(1)(k²+5)
= 4k² + 8k + 4 - 4k² - 20
= 8k - 16
f(x) = 0 有二重根，即判別式 = 0
8k-16 = 0
k-2 = 0
k = 2
ii)解方程x²-2(k+1)+(k²+5)=0
由 (i)，k = 2
f(x) = 0
x² - 2(2+1)x + (2²+5)=0
x² - 6x + 9 = 0
(x - 3)² = 0
x-3 = 0
x = 3
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(b)(i)若二次方程(x/x+1)²-2(k+1)(x/x+1)+(k²+5)=0中k的值是a)(i)的結果,問這個方程是否二重根,為什麼?
是。因為如果設 y = x/(x+1)
(x/x+1)²-2(k+1)(x/x+1)+(k²+5)=0
y² - 2(k+1)y + (k²+5) = 0
f(y) = 0
根據 (a)(i)，f(y) 有二重根。
(ii)解方程(x/x+1)²-2(k+1)(x/x+1)+(k²+5)=0
從 (b)(i) 及 (a)(ii)，
f(y) = 0
y = 3
x/(x+1) = 3
x = 3(x+1)
x = 3x+3
2x = -3
x = -3/2

2006-12-28 15:34:36 補充：
小小補充：ax² + bx + c = 0 的方程，判別式△可判斷出：如果 △＞0，方程有實根。如果 △＜0，方程無實根。如果 △=0，方程有重根。