10-11條 二元一次方程...........

1.
a)
5x² + 4kx + 20 = 0 有一個二重實根，所以判別式 Δ = 0
(4k)² - 4*5*20 = 0
16k² - 400 = 0
k² = 25
k = 5 或 k = -5

b)
kx² + 8x + k = 0 有一個二重實根，所以判別式 Δ = 0
(8)² - 4*k*k = 0
64 - 4k² = 0
k² = 16
k = 4 或 k = -4

c)
x² + kx - (2k + 4) = 0 有一個二重實根，所以判別式 Δ = 0
k² - 4*1*[-(2k + 4)] = 0
k² + 8k + 16 = 0
(k + 4)² = 0
k = -4 或 k = -4

d)
k(x² + 4) = 5x
kx² + 4k = 5x
kx² - 5x + 4k = 0有一個二重實根，所以判別式 Δ = 0
(-5)² - 4*k*(4k) = 0
25 - 16k² = 0
k² = 25/16
k = 5/4 或 k = -5/4


2.
a)
(k + 1) x² = 4x - 1
(k + 1) x² - 4x + 1 = 0 有二個非實數根，所以判別式 Δ < 0
(-4)² - 4*(k + 1)*1 < 0
16 - 4k - 4 < 0
12 - 4k < 0
4k > 12
k > 3

b)
3x(x - 2) = 3x - k
3x² - 6x = 3x - k
3x² - 9x + k = 0 有二個非實數根，所以判別式 Δ < 0
(-9)² - 4*3*k < 0
81 - 12k < 0
12k > 81
k > 27/4


3.
px² + 4px + 2 = 0 有一個二重實根，所以判別式 Δ = 0
(4p)² - 4*p*2 = 0
16p² - 8p = 0
2p² - p = 0
p(2p - 1) = 0
p = 1/2 或 p = 0 (不合、捨棄)


4.
a)
x = 3 或 x = 4
x - 3 = 0 或 x - 4 = 0
(x - 3)(x - 4) = 0
x² - 7x + 12 = 0

b)
x = 1/2 或 x = 2
x - (1/2) = 0 或 x - 2 = 0
2x - 1 = 0 或 x - 2 = 0
(2x - 1)(x - 2) = 0
2x² - 5x + 2 = 0