一元二次方程(5)

9.a. x2 – (2h + 1)x + (h2 – k + 1) = 0

判別式 = [-(2h + 1)]2 – 4(1)(h2 – k + 1)

= (4h2 + 4h + 1) – (4h2 – 4k + 4)

= 4h + 4k – 3

b. 由於方程有二重實根，故判別式 = 0

所以，4h + 4k – 3 = 0

h = (3 – 4k)/4

c. 8h + k + 1 = 0

8[(3 – 4k)/4] + k + 1 = 0

2(3 – 4k) + k + 1 = 0

6 – 8k + k + 1 = 0

7 – 7k = 0

k = 1

h = [3 – 4(1)]/4 = -1/4

d. 方程為: x2 – [2(-1/4) + 1]x + [(-1/4)2 – 1 + 1] = 0

x2 – x/2 + 1/16 = 0

16x2 – 8x + 1 = 0

(4x - 1)2 = 0

x = 1/4


10.a. y = -x2 + (h - 1)x – 2k

由於圖像與x-軸相交於一點

故判別式 = 0

(h - 1)2 – 4(-1)(-2k) = 0

h2 – 2h – 8k + 1 = 0

b. 由於圖像經過( 0, -16)

故-16 = -2k

k = 8

c. h2 – 2h – 8(8) + 1 = 0

h2 – 2h – 63 = 0

(h - 9)(h + 7) = 0

h = 9 或 -7 (捨去)

d. 圓像的方程: y = -x2 + 8x – 16

設P點的座標為(x , 0)

0 = -x2 + 8x – 16

x2 – 8x + 16 = 0

x = 4

故P的座標為(4 , 0)

e. 代(-1 , z)入方程

z = -(-1)2 + 8(-1) – 16

z = -25

故Q點的座標(-1 , -25)