F.4 MATHS!

1)
y = 2x² -------1
3x + y = 2 -------2

將 (1) 代入 (2):

3x + (2x²) = 2

3x + 2x² = 2

2x² + 3x - 2 = 0

(2x-1)(x+2) = 0

2x-1 = 0 或 x+2 = 0

x = 1/2 或 x = -2

若 x = 1/2，y = 2(1/2)² = 1/2

若 x = -2，y = 2(-2)² = 8

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2)
4 = 3(2x²-1) -------1
y = 7x -------2

4 = 3(2x²-1)

4 = 6x² - 3

6x² = 7

x² = 7/6

x = ±√(7/6)

x = ±(√42)/6

若 x = (√42)/6，y = 7[(√42)/6]

若 x = -(√42)/6，y = -7[(√42)/6]

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3)y = x² - 2x + 3 ------------- 1
y = 3x - 1 --------------2

將 (2) 代入 (1):

3x - 1 = x² - 2x + 3

x² - 5x + 4 = 0

(x-1)(x-4) = 0

x-1 = 0 或 x-4 = 0

x = 1 或 x = 4

若 x = 1，y = 3(1)-1 = 2

若 x = 4，y = 3(4)-1 = 11

解方程?
1) y = 2x^2 -------1
3x + y = 2 -------2
代1入2: 3x+2x^2=2
2x^2+3x-2=0
(x+2)(2x-1)=0
x=-2 or x=1/2
當x=-2,y=8
當x=1/2,y=1/2

2)-------1
y = 7x------------------2
1: 4 = 3(2x^2-1)
4/3=2x^2-1
7/6=x^2
x=√7/√6--------------3
代3入2: y=7√7/√6
y=7√42/6
7√42/6=7x
x=√42/6

3)y = x^2 - 2x + 3 ------------- 1
y = 3x - 1----------------------2
1=2: x^2 - 2x + 3 =3x - 1
x^2-5x+4=0
(x-4)(x-1)=0
x=4 or x=1
當x=4, y=3(4)-1=11
當x=1, y=3(1)-1=2

2006-11-28 22:14:54 補充：
2)另一情況:x=-√7/√6--------------3代3入2: y=-7√7/√6y=-7√42/6-7√42/6=7xx=-√42/6