F.4 一元二次方程..超急

一﹑
1.
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
2x(x - 2) - 1(x - 2) = 0
2x² - 4x - x + 2 = 0
2x² - 5x + 2 = 0


2.
x = -3/5 或 x = -3/5
x + (3/5) = 0 或 x + (3/5) = 0
5x + 3 = 0 或 5x + 3 = 0
(5x + 3)² = 0
25x² + 30x + 9 = 0


=====
二
5x + 1 = -2x²
2x² + 5x + 1 = 0
α 和 β 為以上方程的兩根。
兩根之和： α + β = -5/2
兩根之積： αβ = 1/2


1.
(α- 2)(β - 2)
= α(β- 2) - 2(β - 2)
= αβ- 2α - 2β + 4
= αβ- 2(α + β) + 4
= (1/2) - 2 ´(-5/2) + 4
= 19/2


2.
(α -β)²
= α² - 2αβ + β²
= (α² +2αβ + β²) - 4αβ
= (α + β)² - 4αβ
= (-5/2)² - 4(1/2)
= (25/4) - 2
= 17/4


3.
由於 (α - β)² =17/4 及α> β
故此α - β = √(17/4)
α -β = (√17)/2

(α+ 2)(β - 2)
= α(β - 2) + 2(β - 2)
= αβ- 2α + 2β - 4
= αβ - 2(α - β) - 4
= (1/2) - 2(√17)/2 - 4
= (7/2)- (√17)
= (7 - 2√17)/2


4.
α² - β²
= (α + β)( α - β)
= (-5/2) ´ [(√17)/2]
= -5(√17)/4

第一題明明要求寫方程，知識長的002答案明明不是方程，是不是打得靚或知識長的答案，就算錯了也可以得票？

您好，我是 lop，高興能解答您的問題。

一

(1)

(x-1/2)(x-2)
= x^2 - 3x/2 + 1

(2)

(x+3/5)(x+3/5)
= x^2 + 6x/5 + 9/25

二

5x + 1 = -2x^2
2x^2 + 5x + 1 = 0
x = (-5±√17)/4

α = ( - 5 + √17 ) / 4
β = ( - 5 - √17 ) / 4

(1)

(α-2)(β-2)
= αβ - 2α - 2β + 4
= αβ - 2(α+β) + 4
= 1/2註(1) - 2(-5/2)註(1) + 4
= 1/2 + 5 + 4
= 19/2

(2)

(α-β)^2
= α^2 - 2αβ + β^2
= α^2 + β^2 - 2(1/2)註(1)
= α^2 + β^2 - 1
= [(-5+√17)/4]^2 + [(-5-√17)/4]^2 - 1
= (25-10√17+17)/16 + (25+10√17+17)/16 - 1
= (84)/16 - 1
= 21/4 - 1
= 17/4

(3)

(α+2)(β-2)
= αβ - 2α + 2β + 4
= 1/2註(1) + 2(β-α) + 4
= 1/2 + 2{[(-5-√17)/4]-[(-5+√17)/4]} + 4
= 9/2 - 2(2√17)/4
= 9/2 - √17

(4)

α^2-β^2
= (α+β)(α-β)
= (-5/2)註(1)(√17/4)註(2)
= -5√17/8

註：

(1) 兩根積 = c/a , 兩根和 = -b/a
(2) α - β = √17/4 是據 (2) 和 (3) 得出

2011-10-29 18:17:02 補充：
對不起我是錯了，所以最佳解答不會是我。

1.1. (2x-1)(x-2)
=2x^2-5x+2
1.2.(5x+3)^2
=25x^2+30x+9

2x^2+5x+1=0
a+b=-5/2
ab=1/2

2.1.(a-2)(b-2)
=ab-2(a+b)+4
=1/2+5+4
=19/2
2.2. (a-b)^2
=a^2+b^2-2ab
=(a+b)^2-4ab
=17/4
2.3. (a+2)(b-2)
=ab-2a+2b-4
=ab-2(a-b)-4
Since (a-b)^2=17/4
a-b=square root(17/4)
=0.5-2square root(17/4)-4
=-7.62
2.4.a^2-b^2
=(a+b)(a-b)
=square root(17/4) (-5/2)
=-5.15