Quadratic Equations, 高手請進

Solution

1.[4/(x+2)]-4=3/(x+3)(x≠-2, x≠-3)
[4/(x+2)]-4(x+2)/(x+2)=3/(x+3)
[4 -4(x+2)]/(x+2) =3/(x+3)
[4 -4x - 8]/(x+2) =3/(x+3)
(-4x – 4)/(x+2) =3/(x+3)
(-4x – 4) (x+3) =3(x+2)
-4x^2 – 4x – 12x -12 = 3x + 6
0 = 3x + 6 +4x^2 + 4x + 12x +12
0 = 4x^2 + 19x +18
4x^2 + 19x +18 = 0
x = {(-19) ±√ [19^2 – 4 (4)(18)]}/[(2)(4)]
x = [-19 ±√ (361 – 288)]/8
x = [-19 ±√ (361 – 288)]/8
x = (-19 ±√73)/8
x = (-19 +√73)/8 or x = (-19 -√73)/8
x = -1.307 or x = -3.443

2.
1/(x+1)=(x-1)/(x+5) (x≠-1, x≠-5)
1(x+5) =(x-1) (x+1)
x + 5 = x^2 – 1
x^2 – x – 1 – 5 = 0
x^2 – x – 6 = 0
(x + 2) (x – 3) = 0
x = -2 or x = 3

1. [4/(x+2)]-4=3/(x+3)
4(x+3) - 4(x+3)(x+2) = 3(x+2) . [multiply both side by (x+3)(x+2)]
4x+12 - 4(x^2+5x+6) = 3x+6
4x^2+19x-18=0
x=-1.307 or -3.443

2. 1/(x+1)=(x-1)/(x+5)
x+5 = (x-1)(x+1) [multiply both side by (x+1)(x+5)]
x+5 = x^2 - 1
x^2 - x -6 = 0
(x-3)(x+2) = 0
x=3 or -2

2006-10-17 09:52:05 補充：
typing error, the fourth line should be : 4x^2 19x 18=0

2006-10-17 09:53:09 補充：
typing error again, the fourth line should be : 4x^2 ┼ 19x ┼ 18=0

1)
4/(x+2) - 3/(x+3) =4
4(x+3)-3(x+2)=4(x+2)(x+3)
x+6=4(x^2+19x+6)
4X^2+19x+18=0
x=-1.307 or x=-3.443

2)
1/(x+1)= (x-1)/(x+5)
x+5 = (x+1)(x-1)
x+5=x^2-1
x^2 -x -6 =0
x=-2 or x=3