一條因式分解同一條解方程

第一題是 factorization
3x^2 - x -2 1
(3x 2)(x-1) 1 有餘數既....


√x 1=5-x
x √x - 4 = 0

√x = [-1 /- √(1 16)]/2
x = [-1 /- √(17)]^2/4
x = 18-2√(17) / 4 = [9-√(17)]/2
or
x = 18 2√(17) / 4 = [9 √(17)]/2

2008-07-19 10:46:31 補充：
√(x+1)=5-x
x+1 = (5-x)^2
x+1 = 25- 10x + x^2
x^2 -11x + 24 = 0
(x-8)(x-3) = 0
x=3 or 8

√(x+1) = √(3+1)= 2 or -2(If necessary)
√(x+1) = √(8+1)= 3 or -3 (necessary)

5-3 = 2 = LHS
5-8 = -3 = LHS

1)
3x2-x-1
=x(3x-1)-1
2)
√(x+1)=5-x
[√(x+1)] 2 =(5-x) 2
x+1=25-10x+x2
0=25-1-10x-x+x2
x2 -11x + 24=0
(x-3) (x-8)=0
X=8or 3

√(x+1) = (5 - x)
兩邊take平方，
[√(x+1)]^2 = (5 - x)^2
x + 1 = 25 - 10x + x^2
x^2 - 11x + 24 = 0
(x - 3)(x - 8) = 0
x = 3 or x = 8
但是當x = 8時，
L.H.S. = √(8+1) = 3
R.H.S. = (5 - 8) = -3
即是L.H.S. 不等於R.H.S.
所以x = 8是要捨去的