Quadratic Equations

1)
(3x+1)²=5
9x² 6x 1-5=0
9x² 6x-4=0
x=(-6 6√5)/18 or x=(-6-6√5)/18
x=(-1 √5)/3 or x=(-1-√5)/3

2)
As -3 is a root, (x 3)(x² 6x 8)=0
(x 3)(x 4)(x 2)=0
x=-4,-3or-2

3)
obviously 1 is a root, (x-1)(x² 7x 10)=0
(x-1)(x 2)(x 5)=0
x=-5,-2or1

4)
x²-8x-9=0
(x-9)(x 1)=0

2008-03-01 22:24:16 補充：
1)
(3x+1)²=5
9x²+6x+1-5=0
9x²+6x-4=0
x=(-6+6√5)/18 or x=(-6-6√5)/18
x=(-1+√5)/3 or x=(-1-√5)/3

2)
As -3 is a root, (x+3)(x²+6x+8)=0
(x+3)(x+4)(x+2)=0
x=-4,-3or-2

3)
obviously 1 is a root, (x-1)(x²+7x+10)=0
(x-1)(x+2)(x+5)=0
x=-5,-2or1

4)
x²-8x-9=0
(x-9)(x+1)=0

冇晒d加號= =

2008-03-03 06:48:03 補充：
睇個constant term，24可以分解做-24,-12,-8,,-6,-4-3,-2,-1,1,2,3,4,6,8,12,24等數，你可以試下邊個代人條式到之後會等於零

代到-3會等於零即係(x+3)係其中一個數啦
咁就可以寫成(x+3)(Ax²+Bx+C)=0
C最易搵，因為constant term 係24，而3C係constant term，所以C=8
再睇x-cube個term，呢個coefficient係1，所以A係1
所以(x+3)(x²+Bx+8)=0
剩B，可以睇x-term ge coefficient，即係8+3B=26，所以B係6