Maths F4...14

1. By Viète's Laws, let x and y be the 2 roots of
ax²+bx+c=0,
x+y=-b/a, xy=c/a.
So,
a+b=-p
ab=q
(a+3)+(b+3)=-(3+p)
(a+3)(b+3)=2
So, after solving these 4 equation,
a=-1, b=-1, p=2, q=1.
2a=-2,2b=-2.
So, the equation is (x+a)²=0

圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701108230142413873448070.jpg




2011-08-25 13:49:08 補充：
which part dont u understand?
Viète's Laws or process of the 4 equation??

also,do u know the answer?
am i right?

2011-08-25 13:51:27 補充：
OPPS, CORRECTION!!
wrong->So, the equation is (x+a)²=0
right->So, the equation is (x+2)²=0

2011-08-26 13:35:43 補充：
OPPS, CORRETION!!!

wrong->(a+3)+(b+3)=-(3+p)
right->(a+3)+(b+3)=-(3+q)

wrong->(a+3)(b+3)=2
right->(a+3)(b+3)=2p

im so sorry for the mistake...
is it ok now?