數學問題1條(須列式)

x^2 + 2x - 7 ≡ ( x + m )^2 + n
≡ x^2 + 2mx + ( m^2 + n )
於是,
2x = 2mx
m = 1
m^2 + n = -7
( 1 )^2 + n = -7
n = -8
所以m = 1, n = -8。

2008-08-15 16:14:03 補充：
n應等於-8, 這樣R.H.S. = ( x + 1 )^2 - 8 = x^2 + 2x + 1 - 8 = x^2 + 2x - 7 = L.H.S.