好煩但又好簡單0既F.4數

2 (2x+3)^2+ 6x + 7 = 0
先將 (2x+3)^2 展開:
2 [(2x)^2 + 2(2x)(3) +(3)^2] + 6x + 7 = 0
2 (4x^2 + 12x + 9) + 6x + 7 = 0
8x^2 + 24x + 18 + 6x + 7 = 0
8x^2 + 30x + 25 = 0
利用因式分解得出:
(4x + 5) (2x + 5) = 0
跟著:
4x + 5 = 0 或 2x + 5 = 0
x = -5/4 或 x = -5/2

有錯請指正!

2007-10-29 22:42:41 補充：
你所指的(4x + 5) (2x + 5) = 0 是用十字相乘去進行因式分解.我其實不太清楚你所指. 不過, 我嘗試用第二個方法做給你一看.8x^2 + 30x + 25 = 08x^2 + 10x + 20x + 25 = 02x(4x + 5) + 5(4x + 5) = 0抽出公因式:(4x + 5)(2x + 5) = 0如果仍有問題, 歡迎提出.

2 ( 2x + 3 )2+ 6x + 7 = 0
2 ( 4x2 + 12x + 9 ) + 6x + 7 = 0
8x2 + 30x + 25 = 0
( 4x + 5 ) ( 2x + 5 ) = 0
x = -5/4 or x = -5/2

2(4x^2+12x+9)+6x+7=0
8x^2+30x+25=0
(4x+5)(2x+5)=0
x=-5/4 or x=-5/2