yanahv7的部落格

標題:

F4quatratic equation

發問:

x*(x+2)*(x+4)*(x+6) - 21 = 0 x=?

• 有無歐洲 40 天自由行路線圖及費用詳情@1@
• Kelly Clarson's Profile
• 15歲jj長度問題@1@

• 有無歐洲 40 天自由行路線圖及費用詳情@1@
• Kelly Clarson's Profile
• 15歲jj長度問題@1@

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

x (x + 2) (x + 4) (x + 6) - 21 = 0x(x + 6) (x + 2)(x + 4) - 21 = 0(x2 + 6x) (x2 + 6x + 8) - 21 = 0(x2 + 6x + 4 - 4) (x2 + 6x + 4 + 4) - 21 = 0(x2 + 6x + 4)2 - 42 - 21 = 0(x2 + 6x + 4)2 - 37 = 0(x2 + 6x + 4 - √37) (x2 + 6x + 4 + √37) = 0(x2 + 6x + 4 - √37) = 0 or (x2 + 6x + 4 + √37) = 0x = ( - 6 ± √ (62 - 4(4 - √37)) ) / 2 or x = ( - 6 ± √ (62 - 4(4 + √37)) ) / 2 x = - 3 ± √ (5 + √37) or x = - 3 ± √ (5 - √37) = - 3 ± i √ (√37 - 5) 2011-10-20 19:08:31 補充： Chak Pui( 小學級 4 級 ) 你即管抄吧,你很想被檢舉吧?

其他解答:

x (x + 2) (x + 4) (x + 6) - 21 = 0x(x + 6) (x + 2)(x + 4) - 21 = 0(x2 + 6x) (x2 + 6x + 8) - 21 = 0(x2 + 6x + 4 - 4) (x2 + 6x + 4 + 4) - 21 = 0(x2 + 6x + 4)2 - 42 - 21 = 0(x2 + 6x + 4)2 - 37 = 0(x2 + 6x + 4 - √37) (x2 + 6x + 4 + √37) = 0(x2 + 6x + 4 - √37) = 0 or (x2 + 6x + 4 + √37) = 0x = ( - 6 ± √ (62 - 4(4 - √37)) ) / 2 or x = ( - 6 ± √ (62 - 4(4 + √37)) ) / 2 x = - 3 ± i √ (√37 - 5) or - 3 ± √ (5 + √37)