Feb 17th, 2004, 21:34 | #1 |
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注册日期: Jul 2004
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如何求ax2+bx+c=0的根? (2表示x的平方) 谢谢 |
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Feb 18th, 2004, 07:57 | 只看该作者 #4 |
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注册日期: Jul 2004
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要考虑b^2 - 4ac <0 =0 >0 的不同情况 >0:两个实根,同六里台之解 =0:一个实根, X= - b/2a <0:两个虚根, X = - b/2a +/- (4ac - b^2)^0.5 i <BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by 六里台: X1 = - b/2a + (b^2 - 4ac)^0.5 X2 = - b/2a - (b^2 - 4ac)^0.5 ********************************** 愁谁似我? 致杜鹃长啼而泣血,残月难堪而云遮? 几人痴情?几人豪杰?几人不老?几人谈笑几多年? 纵使两心缭绕,芳情缠绵,白头谐老古亦难! 便有冲天豪气,胸藏天机,于国于家也难全。 诸葛六出,难扶刘禅,鹏举英雄,铮铮铁骨风波断! 叹世事难料,任尔韬略也枉然。<HR></BLOCKQUOTE> |
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