fksnlix的部落格

標題:

• 7-11真係有Q幣買-10點@1@
• 代入消元法&加減消元法
• 這到底是為甚麼....Q_Q@1@

 

此文章來自奇摩知識+如有不便請留言告知

X = 什麼？

發問:

在以下方程式中, 請求Ｘ＝什麼？ aX(square)+ bX + C = 0

最佳解答:

X = [-b + sqrt(b^2-4aC)]/(2a) or X = [-b - sqrt(b^2-4aC)]/(2a) 2006-11-06 22:30:27 補充： aX^2 bX C = 0(aX)^2 b(aX) aC = 0(aX)^2 b(aX) (b/2)^2 - (b/2)^2 aC = 0(aX b/2)^2 - (b/2)^2 aC = 0(aX b/2)^2 = (b/2)^2 - aC 2006-11-06 22:30:53 補充： aX b/2 = sqrt((b/2)^2 - aC) or aX b/2 = sqrt((b/2)^2 - aC) X = -b/2a sqrt((b/2)^2 - aC)/a or X = -b/2a sqrt((b/2)^2 - aC)/a X = [-b 2sqrt((b/2)^2 - aC)]/(2a) or X = [-b 2sqrt((b/2)^2 - aC)]/(2a)X = [-b sqrt(b^2 - 4aC)]/(2a) or X = [-b sqrt(b^2 - 4aC)]/(2a) 2006-11-06 22:32:30 補充： aX^2 bX C = 0(aX)^2 b(aX) aC = 0(aX)^2 b(aX) (b/2)^2 - (b/2)^2 aC = 0(aX b/2)^2 - (b/2)^2 aC = 0(aX b/2)^2 = (b/2)^2 - aC 2006-11-06 22:33:11 補充： aX b/2 = sqrt((b/2)^2 - aC) or aX b/2 = sqrt((b/2)^2 - aC) X = -b/2a sqrt((b/2)^2 - aC)/a or X = -b/2a sqrt((b/2)^2 - aC)/a X = [-b 2sqrt((b/2)^2 - aC)]/(2a) or X = [-b 2sqrt((b/2)^2 - aC)]/(2a)X = [-b sqrt(b^2 - 4aC)]/(2a) or X = [-b sqrt(b^2 - 4aC)]/(2a)

其他解答:

ax2 + bx + c = 0 係叫做 Quadratic Equation 佢有個標準嘅答案: x = [-b ± √(b2 -4ac) ] / 2a|||||ax2+ bx + C = 0 x=[-b±√(b2-4ac)]/2a|||||X= [- b ± √(b^2-4aC) ]/ 2a