X = 什麼?

2006-11-07 6:12 am
在以下方程式中, 請求X=什麼?

aX(square)+ bX + C = 0

回答 (4)

2006-11-07 6:19 am
✔ 最佳答案
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)
2006-11-07 6:23 am
ax² + bx + c = 0 係叫做 Quadratic Equation

佢有個標準嘅答案:
x = [-b ± √(b² -4ac) ] / 2a
2006-11-07 6:21 am
ax²+ bx + C = 0

x=[-b±√(b²-4ac)]/2a
2006-11-07 6:20 am
X= [- b ± √(b^2-4aC) ]/ 2a


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