標題:
F.4 a math 問題(new)
發問:
if one root of the equation 6x^2 - (8a - 3)x + (2a^2 - a)=0 is the reciprocal of the other ,find the real values of a and the roots ps.因為個呀sir既關係...a math一直掌握得唔好,有的咩方法可以改善下?(好慘呀!!讀右唔係,唔讀右唔係~~學校要修...可恨!!)
let one root be p, other root is 1/p sum of roots = p + 1/p = (8a - 3) / 6 product of roots = p * 1/p = (2a^2 - a) / 6 = 1 2a^2 - a - 6 = 0 (2a + 3) (a - 2) = 0 a = -3/2 or a = 2 a = -3/2 p + 1/p = -5/2 2p^2 + 5p + 2 = 0 (2p + 1) ( p + 2) = 0 p = -1/2 or p = -2 a = 2 p + 1/p = 13/6 6p^2 - 13p + 6 = 0 p = 3/2 or p = 2/3 so, a = -3/2, roots are -1/2, -2 a = 2, roots are 3/2, 2/3
其他解答:
F.4 a math 問題(new)
發問:
if one root of the equation 6x^2 - (8a - 3)x + (2a^2 - a)=0 is the reciprocal of the other ,find the real values of a and the roots ps.因為個呀sir既關係...a math一直掌握得唔好,有的咩方法可以改善下?(好慘呀!!讀右唔係,唔讀右唔係~~學校要修...可恨!!)
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最佳解答:let one root be p, other root is 1/p sum of roots = p + 1/p = (8a - 3) / 6 product of roots = p * 1/p = (2a^2 - a) / 6 = 1 2a^2 - a - 6 = 0 (2a + 3) (a - 2) = 0 a = -3/2 or a = 2 a = -3/2 p + 1/p = -5/2 2p^2 + 5p + 2 = 0 (2p + 1) ( p + 2) = 0 p = -1/2 or p = -2 a = 2 p + 1/p = 13/6 6p^2 - 13p + 6 = 0 p = 3/2 or p = 2/3 so, a = -3/2, roots are -1/2, -2 a = 2, roots are 3/2, 2/3
其他解答:
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