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2003 HKCEE Maths MC Q.41

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發問:

Q.41) let k be a constant if a and b are the roots of the equation x^2 - 3x + k = 0 then, a^2 + 3b = ?? (Correct ans: 9 - k) could anyone show me the steps of calculation? many thzzz~

最佳解答:

Since a and b are roots of x^2 - 3x + k = 0, a+b=0 ab=k Also a is a root of x^2 - 3x + k = 0, a^2 - 3a + k = 0 ∴a^2=3b-k ∴a^2+3b=3a-k+3b=3(a+b)-k=3(3)-k=9-k 這是舊課程題目 2007-04-09 00:36:18 補充: 剛才打錯Since a and b are roots of x^2 - 3x + k = 0,a+b=3ab=kAlso a is a root of x^2 - 3x + k = 0,a^2 - 3a + k = 0∴a^2=3b-k∴a^2+3b=3a-k+3b=3(a+b)-k=3(3)-k=9-k這是舊課程題目 2007-04-09 00:40:00 補充: 實在抱歉,又打錯,這次一定不會錯Since a and b are roots of x^2 - 3x k = 0,a b=3ab=kAlso a is a root of x^2 - 3x k = 0,a^2 - 3a k = 0∴a^2=3a-k∴a^2 3b=3a-k 3b=3(a b)-k=3(3)-k=9-k這是舊課程題目 2007-04-09 00:41:42 補充: 唉,怎麼我又漏了加號Since a and b are roots of x^2 - 3x k = 0,a b=3ab=kAlso a is a root of x^2 - 3x k = 0,a^2 - 3a k = 0∴a^2=3a-k∴a^2 3b=3a-k 3b=3(a b)-k=3(3)-k=9-k這是舊課程題目 2007-04-09 00:42:41 補充: Since a and b are roots of x^2 - 3x k = 0,a+b=3ab=kAlso a is a root of x^2 - 3x k = 0,a^2 - 3a k = 0∴a^2=3a-k∴a^2 3b=3a-k 3b=3(a b)-k=3(3)-k=9-k 2007-04-09 00:47:04 補充: Since a and b are roots of x^2 - 3x + k = 0,a+b=3ab=kAlso a is a root of x^2 - 3x + k = 0,a^2 - 3a + k = 0∴a^2=3a-k∴a^2+3b=3a-k+3b=3(a+b)-k=3(3)-k=9-k

其他解答:

要hkcee maths 詳解可到 http://www.takwing.idv.hk/
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