Re: [中學] 請問一題高中數學
: 請問一下這題不知道如何算
兩邊平方後
a^2x^2+4ax+(4-b^2)>0
依題意可知
(x-1)(x-7)>0,x^2-8x+7>0
比較係數後
a^2/1=4a/(-8)=(4-b^2)/7
交叉相乘後
-8a^2=4a,-2a^2=a,a(2a+1)=0
由於a不等於0,減根後
a=-1/2 , b=3/2 or -3/2
但b=-3/2時
|(-1/2)x+2|>-3/2
|x-4|>-3
x>1 or x<7不合
因而
a=-1/2,b=3/2...ans
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※ 編輯: wayne2011 (61.58.103.35), 03/22/2017 15:46:47
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