Re: [請益] 二元一次不等式
看板CS_TEACHER (補教老師)作者stevenyenyen (steven)時間9年前 (2015/11/06 00:42)推噓21(21推 0噓 97→)留言118則, 4人參與討論串2/2 (看更多)
※ 引述《coco100 (童話故事的最後)》之銘言:
: 2
: 設二次不等式 ax-2ax+2a-3<0,若無實數解,求a的範圍?
: [答 案] a≧3
代表說ax^2-2ax+2a-3 >= 0
表示恆在x軸(包含x軸)之上
所以開口必定朝上
所以二次項 a >= 0----條件1
因此要開口朝上且無雙根,只能有重複根 因此判別式D <= 0
D=(-2a)^2-4a(2a-3) <= 0
所以a >= 3 或是a <= 0----條件2
結合條件1.2 可得a >= 3
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補上最直白的說法
原式可化為a(x-1)^2+a-3 < 0 無實數解
所以此式代表 a(x-1)^2+a-3 >= 0
要此式成立
表示此圖形-拋物線要符合下列條件
條件1開口朝上 條件2頂點(1,a-3)在x軸之上
a>0 a >= 3
結合條件1.2 可得a >= 3
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※ 編輯: stevenyenyen (180.176.72.116), 11/06/2015 00:44:13
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※ 編輯: stevenyenyen (180.176.72.116), 11/06/2015 23:26:58
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