[請益] 一元二次不等式
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
[解法1] 無實數解表示跟x軸沒有交點,故判別式<0
2
(-2a)-4a(2a-3)<0
得 a>3或a<0
2
[解法2] 無實數解表示代入任意x皆不滿足ax-2ax+2a-3<0
2
表示代入任意x皆滿足 ax-2ax+2a-3≧0
所以開口向上a>0且和x軸最多一個交點
故 判別式≦0
2
(-2a)-4a(2a-3)≦0
得 a≧3或a≦0(不合)
[解法1]是錯在哪裡?? 感恩!!
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※ 編輯: coco100 (114.46.49.242), 11/05/2015 21:28:46
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