prove that
i,(α+β) 2 - 4αβ≧0
ii,(α-β)2 - 4αβ≧0
for any real numbers, if possible
prove numbers
2008-02-12 7:16 am
回答 (1)
2008-02-12 7:22 am
✔ 最佳答案
(i)(α+β) ^2 - 4αβ
=α^2+β^2 +2αβ - 4αβ
=α^2+β^2- 2αβ
=(α-β)^2
>=0
(ii)
(α-β) ^2 - 4αβ
=α^2+β^2 -2αβ - 4αβ
=α^2+β^2 -6αβ
So it may not be >=0
For example take α=β=1
Then (α-β) ^2 - 4αβ=-4<0
收錄日期: 2021-04-25 16:54:41
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