[中學] 方程式
(一) 2 ╴ ╴
設m屬於實數，若x - (3+√2 )x + √2m-4 =0 恰有一整數根，m=_____
解：
設此整數根為α
2 ╴ ╴
α - (3+√2)α + √2m - 4 =0
2 ╴
α - 3α + √2m - 4 =0
整數放一起
2 ╴
(α - 3α - 4) + √2 (- α + m) =0
------------- --------
↓ ↓
屬於整數 屬於整數
若成立 則
↓ ↓
=0 =0
2
┌α -3α -4 =0
┤
└-α + m =0
┌(α-4)(α+1)=0
┤
└α = m
得α=4 或 -1 即m=4 或 -1
----------------------------------------------------------
問題來了= =
為什麼要設根為α?? 我看好像只是把x代換成α而已 不設α的話會造成什麼結果???
還有~~~恰有一整數根 這是什麼意思????
b^2-4ac=0嗎????
實在是看不懂一整數根跟方程式的關係…
是要設x=整數嗎?
然後在x=整數 m=整數的條件下 推論出二個括號都是=0 ??
這算法真的好奇怪
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