1. 若α和β為方程2x^2 - 3x + 4 = 0的根 , 試求下列各式之值:
a) α^2 ＋ β^2
b) α/β ＋ β/α
F.4 a.maths
2007-10-22 5:03 am
回答 (2)
2007-10-22 5:23 am
✔ 最佳答案
1a) α + β = 3 / 2 αβ = 4 / 2 = 2
α^2 ＋ β^2 = ( α ＋ β )^2 - 2αβ
= ( 3 / 2 )^2 - 2 ( 2 )
= -1.75
b) α/β ＋ β/α
= ( α^2 ＋ β^2 ) / αβ
= ( - 1.75 ) / 2
= -0.875
參考: My Maths Knowledge
2007-10-22 5:23 am
2x^2 - 3x + 4 = 0
α+β=-(-3)/2
=3/2
αβ=4/2
=2
a) α^2 ＋ β^2
=(α+β)² - 2αβ
=(3/2)² -2(2)
=9/4 - 4
b)α/β ＋ β/α
=(αβ+αβ)/αβ
=﹝(3/2)+(3/2)﹞/ 2
=4/2
=2
2007-10-21 21:25:51 補充：
唔好意思第2題睇錯左..係上面果位仁兄岩~!
α+β=-(-3)/2
=3/2
αβ=4/2
=2
a) α^2 ＋ β^2
=(α+β)² - 2αβ
=(3/2)² -2(2)
=9/4 - 4
b)α/β ＋ β/α
=(αβ+αβ)/αβ
=﹝(3/2)+(3/2)﹞/ 2
=4/2
=2
2007-10-21 21:25:51 補充：
唔好意思第2題睇錯左..係上面果位仁兄岩~!
參考: myself
收錄日期: 2021-04-13 14:04:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071021000051KK05089