Maths(HKMO)
2010-01-01 9:37 pm
Consider the quadratic equation x^2 - (a - 2)x - a - 1 = 0, where a is a real number. Let α and β be the roots of the equation. Find the value of a such that the value of α^2+β^2 will be the least.
回答 (3)
2010-01-01 9:40 pm
✔ 最佳答案
α^2+β^2 =(α+β)^2-2αβ=(a-2)^2+2(a+1)=a^2-4a+4+2a+2=a^2-2a+6=(a-1)^2+5 So the value of a is 1
2010-01-02 2:35 am
α + β = a-2
α β = -a-1
α^2+ β^2
=(α + β)^2 -2αβ
=(a-2)^2-2(-a-1)
=a^2-4a+4+2a+2
=a^2-2a+6
=(a-1)^2+5
Since α^2 + β^2 will be the least when (a-1)^2=0
So, the value of a is 1.
α β = -a-1
α^2+ β^2
=(α + β)^2 -2αβ
=(a-2)^2-2(-a-1)
=a^2-4a+4+2a+2
=a^2-2a+6
=(a-1)^2+5
Since α^2 + β^2 will be the least when (a-1)^2=0
So, the value of a is 1.
2010-01-01 9:52 pm
α+β =a-2
αβ=-a-1
α^2+β^2
=(α+β)^2-2αβ
=(a-2)^2-2(-a-1)
=a^2-4+2a+2
=a^2+2a-2
根住仲有冇得做?
唔係幾識...
αβ=-a-1
α^2+β^2
=(α+β)^2-2αβ
=(a-2)^2-2(-a-1)
=a^2-4+2a+2
=a^2+2a-2
根住仲有冇得做?
唔係幾識...
收錄日期: 2021-04-23 18:26:03
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https://hk.answers.yahoo.com/question/index?qid=20100101000051KK02388