香港必買2017

標題:

F.4 A.maths Discriminant2

發問:

Determine the value of a for which each equation has real solution(D≧0) 1. x^2-(2a-1)x+1+a^2=0 2. x^2-2ax+(a+1)^2=0 3. (a+1)x^2-2(a+1)x+(a-1)=0 (a≠-1) 更新: (D≧0) 更新 2: (D>-0)

最佳解答:

Determine the value of a for which eachequation has real solution(D>=0) 1. x^2－( 2a－1)x+1+a^2=0 Sol D=( 2a－1)^2－4*1*(1+a^2) = 4a ^2－4a+1－4－4a^2 =－4a－3>=0 －3>= 4a －3/4>=a 2. x^2－2ax+(a+1)^2=0 D=(－2a)^2-4*1*(a+1)^2>= 0 a^2－a+1)^2>= 0 a^2－a^2－2a－1>=0 －2a－1>=0 －1>= 2a －1/2>=a 3. (a+1)x^2－2(a+1)x+(a－1)=0 (a≠－1) D=(－2a－2)^2－4*(a+1)(a－1)>=0 (a+1)^2－(a+1)(a－1)>= 0 a^2+ 2a +1－a^2+1>=0 2a +2>= 0 a>－1

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