設方程式x^2+ax-a+1=0之兩根均為整數....

則a=?

1 個解答

評分
  • 匿名使用者
    2 0 年前
    最佳解答

    令兩根為b, c皆為整數且b > c

    x^2+ax-a+1=(x- b)(x- c)= x^2 - (b+c)x+ bc

    比較係數

    b+ c = -a..............(1)

    bc = - a+ 1 ..........(2)

    (1)代入(2)

    =>bc = b+ c + 1

    =>bc - b - c +1 =2

    =>(b-1)(c-1) =2

    因為b, c皆為整數,所以b - 1, c - 1皆為整數

    b - 1 = 2 => b = 3

    c - 1 = 1 => c = 2

    由(1) a = - (b + c) = -5

    b - 1 = - 1 => b = 0

    c - 1 = - 2 => c = -1

    由(1) a = - (b + c) = 1

    所以 a = 1 或 - 5

還有問題?馬上發問,尋求解答。