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Math Help - algebraic equation problem

  1. #1
    Junior Member
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    algebraic equation problem

    There are 2 pairs of numbers: (a,c) & (b,d).

    (a,c) is the root of x^2 + ax - b = 0

    (b,d) is the root of x^2 + cx + d = 0

    Find all possible real values for a, b, c and d.
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  2. #2
    Senior Member I-Think's Avatar
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    Systems

    It's the undertaker. Hi

    a and c are the roots of the equation x^2+ax-b=0
    Then (x-a)(x-c)=x^2+ax-b
    x^2-(a+c)x+ac=x^2+ax-b
    Equating coefficients
    -(a+c)=a ac=-b....(1)
    c=-2a.....(2)

    b and d are the roots of the equation x^2+cx+d=0
    Then (x-b)(x-d)=x^2+cx+d
    x^2-(b+d)+bd=x^2+cx+d
    Equating coefficients
    -(b+d)=c...(3) bd=d.....(4)

    We have a system of 4 equations
    ac=-b....(1) c=-2a.....(2)
    -(b+d)=c...(3)  bd=d.....(4)


    From (4):  bd=d
     b=1

    Into (3): -1-d=c and into (1): ac=-1
    (2) into (1): -2a^2=-1\longrightarrow a^2=1/2
    a=\pm\sqrt\frac{1}{2}

    Hence c =-2(\pm\sqrt\frac{1}{2})

    Into (4): -(1+d)=-2(\pm\sqrt\frac{1}{2})
    d=2(\pm\sqrt\frac{1}{2})-1

    N.B. When choosing values to input into the equation, if you let one of a,c,d be positive, then all the others have to be positive and vice-versa.
    Possible values

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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by I-Think View Post
    It's the undertaker. Hi

    a and c are the roots of the equation x^2+ax-b=0
    Then (x-a)(x-c)=x^2+ax-b
    x^2-(a+c)x+ac=x^2+ax-b
    Equating coefficients
    -(a+c)=a ac=-b....(1)
    c=-2a.....(2)

    b and d are the roots of the equation x^2+cx+d=0
    Then (x-b)(x-d)=x^2+cx+d
    x^2-(b+d)+bd=x^2+cx+d
    Equating coefficients
    -(b+d)=c...(3) bd=d.....(4)

    We have a system of 4 equations
    ac=-b....(1) c=-2a.....(2)
    -(b+d)=c...(3)  bd=d.....(4)


    From (4):  bd=d
     b=1

    Into (3): -1-d=c and into (1): ac=-1
    (2) into (1): -2a^2=-1\longrightarrow a^2=1/2
    a=\pm\sqrt\frac{1}{2}

    Hence c =-2(\pm\sqrt\frac{1}{2})

    Into (4): -(1+d)=-2(\pm\sqrt\frac{1}{2})
    d=2(\pm\sqrt\frac{1}{2})-1

    N.B. When choosing values to input into the equation, if you let one of a,c,d be positive, then all the others have to be positive and vice-versa.
    Possible values
    a=b=c=d=0?

    CB
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