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Math Help - Roots of a quadratic equation

  1. #1
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    Roots of a quadratic equation

    Can someone please help me with this?





    Given that the roots of the equation x^2+ax+(a+2)=0 differ by two, find the possible values of the constant a?





    I used the symmetry axis of the parabola \frac{-b}{2a}=1to find one of the possible values =-2, but my teacher said, when we were checking this that 6 is also a possible answer. How is this possible? Is my method of solving this correct?











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  2. #2
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    Quote Originally Posted by Coach View Post
    Can someone please help me with this?





    Given that the roots of the equation x^2+ax+(a+2)=0 differ by two, find the possible values of the constant a?
    Let r_1,r_2 be roots. Then r_1+r_2 = a and r_1r_2 = (a+2). Thus, r_1r_2 = r_1+r_2+2. But we also know that r_1-r_2=2\implies r_1 = r_2+2 thus (r_2+2)r_2 = (r_2+2)+r_2+2. Now solve.
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  3. #3
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    I solved for r and I got 2 or -2, why is only -2 a possible solution?

    I wonder how my teacher then said that also six was a possible answer?




    I know this is probably a stupid question, and that I am breaking all internet etiquette rules by pleading for an answer, but I really need this since I have a test tomorrow.

    Thank you.
    Last edited by Coach; November 20th 2007 at 11:37 AM.
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  4. #4
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    Quote Originally Posted by Coach View Post
    Given that the roots of the equation x^2+ax+(a+2)=0 differ by two, find the possible values of the constant a?
    Hello,

    the text of your problem indicates that there are real roots:
    x_1=s~\vee~x_2=t
    Then you can write the equation as:

    (x-s)(x-t)=x^2+ax+(a+2)=0 Expand

    x^2-(s+t)x+st=x^2+ax+(a+2)
    Use the property that the roots should differ by 2. You'll get a system of equations:

    \left \{\begin{array}{l}I:\ -(s+t)=a \\ II: s-t=2 \\III:\ s \cdot t = a+2\end{array} \right.

    Plug a from I into III:

    \left \{\begin{array}{l}II: s-t=2~\iff~s=t+2 \\III:\ s \cdot t = -(s+t)+2\end{array} \right.

    Plug s from II into III:

    (t+2)t=-t-2-t+2~\iff~ t^2+2t=-2t~ \iff~t^2+4t=0~ \iff~t=0~\vee~t=-4

    Then s = 2~\vee~ s= -2 and a=-2~\vee~a=6
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  5. #5
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    Thank you so much!
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