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Math Help - nature of roots

  1. #1
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    nature of roots

    Find the nature of roots for
    (x-a)(x-b) + (x-c)(x-a) + (x-b)(x-c) = 0
    I started solving the above equation.
    This was deduced to
    3x2 2(a + b+c)x +(ab+bc+ca) =0
    from this, the discreminant
    4(a + b+c)2 -4(3)( ab+bc+ca)
    4[a2+b2+c2+2(ab+bc+ca)-3(ab+bc+ca)]
    4[a2+b2+c2-(ab+bc+ca)]
    From this I am not able to say what type are the roots.
    Kindly guide me.
    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: nature of roots

    Completing squares we can write the discriminant D in the form

    D=\ldots=(2a-b-c)^2+3(b-c)^2

    This implies D\geq 0 for all a,b,c\in\mathbb{R} i.e. all the roots are real. Besides, the roots are equal if and only if (2a-b-c=0)\;\wedge\; (b-c=0) or equivalently if and only if a=b=c .
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  3. #3
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    Re: nature of roots

    Thanks Mr.Fernando Revilla.
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