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Math Help - Real roots

  1. #1
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    Real roots

    Hi,

    Prove that the roots of the following equation are real:

     (a-b+c)x^2 + 4(a-b)x + (a-b-c) = 0

    Thanks,
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  2. #2
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    Hi

    Quote Originally Posted by sancellX View Post
    Hi,

    Prove that the roots of the following equation are real:

     (a-b+c)x^2 + 4(a-b)x + (a-b-c) = 0

    Thanks,
    This equation has real solutions iff \Delta \ge 0, where \Delta is the discriminant.

    mx^2+nx+p=0 \implies \Delta=n^2-4mp

    Therefore :

    \Delta=[4(a-b)]^2-4(a-b+c)(a-b-c)

    \Delta=16(a-b)^2-4{\color{red}[(a-b)+c][(a-b)-c]}


    The red part is something like (x+y)(x-y)=x^2-y^2, with x=(a-b) and y=c.


    So \Delta=16(a-b)^2-4[(a-b)^2-c^2]=16(a-b)^2-4(a-b)^2+4c^2=12(a-b)^2+4c^2


    Knowing that a squared number is always positive, what can you conclude ?
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  3. #3
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    evaluate the discriminate b^2 - 4ac

    4^2(a-b)^2 - 4(a-b+c)(a-b-c)
    4^2(a-b)^2 - 4(a^2 -2ab +b^2 -c^2)
    4^2(a-b)^2 - 4(a-b)^2 +4c^2
    12(a-b)^2+4c^2

    The sum of two square is clearly positive.

    Bobak

    Edit: I got beaten by a matter of seconds.
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  4. #4
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    Quote Originally Posted by bobak View Post
    evaluate the discriminate b^2 - 4ac

    4^2(a-b)^2 - 4(a-b+c)(a-b-c)
    4^2(a-b)^2 - 4(a^2 -2ab +b^2 -c^2)
    4^2(a-b)^2 - 4(a-b)^2 +{\color{red}4}c^2
    12(a-b)^2+{\color{red}4}c^2

    The sum of two square is clearly positive.

    Bobak
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