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Math Help - Proof using Discriminants

  1. #1
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    Proof using Discriminants


    Prove that the equation x + (a-2d)x + (ab - ad - b) = 0 has real roots for all values of a, b and d.

    Here's what I've done so far (correct me if I've done something wrong btw) ...


    a= 1 b = a-2d c= ab - ad - b
    b - 4ac ≥ 0
    (a-2d) - 4(1)(ab-ad-b) ≥ 0

    so if I'm thinking right, I have to prove the left side is positive? I've been trying different things but I'm stuck ... can somebody help me please?
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  2. #2
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    Re: Proof using Discriminants

    Quote Originally Posted by HelenMc9 View Post
    Prove that the equation x + (a-2d)x + (ab - ad - b) = 0 has real roots for all values of a, b and d.
    Here's what I've done so far (correct me if I've done something wrong btw) ...
    a= 1 b = a-2d c= ab - ad - b
    b - 4ac ≥ 0
    (a-2d) - 4(1)(ab-ad-b) ≥ 0

    You can show that (a-2d)^2-4(ab-ad-b^2)=(a-2b)^2+4d^2~?
    Thanks from MarkFL
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    Re: Proof using Discriminants

    where does the 4d^2 come from?
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    Re: Proof using Discriminants

    Quote Originally Posted by HelenMc9 View Post
    where does the 4d^2 come from?

    Simple algebra: (a-2d)^2=a^2-4ad+4d^2~.
    Thanks from MarkFL
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    Re: Proof using Discriminants

    hmm ... really sorry, but still a bit confused!
    How do you get to ' (a - 2d) - 4(ab - ad -b) = (a -2b) + 4d ' ??
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    Re: Proof using Discriminants

    Quote Originally Posted by HelenMc9 View Post
    hmm ... really sorry, but still a bit confused!
    How do you get to ' (a - 2d) - 4(ab - ad -b) = (a -2b) + 4d ' ??
    If one cannot do simple algebra, then this question is beyond doing.

    (a-2d)^2-4(ab-ad-b^2)=a^2-4ad+4d^2-4ab+4ad+4b^2.
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    Re: Proof using Discriminants

    Well that's what I needed - someone to help me figure out where I was going wrong with the 'simple algebra'!! I can see now it was quite obvious but for me, a secondary school student who's not particularly mathematically minded, it's sometimes the 'obvious' stuff that catches me out! So thank-you for helping, I understand it now!
    so now to finish off the question: I can say (a-2b)^2 - 4d^2 is greater than zero because anything squared has to be positive?
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  8. #8
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    Re: Proof using Discriminants

    Not really if 4d^2 is bigger than (a-2b)^2, then its smaller than zero.
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  9. #9
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    Re: Proof using Discriminants

    Quote Originally Posted by HelenMc9 View Post
    so now to finish off the question: I can say (a-2b)^2 - 4d^2 is greater than zero because anything squared has to be positive?
    Quote Originally Posted by Jesstess123 View Post
    Not really if 4d^2 is bigger than (a-2b)^2, then its smaller than zero.

    The point is that you both are incorrect.
    Because it is (a-2b)^2{~\color{blue}+~}4d^2.
    The sum of two squares is non-negative.
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