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Math Help - Real roots

  1. #1
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    Real roots

    Hi, I worked out a solution for this problem but I need someone to check it:

    Prove that the roots of  (x-a)(x-b) = h^2 are always real:

     x^2 -bx-ax+ab-h^2 =0
     A= 1, B= -(a+b), C= ab-h^2 where  B^2 - 4AC = positive or zero for the roots to be real.

     B^2-4AC = [-(a+b)]^2 -4(1)(ab-h^2)

     a^2+2ab+b^2-4ab+4h^2 = a^2-2ab+b^2 +4h^2

    = (a-b)^2 + 4h^2

     (a-b)^2+4h^2 consists of two perfect squares, which can never be negative. Specifically, the roots are rational because  B^2- 4AC is a perfect square.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Recklessid View Post
    Hi, I worked out a solution for this problem but I need someone to check it:

    Prove that the roots of  (x-a)(x-b) = h^2 are always real:

     x^2 -bx-ax+ab-h^2 =0
     A= 1, B= -(a+b), C= ab-h^2 where  B^2 - 4AC = positive or zero for the roots to be real.

     B^2-4AC = [-(a+b)]^2 -4(1)(ab-h^2)

     a^2+2ab+b^2-4ab+4h^2 = a^2-2ab+b^2 +4h^2

    = (a-b)^2 + 4h^2

     (a-b)^2+4h^2 consists of two perfect squares, which can never be negative. Specifically, the roots are rational because  B^2- 4AC is a perfect square.
    yes, that works. just say
    (a - b)^2 \ge 0 and 4h^2 = (2h)^2 \ge 0 for all a,b,h \in \mathbb{R} (since they are squares), so that the discriminant b^2 - 4ac \ge 0 and we have real roots
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  3. #3
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    That looks perfect to me nothing I can see wrong with it
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