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Math Help - Exist 2 numbers or not (how ??)

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    Exist 2 numbers or not (how ??)

    Is there exist 2 +ve numbers so that:

     \frac {1}{a-b} = \frac{1}{a} + \frac{1}{b} ? show how.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by razemsoft21 View Post
    Is there exist 2 +ve numbers so that:

     \frac {1}{a-b} = \frac{1}{a} + \frac{1}{b} ? show how.
    you have \frac 1{a - b} - \frac 1a - \frac 1b

    \Rightarrow \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0

    now, can you actually solve this? will the solutions be positive?
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    Quote Originally Posted by Jhevon View Post
    you have \frac 1{a - b} - \frac 1a - \frac 1b

    \Rightarrow \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0

    now, can you actually solve this? will the solutions be positive?
     \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0

    ab-ab+b^2-a^2+ab=0

     <br />
b^2-a^2+ab=0<br />

    DETERMINATE is +ve but how can i solve the rest ?
    Last edited by razemsoft21; April 29th 2009 at 12:11 AM.
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    Quote Originally Posted by razemsoft21 View Post
     \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0

    ab-ab+b^2-a^2+ab=0

     <br />
b^2-a^2+ab=0<br />

    DETERMINATE is +ve but how can i solve the rest ?
    Treat the equation as a quadratic in b:

    b^2 - ab - a^2 = 0 \Rightarrow b = \frac{-a \pm |a| \sqrt{5}}{2}.
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    Thumbs up

    Quote Originally Posted by mr fantastic View Post
    Treat the equation as a quadratic in b:

    b^2 - ab - a^2 = 0 \Rightarrow b = \frac{-a \pm |a| \sqrt{5}}{2}.
    Thanks, it works
    but

    b^2 - ab - a^2 = 0 should be:

    b^2 + ab - a^2 = 0

    typing mistake thanks again
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