Is there exist 2 +ve numbers so that: $\displaystyle \frac {1}{a-b} = \frac{1}{a} + \frac{1}{b}$ ? show how.
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Originally Posted by razemsoft21 Is there exist 2 +ve numbers so that: $\displaystyle \frac {1}{a-b} = \frac{1}{a} + \frac{1}{b}$ ? show how. you have $\displaystyle \frac 1{a - b} - \frac 1a - \frac 1b$ $\displaystyle \Rightarrow \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0$ now, can you actually solve this? will the solutions be positive?
Originally Posted by Jhevon you have $\displaystyle \frac 1{a - b} - \frac 1a - \frac 1b$ $\displaystyle \Rightarrow \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0$ now, can you actually solve this? will the solutions be positive? $\displaystyle \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0 $ $\displaystyle ab-ab+b^2-a^2+ab=0$ $\displaystyle b^2-a^2+ab=0 $ DETERMINATE is +ve but how can i solve the rest ?
Last edited by razemsoft21; Apr 29th 2009 at 12:11 AM.
Originally Posted by razemsoft21 $\displaystyle \frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0 $ $\displaystyle ab-ab+b^2-a^2+ab=0$ $\displaystyle b^2-a^2+ab=0 $ DETERMINATE is +ve but how can i solve the rest ? Treat the equation as a quadratic in b: $\displaystyle b^2 - ab - a^2 = 0 \Rightarrow b = \frac{-a \pm |a| \sqrt{5}}{2}$.
Originally Posted by mr fantastic Treat the equation as a quadratic in b: $\displaystyle b^2 - ab - a^2 = 0 \Rightarrow b = \frac{-a \pm |a| \sqrt{5}}{2}$. Thanks, it works but $\displaystyle b^2 - ab - a^2 = 0 $ should be: $\displaystyle b^2 + ab - a^2 = 0$ typing mistake thanks again
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