Exist 2 numbers or not (how ??)

**razemsoft21** · April 28th 2009, 11:01 PM

Is there exist 2 +ve numbers so that:

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

**Jhevon** · April 28th 2009, 11:07 PM

Originally Posted by razemsoft21

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?

**razemsoft21** · April 28th 2009, 11:37 PM

Originally Posted by Jhevon

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 ?

**mr fantastic** · April 29th 2009, 01:35 AM

Originally Posted by razemsoft21

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

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

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

DETERMINATE is +ve but how can i solve the rest ?

Treat the equation as a quadratic in b:

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

**razemsoft21** · April 29th 2009, 05:44 AM

Originally Posted by mr fantastic

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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