# Exist 2 numbers or not (how ??)

• April 28th 2009, 11:01 PM
razemsoft21
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.
• April 28th 2009, 11:07 PM
Jhevon
Quote:

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?
• April 28th 2009, 11:37 PM
razemsoft21
Quote:

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?

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

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

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

DETERMINATE is +ve but how can i solve the rest ?
• April 29th 2009, 01:35 AM
mr fantastic
Quote:

Originally Posted by razemsoft21
$\frac {ab - b(a - b) - a(a - b)}{ab(a - b)} = 0$

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

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

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

Treat the equation as a quadratic in b:

$b^2 - ab - a^2 = 0 \Rightarrow b = \frac{-a \pm |a| \sqrt{5}}{2}$.
• April 29th 2009, 05:44 AM
razemsoft21
Quote:

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:

$b^2 + ab - a^2 = 0$ (Clapping)

typing mistake thanks again