# Thread: Exist 2 numbers or not (how ??)

1. ## Exist 2 numbers or not (how ??)

Is there exist 2 +ve numbers so that:

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

2. 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?

3. 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 ?

4. 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:

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

5. 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