# Find a/b

#### Albert

##### Well-known member
$a,b\in R$

$if:\,\, a^2+ab-b^2=0$

$find:\,\, \dfrac {a}{b}=?$

#### anemone

##### MHB POTW Director
Staff member
Re: find a/b

My solution:

If we let $a=bk$, then $a^2+ab-b^2=0$ becomes $(bk)^2+(bk)b-b^2=0$ or simply $b^2(k^2+k-1)=0$ but we're told that $a,b\in R$, thus $b \ne 0$ but $k^2+k-1=0$ or $k=\dfrac{-1\pm\sqrt{5}}{2}$, i.e. $\dfrac{a}{b}=\dfrac{-1\pm\sqrt{5}}{2}$.

#### mente oscura

##### Well-known member
Re: find a/b

$a,b\in R$

$if:\,\, a^2+ab-b^2=0$

$find:\,\, \dfrac {a}{b}=?$
Hello.
$$a=\dfrac{-b \pm \sqrt{b^2+4b^2}}{2}= \dfrac{-b \pm b \sqrt{5}}{2}$$

$$\dfrac{a}{b}=\dfrac{-b \pm b \sqrt{5}}{2b}= \dfrac{-1 \pm \sqrt{5}}{2}$$

Regards.

#### Albert

##### Well-known member
Re: find a/b

$a^2+ab-b^2=0---(1)$
from (1) we have :$\dfrac{a}{b}=\dfrac{b}{a}-1 ---(2)$
let $x=\dfrac{a}{b}$
$\therefore x^2+x-1=0$
$x=\dfrac{-1\pm\sqrt{5}}{2}$

as others have pointed If we put $x = \frac{a}{b}$ we get $x^2 +x -1= 0$
Now if we put y = -x we get $y^2 = 1 + y$ so solutions are $\phi$ and $-1/\phi$ where $\phi$ is the golden ratio
This gives solution x = -$\phi$ and $1/\phi$