Finding the Ratio of $a$ to $b$ for $a^2+ab-b^2=0$

[Albert1]

$a,b\in R$

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

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

 

[anemone]

Re: find a/b

My solution:

Spoiler

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]

Re: find a/b

$a,b\in R$

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

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

Hello.

Spoiler

[tex]a=\dfrac{-b \pm \sqrt{b^2+4b^2}}{2}= \dfrac{-b \pm b \sqrt{5}}{2}[/tex]

[tex]\dfrac{a}{b}=\dfrac{-b \pm b \sqrt{5}}{2b}= \dfrac{-1 \pm \sqrt{5}}{2}[/tex]

Regards.

 

[Albert1]

Re: find a/b

Spoiler

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

 

[kaliprasad]

Re: find a/b

Good question

Spoiler

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 $1/\phi$