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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$a,b\in R$

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

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

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

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

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$a,b\in R$

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

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

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

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

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- #2

- Feb 14, 2012

- 3,755

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

- Nov 29, 2013

- 172

Hello.$a,b\in R$

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

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

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

Regards.

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- #4

- Jan 25, 2013

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$a^2+ab-b^2=0---(1)$

from (1) we have :$\dfrac{a}{b}=\dfrac{b}{a}-1 ---(2)$

let $x=\dfrac{a}{b}$

$\therefore x^2+x-1=0$

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

- Mar 31, 2013

- 1,322

Good question

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 $1/\phi$