# Defining equation for golden ratio; quadratic fields

• Sep 17th 2008, 11:02 PM
Pn0yS0ld13r
Defining equation for golden ratio; quadratic fields
Definition: If $\displaystyle \alpha$ is an irrational number in $\displaystyle \mathbf{Q}\left(\sqrt{d}\right)$, then the equation $\displaystyle ax^{2}+bx+c=0$ is called the defining equation for $\displaystyle \alpha$ if $\displaystyle \alpha$ satisfies the equation and a, b, and c are integers, $\displaystyle (a,b,c)=1$, and $\displaystyle a>0$.

Find a defining equation for the golden ratio $\displaystyle \dfrac{1+\sqrt{5}}{2}$.

How do you approach this problem?
• Sep 17th 2008, 11:14 PM
Moo
Hello,
Quote:

Originally Posted by Pn0yS0ld13r
Definition: If $\displaystyle \alpha$ is an irrational number in $\displaystyle \mathbf{Q}\left(\sqrt{d}\right)$, then the equation $\displaystyle ax^{2}+bx+c=0$ is called the defining equation for $\displaystyle \alpha$ if $\displaystyle \alpha$ satisfies the equation and a, b, and c are integers, $\displaystyle (a,b,c)=1$, and $\displaystyle a>0$.

Find a defining equation for the golden ratio $\displaystyle \dfrac{1+\sqrt{5}}{2}$.

How do you approach this problem?

The solutions to such an equation are :

$\displaystyle x=\frac{-b {\color{red}\pm} \sqrt{b^2-4ac}}{2a}$

Since a and b are integers, the only way to get this $\displaystyle \sqrt{5}$ is $\displaystyle \sqrt{b^2-4ac}$.
So you can see that $\displaystyle \frac{1{\color{red}-}\sqrt{5}}{2}$ will also be a root. (The reasoning may be similar to the ones involving complex roots).

Develop $\displaystyle \left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$ and identify a,b and c :p
• Sep 19th 2008, 01:19 AM
Pn0yS0ld13r
Thank you Moo.

I can't believe I didn't get this before... (Headbang)