\(\frac{l+w}{l}=\frac{l}{w}\)

Find \(\frac{l}{w}\), your answer should be the golden ratio that is a fraction with radicals in simplest form.

(BTW, the answer is definitely NOT \(\frac{l+w}{l}\))

This is a fun math problem I solved yesterday night.

CalculatorUser Apr 28, 2019

#1**0 **

multiply both by lw

l^{2}+w^{2}=l^{2}^w

subtract l^{2}

w^{2}=w

subtract w

w=0

??

Guest Apr 29, 2019

#2**+2 **

You incorrectly multiplied by lw,

it should be lw+w^2=l^2

You also CAN'T subtract l^2

CalculatorUser
Apr 29, 2019

#3**+2 **

**Easy Golden Ratio Quadratic**

\(\dfrac{l+w}{l}=\dfrac{l}{w}\)

I assume:

\(\text{Let }\textbf{Golden Ratio} =\varphi\)

\(\begin{array}{|rcll|} \hline \dfrac{l+w}{l} &=& \dfrac{l}{w} \\\\ \dfrac{l}{l}+ \dfrac{w}{l} &=& \dfrac{l}{w} \\\\ 1 + \dfrac{w}{l} &=& \dfrac{l}{w} \quad | \quad \dfrac{l}{w} = \varphi,\ \dfrac{w}{l} = \dfrac{1}{\varphi} \\\\ 1 + \dfrac{1}{\varphi} &=& \varphi \quad |\quad \cdot \varphi \\\\ \varphi + 1 &=& \varphi^2 \\ \varphi^2 -\varphi - 1 &=& 0 \\\\ \varphi &=& \dfrac{1\pm \sqrt{1-4\cdot (-1) }}{2} \\ \varphi &=& \dfrac{1\pm \sqrt{5}}{2} \\\\ \mathbf{\varphi} &\mathbf{=}& \mathbf{\dfrac{1+ \sqrt{5}}{2}} \\ \hline \end{array}\)

heureka Apr 29, 2019