How is quadratic equation derived from Euclid? Referring to Book 2 of Elements, given a segment AB, find a point C between A and B such that the lengths of segments satisfy AC > CB as well as the following proportionality relation: AC / AB = BC / AC Please show me how exactly this can be translated into the quadratic equation for x = AC / AB: x^2 + x  1 = 0 
What translation of the elements do you have? What proposition are you referring to? 
Hi, please explain to me how the following can be translated to the quadratic equation below, if you can, thanks. Given a segment AB, find a point C between A and B such that the lengths of segments satisfy AC > CB as well as the following proportionality relation: AC / AB = BC / AC Please show me how exactly this can be translated into the quadratic equation for x = AC / AB: x^2 + x  1 = 0 
It's explained here. 
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Dan 
We can, without loss of generality, let AB= 1. Then x= AC/AB= AC. CB= AB AC= 1 x. $\displaystyle \frac{AC}{AB}= x= \frac{BC}{AC}= \frac{1 x}{x}$. Multiplying both sides by x, $\displaystyle x^2= 1 x$. Adding x to both sides and subtracting 1 from both sides, $\displaystyle x^2+ x 1= 0$. 
That's wonderful how you did that Country Boy, first defining AB = 1. Might you recommend material to learn how to translate geometry to algebra like this? Thanks! 
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That's the one, thanks. 

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