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 February 25th, 2019, 09:42 AM #1 Newbie   Joined: Aug 2015 From: USA Posts: 17 Thanks: 0 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
 February 25th, 2019, 12:08 PM #2 Newbie     Joined: Feb 2019 From: United Kingdom Posts: 28 Thanks: 3 What translation of the elements do you have? What proposition are you referring to?
 February 25th, 2019, 04:49 PM #3 Newbie   Joined: Aug 2015 From: USA Posts: 17 Thanks: 0 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
 February 25th, 2019, 05:08 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 It's explained here. Thanks from topsquark, paulm and SenatorArmstrong
February 25th, 2019, 05:49 PM   #5
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Quote:
 Originally Posted by paulm 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
Why did you repeat your original post?

-Dan

 February 26th, 2019, 07:00 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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, $\displaystyle x^2+ x- 1= 0$. Thanks from paulm, SenatorArmstrong, idontknow and 1 others Last edited by Country Boy; February 26th, 2019 at 07:04 AM.
 February 26th, 2019, 06:40 PM #7 Newbie   Joined: Aug 2015 From: USA Posts: 17 Thanks: 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!
February 27th, 2019, 06:21 AM   #8
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Quote:
 Originally Posted by paulm 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!
It used to be studied under the name of "analytic geometry" in English. I do not think it is taught in the US public schools any longer, but look for that phrase in the title of a a book.

 February 27th, 2019, 11:46 AM #9 Newbie   Joined: Aug 2015 From: USA Posts: 17 Thanks: 0 That's the one, thanks.
 March 1st, 2019, 01:42 AM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond

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