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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 Tags derived, equation, euclid, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Elinaldo Calculus 3 October 20th, 2015 04:19 PM hyperbola Physics 1 June 25th, 2015 03:19 AM WWRtelescoping Advanced Statistics 1 April 15th, 2014 12:00 PM jackson_wang Calculus 1 November 30th, 2011 01:13 PM dgoodman Algebra 0 September 30th, 2011 06:46 AM

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