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February 25th, 2019, 09:42 AM   #1
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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
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February 25th, 2019, 12:08 PM   #2
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What translation of the elements do you have? What proposition are you referring to?
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February 25th, 2019, 04:49 PM   #3
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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
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February 25th, 2019, 05:08 PM   #4
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It's explained here.
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February 25th, 2019, 05:49 PM   #5
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Quote:
Originally Posted by paulm View Post
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
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February 26th, 2019, 07:00 AM   #6
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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, $\displaystyle x^2+ x- 1= 0$.

Last edited by Country Boy; February 26th, 2019 at 07:04 AM.
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February 26th, 2019, 06:40 PM   #7
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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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February 27th, 2019, 06:21 AM   #8
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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.
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February 27th, 2019, 11:46 AM   #9
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That's the one, thanks.
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March 1st, 2019, 01:42 AM   #10
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