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-   -   How is quadratic equation derived from Euclid? (http://mymathforum.com/algebra/345859-how-quadratic-equation-derived-euclid.html)

paulm February 25th, 2019 09:42 AM

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

NineDivines February 25th, 2019 12:08 PM

What translation of the elements do you have? What proposition are you referring to?

paulm February 25th, 2019 04:49 PM

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

skipjack February 25th, 2019 05:08 PM

It's explained here.

topsquark February 25th, 2019 05:49 PM

Quote:

Originally Posted by paulm (Post 606268)
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

Country Boy February 26th, 2019 07:00 AM

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, $\displaystyle x^2+ x- 1= 0$.

paulm February 26th, 2019 06:40 PM

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!

JeffM1 February 27th, 2019 06:21 AM

Quote:

Originally Posted by paulm (Post 606323)
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.

paulm February 27th, 2019 11:46 AM

That's the one, thanks.

greg1313 March 1st, 2019 01:42 AM

https://en.wikipedia.org/wiki/Golden_ratio


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