Friday, January 26, 2007

Special Challenge Week of 1-29-07

The problem below (designed for Algebra 2 and beyond) was part of a weekly online challenge I established a little over a year ago. It ran for 2 1/2 months and the response from students was overwhelming. I posted the problem on our dept web page at precisely 6 PM on Mondays. Students had until 6 PM on Wed to email me their detailed solutions. One student would always wait until 5:59 PM! I would post the answers, comments and the best 2-3 student solutions. I rarely had to give my solutions - theirs were usually better! I may do this on weekends for a Monday challenge here and give you time to play with it for a day. Of course, I know some of you will have it solved in a heartbeat, but I'm hoping you will see why I created this for the students -- an opportunity to delve more deeply and learn how to deal with more sophisticated problems having several layers.

This week's problem is algebraic and not that bad. If you think you have all the answers, you can post them and your comments but hold off on detailed solutions to give others a chance! I will respond to the answers submitted at first. Sorry, it's still in hard to read format...

mrc said...

a) 16/25
b) domain of f: 2/(1+sqrt(5)) < x < 1
c) the lower bound of the domain is the golden ratio conjugate

Dave Marain said...

mrc: looks good but check your inequality

I will post 'official' answers and comments later on Sun or Mon...

Students always seem amazed when i tell them that the conjugate of the Golden ratio is the same as one less than that number but of course 1/x = x-1 or
1 = x^2-x is how one determines the number!

jonathan said...

I like those equations like this:

x+1 = x^2
x = x
x-1 = 1/x

mathmom said...

I assume we're sticking with real numbers here?

mrc said...

Busted! Of course there's no reason you can't have a zero inside your square root. So all my < should be <=.

Dave Marain said...

right on, mrc!
yes, unless i specify otherwise, assume from now on that the domain in all my problems is real!

jonathan--
why do you like those eqns like that? consistency of form, i.e., linear expressions on l.h.s?

Darmok said...

What an interesting problem! I have no idea how to solve it.

jonathan said...

Yes, I like the consistent format. In words:

"phi is one more than its reciprocal and one less than its square."

The equations make the point more simply, more powerfully.