Date: Sep 29, 2012 6:50 AM
Author: Robert Hansen
Subject: Re: An Algebra 2 Test
What can I say? I didn't know the "hidden trick".
On Sep 29, 2012, at 1:23 AM, Wayne Bishop <email@example.com> wrote:
> My guess is that the approach Dave had in mind was more obvious:
> a^2 - (a-1)^2 = (a - (a-1)) (a + (a-1)) = (1)(2a - 1)
> At 08:36 PM 9/28/2012, Robert Hansen wrote:
>> a^2 - (a-1)^2 = a^2 - (a^2 - 2a + 1) = 2a - 1 = 13.04822...
>> This is a very good point, work the algebra FIRST. I wonder how many algebra teachers are even capable of concocting such a problem? Working backwards and making sure that 2*a doesn't require any more than single digit math (no digit greater than 4). I have to think about this. This works well with large integers as well, with the same condition on the digits.
>> Lou, what do you say to problems like this with regards to our prior art discussion?
>> Ha, calculators are allowed. Calculators with 70 digits of precision I suppose.:)
>> Bob Hansen
>> On Sep 28, 2012, at 5:51 PM, "Dave L. Renfro" <firstname.lastname@example.org> wrote:
>> > Robert Hansen wrote:
>> > http://mathforum.org/kb/message.jspa?messageID=7897638
>> >> I want to try something different. I want everyone to contribute
>> >> problems for a hypothetical algebra 2 exam. You can contribute
>> >> just topics if you wish though I would like see examples as well.
>> >> I am going with algebra 2 rather than algebra 1 because I think
>> >> the line is more well defined.
>> > I only have a few moments before I need to leave to tutor
>> > someone, but here's a somewhat silly one off the top of
>> > my head:
>> > Determine the exact value of a^2 - b^2 if
>> > a = 7.0241132301442003123012230341430201
>> > b = 6.0241132301442003123012230341430201
>> > Calculators are allowed.
>> > Dave L. Renfro