## Wednesday, January 24, 2007

### Problems for 1-25-07

Important note: The problems are stored as an image and it is currently appearing fuzzy and too small. Guess that makes this fuzzy math! TRY CLICKING ON THE IMAGE TO MAGNIFY IT. I will try to make the image larger but no promises. Blogger was down for awhile and they're having problems with images right now.

PLS READ COMMENTS FOR EACH DAY'S PROBLEMS. ANSWERS APPEAR IN THE COMMENTS AND I USUALLY POST THOSE MUCH LATER IN THE EVENING IF I REMEMBER! ALSO, SOME OF THE BEST CONVERSATIONS ABOUT DIFFERENT WAYS OF SOLVING THESE AS WELL AS TEACHING STRATEGIES, CONTENT, INSTRUCTION AND ASSESSMENT TAKE PLACE THERE.

Darmok said...

OK, 1 is obviously 10md dollars squared. Just kidding. Actually I’m slightly confused by the wording, but the way I read it, after 10 weeks she’ll be in her eleventh week of work and so she will be earning d + 10m.

2 is 2; if half the number = 2*.5 =1, the number's obviously 2.

3, hmmm, -16 < n < 3, so 15 + 1 + 2 = 18, right?

4, we need to divide by the smallest factor (not including 1), so divide by 2 to get 2^19*3^18. Requiring the factor to be even is superfluous.

5, a^4 = 5, so 4*5^3=500. But are we assuming is a real number? a^4 could just as easily = -5, in which case it would be -500. (If a^4 = -5, then a^2 = iR5 or -iR5, and I'm ashamed to admit I don't remember how to take the square root of i, not that it's relevant to solving the problem).

For 6, I'm going to need paper. Some of the choices are cut off for me. Let's see, if PRS = PRQ, then PR bisects SRQ, If S = Q then we have two similar triangles since the third (P) angles must be congruent, and since they share a side, the triangles are congruent and form a kite. PQ is not necessarily = RS, but the other pairs must be congruent, so II and III.

Anonymous said...

For 6, the diagram is deceptive, since it looks like a parallelogram, but in fact, given the congruences stated in the problem, it probably isn't. (If it is, it's a square and all 3 statements are true.) Assuming it does not have to be a parallelogram, I agree with darmok's reasoning.

Darmok said...

OK, so Google was kind enough to tell me that the square root of i is (1/r2)(1+i). Can anyone explain that to me in an intuitive way? I mean I can see that squaring it would yield i, but I was curious if there was a way to easily visualize it.

Anyway, if a^8=25, a has 8 possible values. It could be ±5^(1/4) or ±5^(1/4)*i, in which case the answer is 500. Or it could be...I think... ±5^(1/4) / r2 * (i±1) [those are independent ±s), in which case the answer is -500.

Darmok said...

OK, so I've been fascinated by the 8 roots of a^8 = b. There was no obvious pattern to me and I was having trouble grasping them, so I sketched them on the complex plane, and assuming I did it right, they all fall on a circle of radius b^(1/8), and they're evenly spaced at 0, 45, 90, and so on. That is amazing, though I can't really grasp the significance of it or why it should be that way. Does that hold true for other relations?

Darmok said...

OK, it holds true at least for a^2 and a^3. I'm really amazed; it's so unexpectedly elegant (unexpected to me, that is), kind of like the way I see Euler's identity: beautiful, though I don't really grasp it.

Darmok said...

And now I officially feel like a fool, I just read that you can basically use Euler's formula to find all roots of a number. So of course they will lie on a circle. Too bad I didn't remember this/look it up earlier.

Anonymous said...

darmok, here's an alternate:

sqr(i) is going to be a complex number, right? so a + bi

a+bi = sqr(i)
(a+bi)^2 = i
a^2 +2abi -b^2 = 0 + 1i

Let the real part = real part and
let the complex part = complex part

a^2 - b^2 = 0

2ab = 1

Now, we have a quadratic system to solve. (I'll leave that for you)

I prefer this, at least at first, to using a formula.

Dave Marain said...

The few responses suggest it was very hard to read the problems, yes? It couldn't be the problems themselves!

1) (D) d + 9m
2) 2
3) 18
4) E
5) 500 (if 'a' is assumed to be real)
6) D (II, III)

1) As Darmok pointed out, #1 is ambiguous. These kinds of questions always are but mine could have been worded more clearly. Students should be encouraged to make a table to see the relationships. Thus, at the end of week 1 her salary was still \$d or d + (0)m. Assuming she gets paid at the end of each week, her salary at the END of week 2 would be d + (1)m, or we could say 'after 2 weeks.' Thus, after 10 weeks, her salary would be d + (9)m. The problem of course is what is meant by the word 'after'. I guess I could have written it 'at the end of'. But then again, when you're born are you celebrating your FIRST birthday! In some countries, like Korea, you are 1 on the day of your birth!
2) straightforward but some students struggle with anything that looks like a fraction!
3) easy but good skill practice and many will still be off by 1 when counting
4) darmok's approach is beautiful; imagine if our students had the number sense and the skills to go with it
5) if we regard this as an SAT question, then only real values are allowed; if we don't specify domain, anything goes and a case could be made for -500!
6) wonderful analysis, darmok! i love the way you articulate your thought processes...

Dave Marain said...

whoa!
where did all those other comments come from!
i apologize for not acknowledging mathmom and jonathan
nice discussion about complex roots; visualizing them on a circle is beautiful though...

Anonymous said...

Darmok, Dave,

About problem 5, you're trying to do too much, and your students will probably try to do too much too. You don't need to solve for all possible values of a to solve the problem. In fact, the point about whether a must be real is a good one. Yes, finding all the 8th roots of 25 can be a good test of persistence, but you don't need it. Just do:

a^8 = 25,
a^4 = 5 or -5 [if a is allowed to be complex],
4a^12 = 4 (a^4)^3 = 500 or -500 [if ...].

I used to ask my 'intermediate algebra' students questions like:

How many real roots does

3x^2 + x + 1 = 0 have?

Most of the students would use the quadratic formula, write both roots, and tell me whether they were real or not. I would explain that all they needed to do was to check whether the discriminant is positive, zero, or negative. They'd still compute both roots. I would tell them that I would deduct points from their exam scores if they did that. They would still compute both roots.

I think the students were so used to writing everything they could in attempts to earn 'partial credit' that they lost sight of the problem. Perhaps I was wrong to deduct points they way I had done--it was 17 years ago. But I warned them first!

Darmok said...

Thanks, Jonathan, that helps!

Dave, perhaps you would consider something like “how much did she earn her tenth week?” or something like that. Thanks, but I thought 4 was relatively straightforward; in fact, I don't even know how else one could solve it. As far as my thought processes, I have to take some of these problems step-by-step; I can't just see the answer the way you guys do. I am still amazed by the circle; I had no intention of visualizing them that way; it just happened!

Eric, I can understand your confusion, since I posted so many comments. If you look at my first comment, I gave the answer as ±500. Trying to find the roots was a later, side exploration; I sometimes get intrigued by unexpected patterns and such and the others are kind enough to indulge me and help me out where my math is a bit rusty.