Desmos Common Core Activity Linking Circles, Tangents and Linear-Quadr Systems

Detailed investigation with extensive background notes for instructor and step by step outline for students to follow. Students will be asked to use a slider to approximate the position of a tangent line of slope -1 to a circle centered at (0,0). The tangent line, x+y=k, requires use of a parameter.

Students will begin with a particular radius, 3, then solve a linear-quadratic system to determine the exact equation of one of the tangent lines. They will also be asked to enter an expression for the other tangent line of slope -1 using the same parameter k. After approximating the locations of similar tangent lines for other radii, they will be asked to solve a general system using radius r.

There are different systems offered to the instructor, depending on the sophistication of the student. Finally, a geometric solution is suggested using 45-45-90 triangles.



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Parametric/Projectile Motion Simulated in Desmos - A Common Core Activity for Algebra/Precalculus

[Updated using folders to reduce amount of visible text. Click on the arrow next to the Folder icon to see the frames below. Thanks to Desmos team for this helpful hint!]

CLICK ON GRAPH TO ACTIVATE DESMOS...

The Desmos activity above is both an investigation of parametric representation and a tutorial for more advanced use of this remarkable WebApp. The The text in the side frames begins with a detailed background of the activity for the instructor and how Desmos can be used to demonstrate projectile motion using both parametric and rectangular coordinates. Some of the uses of slider 'variables' are demonstrated including animation, a powerful feature of Desmos.

In addition to showing how to use parameters in Desmos, the activity itself asks students to compare two different trajectories, representing an object dropped from some initial height, then a 2nd object two seconds later. The horizontal translation of the first graph is juxtaposed against the algebraic representations of these graphs using both system of coordinates.

The student activity starts about halfway down. There is a series of questions and actions the student needs to take in Desmos.

I'm hoping this will prove useful for both the instructor and the student.  Desmos is powerful but, in my opinion, some of the illustrative examples provided by Desmos do not flesh out the ideas behind the various uses of slider 'variables'. I'm hoping this will fill in some of those gaps.  I'm still a novice here so I'm sure more advanced users will be able to improve upon this...

Your comments and reactions are very helpful to me...







If interested in purchasing my NEW 2012 Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar. 175 problems divided into 35 quizzes with answers at back and DETAILED SOLUTIONS/STRATEGIES for the 1st 8 quizzes. Suitable for SAT I, Math I/II Subject Tests, Common Core Assessments, Math Contest practice and Daily/Weekly Problems of the Day. Includes multiple choice, case I/II/III type and constructed response items. Price is $9.95. Secured pdf will be emailed when purchase is verified.

Using "SAT-Type" Problems to Develop Understanding of Quadratic Functions in Algebra

f(x) = t-2(x+4)² where t is a constant.
If f(-8.3) = f(a) and a > 0, what is the value of a?


This type of question is of the Grid-in type (or short constructed response) that now appears on standardized testing like the SAT-I and ADP Algebra 2.

I administered it to a group of strong SAT students recently and the students who completed Alg II struggled with it. As our president might say, this was a "teachable moment!"

A few thoughts...
Should textbooks include more questions of this type both as examples and regular homework exercises? As you might guess, I'm very much opposed to having questions labeled as Standardized Test Practice in texts or appear in a separate section of the text or in ancillaries.

By the way, by including the label "SAT-type problems" in the title of this post I'm trying to engender both positive and negative response. Those of you who have followed this blog for 2- 1/2 years know that what I'm really referring to are "conceptually-based questions." Some of you react adversely to the idea that standardized test questions should influence our curriculum or how we teach. N'est-ce pas?

Your comments...

A Tale of Two Equations: Balancing Procedures and Conceptual Understanding

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The following is intended for all students in 2nd year algebra. Your stronger students should not find these overly challenging but there is more here than meets the eye. The purpose here is to demonstrate how we can review procedures AND develop deeper understanding of important mathematical ideas in the same lesson. The graphing calculator can be used to enhance the lesson by employing multiple representations (Rule of Four) to reinforce the essential ideas.

Note: Finding the solutions is only the tip of the iceberg. Understanding WHY one equation must have finitely many solutions and the other must have infinitely many solutions is the bigger idea here...

SOLVE EACH OF THE FOLLOWING

Equation 1:
(x-1)(x-2) = (1-x)(2-x)

Equation 2:
(x-1)(x-2)(x-3) = (1-x)(2-x)(3-x)


Click on Read More... for solutions and further discussion.




ANSWERS
Equation 1: All real numbers
Equation 2: {1,2,3}

DISCUSSION
Would most of your students eliminate parentheses in the first equation and solve by traditional methods? Even though the left and right sides of the equation appear similar, it is reasonable to expect they will distribute and solve since that is what they're used to doing. This is fine and the standard procedure should be reviewed.

Assuming students will not make "careless" mechanical errors in distributing, they should obtain:
x² - 3x + 2 = 2 - 3x + x².
This generates a nice discussion of an "identical equation" or identity since the left and right sides are mathematically equivalent (if they recognize that!). The instructor may or may not want to continue the mechanical approach of moving all terms to one side producing 0 = 0 to reinforce that the equation is satisfied by all real numbers. Your stronger student will not have much difficulty with this.

Before moving on to the 2nd equation, we can develop a deeper conceptual understanding by asking students to approach the problem another way. We know that some students will wonder about the form of the original equation. Could we have predicted that the two sides would be identical without removing parentheses? Could we also have determined by inspection that both x = 1 and x = 2 are solutions? Asking them to revisit the original equation to see this is critical. Now what about trying some other real number, say x = 5. This should strongly suggest that all real numbers will satisfy the equation. Using the graphing calculator will also drive this point home visually. Store the left side of the equation Y1 and the right side of the equation in Y2. Change the appearance of Y2 (make it bold for example) and have them observe on the viewscreen that the graphs are identical.

So, how come the 2nd equation only has 3 solutions! I'll leave that to my readers to elaborate on...
How can we generalize this?

These kinds of lessons seem to involve way too much overhead, stealing so much valuable time away from other content. BUT these are precisely the kinds of problems students are expected to grapple with in Japan and other countries. Do you really believe "Less is More?"

...Read more

A Recurring Problem for SATs (Functions)

SAT "Grid-In" Type
Level of Difficulty: 5 (High)
Content: Algebra 2, precalculus

The function F satisfies the condition
F(N + 6) = F(N) + 8, for all integers N.
If F(7) = -2, what is the value of F(25)?

Click on Read more to see the answer, solution, discussion.


Answer: 22

Suggested Solution:

Replace N by 7 since F(7) is known:
Therefore, F(7 + 6) = F(7) + 8 or F(13) = -2 + 8 = 6

Next, replace N by 13 since F(13) is known:
F(19) = F(13) + 8 = 6 + 8 = 14

Finally, F(25) = F(19) + 8 = 14 + 8 = 22.

Comments
(1) Too difficult for the SATs? Not really! A similar problem recently appeared. There aren't that many "hard" questions (Level 5) on the SAT but, if a student wants to score over 700 they will need exposure to these types in practice.

(2) Consider writing some variations of these function-type problems for additional practice. At first, change the constants, then consider changing the operations (from addition to multiplication for example). One could raise the bar even higher by asking the question in reverse:
If F(25) = 22, what is the value of F(7)?

(3) There is considerable advanced theory in functional equations and recurrence relations underlying these problems. However, the student needs only to feel comfortable with the function symbolism (or should I call it "Math Notations!"). Starting by "plugging in' N = 7 seems simple in retrospect but most students are too intimidated to consider it. Even the precalculus student may be able to get started, but, without experience, they will often get lost. This is all about exposure, but isn't it always?

(4) One could rewrite this problem using sequence notation:
a_N+6 = a_N + 8. By expressing the problem in the context of the Nth term of a sequence, students may grasp it a bit better, but, in the end, it's all about interpreting function notation.

...Read more

Another Quadratic Function SAT Problem

Have you noticed the SAT Tips of the Week in the sidebar? These are intended both for math teachers and students.

The "new" SAT has a few 2nd year algebra questions and, typically, there is at least one 'parabola' problem usually expressed in function form. Here are two different versions - one multiple choice and one "grid-in" (student-generated response).
Can you predict which one might give most students more difficulty?
It might be interesting to list all of the skills, knowledge and concepts being tested here. Are all of these typically included in your Algebra 2 course? Do students get enough exposure to these kinds of problems?


Version I
For some constant r, the graph of the quadratic function f(x) = -x² + 2rx is a parabola with x-intercepts at P and Q and vertex V. What is the area of ΔPQV, in terms of r?

(A) r² (B) r³ (C) 2r²

(D) 2r³ (E) 4r³


Version II

For some constant r, the graph of the quadratic function f(x) = -x² + 2rx is a parabola with x-intercepts at P and Q and vertex V. If the area of ΔPQV equals 27, what is the value of r?

Answers, solutions, strategies and comments will appear below the Read more...



Answers/Solutions/Comments/...

Version I
Answer: (B) r³

Possible Solution (no frills):
Factoring, we have f(x) = -x(x-2r); x-intercepts are 0 and 2r. Therefore base of triangle has length 2r.
The x-coordinate of the vertex is r (why?), so y-coord = f(r) = -r(r-2r) = r².
Area of triangle = (1/2)(2r)(r²) = r³.

Version II
Answer: r = 3

Possible Solution:
From Version I, we obtain r³ = 27, so r = 3.

Comments

• These kinds of questions typically appear among the last 3-4 problems on a section, meaning they are of above-average difficulty. Students who are in Algebra 2 or beyond should definitely attempt it. After reviewing it, most students may conclude it's not very hard at all!
  
• Testmakers are more frequently using a parameter like 'r' to make it more difficult to merely punch it into the graphing calculator and read off the intercepts and vertex.
• This kind of question could also appear on standardized tests like the Algebra 2 End of Course Exam from Achieve/ADP or other state tests.
• Pick up a copy of 10 Real SATs from the College Board to find several other practice problems like this.
• One could modify and extend these problems in many ways. For example inscribe the parabolic region (bounded by the cruve and the x-axis) in a rectangle and determine its area, a simple variation. More interestingly is to note that the ratio of the area enclosed by the parabola to the area of the rectangle is 2:3, a famous result proved in Calculus.
• Skills, knowledge required for this question? Worth enumerating in my opinion...
  
• I also believe strongly that our students should be tackling these kinds of problems on a regular basis to deepen their understanding of the relationship among the function, the coordinates of key points and the geometry. This used to be known as Analytic Geometry.
  

...Read more

Using Algebra to Enhance Numerical Accuracy On Your Calculator: When is ZERO really ZERO!

When a calculator displays zero as a result should students assume that is exact or only accurate to the precision the machine can store and/or display?

The next time students ask you why we use the conjugate method to rationalize denominators, here's an example of why we sometimes use the method in "reverse". This happens more frequently in calculus but the following is an apparently trivial numerical computation your students can try on their graphing calculators. The results of this computation depend heavily on the specific technology used (e.g., expect different results between the TI-89 and the TI-84), but hopefully they will get the idea. This numerical issue came up as I was solving an applied problem which required finding the difference between two very large numbers (the difference between distances from the center of the earth to a point slightly above its surface and the radius of the earth). This numerical issue has come up more before on this blog. Look here if you want to see another application.

Here's the computation:
Let R = 2.0916 x 10^7
We need to compute the following expression (denoted by **)



For the Student
(a) Do the calculation directly on your calculator. You will want to store this value of R as a variable for later use:
2.0916x10^7 STO> ALPHA R
Does your calculator display zero? If so, explain this "error."
Note: This display depends on the calculator being used. I experimented with the -84 and -83. Let me know how the display appears on other machines. Of course, one would expect a very different outcome if using Mathematica!

(b) Rewrite the above expression ** by multiplying the numerator and denominator by the conjugate of the expression. (Hint: Put the original expression "over 1").

(c) Recalculate the value of ** using the modified but equivalent form from part (b).
What result do you see this time? Can you explain what may be going on?

(d) Find other numerical expressions that produce an incorrectly displayed result on your calculator! Post these in the comments section pls!

A Traditional Algebra 2 Rational Equation for the Holidays or ...

Solve for x:



Comments:
(1) The traditional procedure is time-consuming but part of the rites of passage for Algebra 2 students. This is the kind of exercise that "trains the brain!"
(2) Note that the form of the equation is a + (1/a) = 2. Does this suggest anything to you?
(3) Would this question be appropriate for SATs? Algebra 2 End of Course Exams?

An Algebra 2 Math Contest Problem and a Warmup Too!

Not the most challenging contest problem but something to give to your students to develop logical careful thinking and some "basic skills." There's a slight 'twist' but nothing that will faze our math experts out there.

SOLVE

(x² - 6x + 9)^{(x2 - 4x + 3)} = 1

Setting the Tone in Precalculus - Another Coordinate Investigation

Note: Read the first comment I posted which suggests a purely Euclidean geometry approach to this problem...

Don't forget our MathAnagram for Aug-Sept. Thus far we have received a couple of correct responses. You are encouraged to make a conjecture!
Look here for directions. Here is the anagram again:

PRINCE? NAH! E-ROI!

Tangent problems are usually the domain of calculus but we can keep them within the reach of geometry and algebra if we restrict our attention to circles. The calculus student spends a considerable amount of time solving a wide variety of "tangent to the curve" exercises. As any calculus instructor will tell you, many of the harder problems ask students to determine the equations of the tangent to a curve from a point not on the curve. The issue there is not the calculus. It's all about an understanding of the interface between the algebra and geometry, the essence of coordinate methods. I developed this investigation specifically to address this issue before students enter calculus. Might be another "fun" problem to start the year off with. If nothing else, it will establish the rigor of your precalculus course early on!

Part I of Investigation
Determine the coordinates of the points of tangency for the tangent lines to the unit circle from the point (0,2).
Note: Unit circle refers to the circle of radius 1, center (0,0).

The remaining parts will all refer to this same circle.

Part II of Investigation
Repeat part I for (0,3) and (0,4).
Write your observations, conjectures.

Part III of Investigation
Show that the y-coordinate of the points of tangency for the tangent lines to the unit circle from the point (0,k) is 1/k, where k ≥ 1.

Notes, Comments...
(1) The result of Part III suggests that as k increases, the y-coordinate of the point of tangency decreases (inverse ratio). Ask students what happens as k approaches 1.
Students should make sense of this visually by sketching tangent lines from various points on the y-axis above the circle.
(2) There are several effective methods for solving the above parts, however, one needs to know the fundamental relationship between a tangent line and the radius drawn to the point of tangency. From that point on, one can represent the slope in two ways or represent the y-coordinate of the point of tangent in two ways. This requires strong understanding of coordinates, graphs and algebraic relationships. You may find other methods -- share them! BTW, one could also use trig methods.
(3) I chose the unit circle and a point on the y-axis for simplicity so that the student could focus on essential ideas. However, one could generalize the result to any circle and any point outside. Have fun with that!
(4) Anyone mildly surprised by the reciprocal relationship between the y-coordinate of the point on the y-axis and the y-coordinate of the point of tangency? Can anyone make sense of that?

When Curves Collide Part II - Quadratic Systems Re-Explored!

One of MathNotations more popular posts (hundreds of views) was published one year ago this week: When Curves Collide.

Here's a variation to review the essential ideas or to use as an assessment problem or just to challenge yourself. Parts (a) thru (d) require some theoretical analysis and algebraic skill. Part (e) is the main challenge...

An Investigation for Algebra 2/Precalculus
Consider the quadratic-quadratic system:

x² + y² = 1
y = ax² -1, a>0

(a) Show that (0,-1) is always a solution to this system.

(b) For what values of the parameter 'a' will there be 3 distinct solutions to the system?
Coordinate Interpretation: For what values of 'a' will the parabola and circle intersect in 3 distinct points?

(c) For what value(s) of the parameter 'a' will two of the points of intersection be above the x-axis? Below the x-axis (in addition to (0,-1))? On the x-axis?

(d) For the case that there are 3 distinct solutions, determine the two solutions, other than (0,-1), in terms of 'a'.

(e) Now for the main problem:

Assume the graph of our system has three points of intersection: P, Q and R(0,-1). If the area of ΔPQR is 32/25, determine the coordinates of P and Q and the value of 'a'.

(f) Can you think of an even more clever variation!

Multiple Representations (Rule of 4) in Algebra 2 or Precalculus

If interested in purchasing my NEW 2012 Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar.  175 problems divided into 35 quizzes with answers at back. Suitable for SAT/Math Contest/Math I/II Subject Tests and Daily/Weekly Problems of the Day. Includes both multiple choice and constructed response items.
Price is $9.95. Secured pdf will be emailed when purchase is verified. DON'T FORGET TO SEND ME AN EMAIL (dmarain "at gmail dot com") FIRST SO THAT I CAN SEND THE ATTACHMENT!

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Did you overlook our Mystery Mathematician for May? I've received two correct responses thus far, but submissions can still be emailed until the 15th of the month. Don't forget to include the info requested in a previous post.



If (A+3) ÷ (B+5) ≥ 10 and B ≥ 7,
what is the least possible value of A?
DISCUSSION

The use of multiple representations of a concept or procedure in mathematics is highly recommended by NCTM and other math education experts. Also known as the Rule of Four, it suggests that instructors use some or all of the following, when introducing a new concept. This requires careful planning and considerable thought on the part of the teacher. Over time and with experience, it will flow. However, it does help to see many models of this heuristic for geometry, algebra, etc.

The Rule of Four suggests that a concept be presented
(a) Using natural language (words)
(b) Numerically (concrete examples, 'plugging in', use of data tables, etc.)
(c) Visually (e.g., using graphs, charts, concrete models)
(d) Symbolically (algebraical mode)

From my experience, many students will approach the problem at the top by ignoring the inequalities and simply plug in 7 for B. They've learned that this strategy usually works on standardized tests. It is our role as educators to challenge them to think more deeply. Create disequilibrium by provoking them with a question like,
"But to make a fraction small, don't you need to make the denominator as large as possible?" Of course this statement does not apply to this problem, but I'll wager that it would cause some to reconsider their initial answer!

Do you think that most students would quickly recognize that the relationship between A and B can be described by a linear inequality, which can be then be approached both algebraically and graphically? Do you think I need strong medication for asking you that question!

To deepen their understanding, one could ask:
How would you have to change the above problem so that one could ask for the greatest possible value of A?

I plan on posting further examples of the Rule of Four. I am aware that I have not fully demonstrated this technique for the problem above. I'm only hinting at it. More will likely come out in the comments...

Coordinate Triangle Problem - Interface between Algebra and Geometry

For Geometry or Algebra 2 students or anyone who wants a diversion...

The vertices of ΔABC are A(m,2k), B(k+11,k-2) and C(2k+6,k-2). The area of the triangle is 15.

(a) What restrictions need to be placed on k to insure there is a triangle.
(b) Given those restrictions, determine all possible values for k.

Comments:
(1) This is not intended to be a highly challenging problem. It can be used as review for a final exam, standardized tests, SATs, etc. Of course, on the SAT, the question would only ask students to grid-in one possible answer and would not generally ask about restrictions.
(2) You may want to ask your students why the value of m is irrelevant.
(3) There are two possible values for k in this problem. Challenge your students to write a revised version that would have more possibilities. Would the coordinates have to involve quadratic expressions in k?
(4) if anyone tries this in the classroom, please let us know how it went, specifically, student reaction. How was it implemented? As a warm-up, extra challenge at end of class, part of homework assignment, extra credit?

SAT-Type Algebra Challenges - How Would You or your Students do on these?

Of course this comes a bit late for all those juniors who took their SATs today but the following questions can be used to prepare for the next one OR for anyone who wants to challenge themselves or their students...


-4 ≤ P ≤ 3 and -5 ≤ Q ≤ 4


(i) What is the greatest possible value of (Q+P)(P-Q)? Explain your reasoning.

(ii) What is the least possible value of (Q+P)(P-Q)? Explain your reasoning.

Comments:
(a) SAT questions don't ask for explanations but this goes beyond that.
(b) These would be known as 'grid-ins' or student-constructed response questions. On a real SAT, answers to these must be greater than or equal to zero.
(c) The intent here is to go beyond the typical 'plug-in' methods most students use. One can apply actual skills and reasoning!
(d) Even the strongest students fall into a 'trap' set in these questions. Try them in your classes!
(e) Do these kinds of questions develop algebraic reasoning and mathematical power OR are they just your typical 'tricky' SAT-type that has little value outside the test?
(f) Make up your own version of one of these or, even better, encourage your students to invent their own!

Beyond Mixed Nuts --A More Challenging Ratio Problem

Readers are strongly encouraged to read the extensive discussion of methods in the comments.


In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?

Comments:
(1) Isn't it a shame that the Jn:Sn ratio isn't 5:7!
(2) Algebraic methods seem to be the most reasonable here, but would some students attempt straight numerical methods using common multiples? Can you find such a way?
(3) Does anyone ever use or teach tree models for ratio problems? We know they are useful for probability problems, but isn't there a similarity here?
(4) Would a Singapore bar model approach work here? I'm counting on those who are proficient with this method! This is not my area of expertise...
(5) Is this question too difficult for the SATs? More appropriate for a math contest question?

A Much Harder Mental Math 'Trick'?? - Algebra May Not Be Optional!

Update: For more background on the investigation below, I strongly encourage readers to visit Sol's wonderful blog wildaboutmath and, specifically, the post he wrote about the algebra behind multiplying 2- and 3-digit numbers. Further, he developed a video (mathcast) to demonstrate the method for 'mentally' mutiplying any two 3-digit numbers. Sol inspired me to develop this activity!


Target audience: Strong Algebra 1 or Algebra 2 students

Ok, I hope you enjoyed challenging yourself or your students to describe a step-by-step procedure for squaring three-digit numbers with a middle digit of zero. I believe many middle school students could devise a method using only operations on the digits, without the need to use algebraic representation or proof. I really believe that recording and organizing the necessary data and the search for patterns make that a valuable activity. Perhaps, even more importantly, the verbalization of the strategy or trick may be the most important benefit for our students! Finally, I strongly believe these kinds of investigations deepen student understanding of place-value and develop number sense.

But now we will move on to a much more significant challenge. I highly doubt that any of your students will be able to find the right combinations of digits to square any 3-digit number! One will still have to record the data and attempt to see a pattern, but, this time, a non-algebraic approach may be too formidable. I am fully aware that there are classical mental math methods for multiplying numbers (Vedic Maths and Trachtenberg to name a couple), but I didn't research those when writing this activity. I simply applied algebraic methods and looked for an algorithm that made sense to me. Thus my method may be similar to those others or not!

This challenge should be frustrating for some and many of you will question whether there is any benefit in looking for some strategy for squaring such numbers by some artificial-looking or convoluted combinations of digits. If the object here were to develop an easy mental math trick, you'd be right. BUT that is not my objective - the idea is to use algebra to motivate an alternate approach to multiplying 3-digit numbers. And, yes, there is some mental math involved, although, I suspect one will need to record intermediate results and do some paper and pencil arithmetic along the way (sorry, no calculator allowed other than to check your answer!). In the end, students should have a much deeper understanding of the meaning of digits, place value that is.

We begin with the algebra prerequisite for this problem, then move on to the real challenge!

READER/STUDENT CHALLENGE/ACTIVITY

(1) Show that
(p+q+r)² =
p² + q² + r² + 2pq + 2pr +2qr

Ok, here goes...
(2) Using the algebraic formula above, devise a method/strategy/algorithm/procedure/'trick' using combinations of digits to square ANY THREE-DIGIT NUMBER 'HTU'.
Note: Again, I reiterate, others have probably invented much cleverer methods for multiplication than this. The key here is to use the above algebraic representation!

In the end, you should be able to do an example 'mentally' like

463² = 214,369 (or should I write it as 21-4-3-6-9).
Lots of 'carrying' here or 'regrouping' as we say these days! If I give you the intermediate step, it may give it away. If I don't, you may give up! Let's see what happens...

An Exceptional Math Contest Problem

Update: The solution I emailed to the director of the Math League is at the bottom after you scroll down a ways. Note how this ties into the difference of squares problem I posted the other day!

A very good friend of mine who runs one of the most successful math league contests in the US has given me permission to reprint the question below with the following attribution:
“Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.”

This question was the last question on a recent contest. It certainly made me think! I felt the need to be rigorous about justifying one of the steps and that was the reason for the
M² - N² = 12 post from the other day. There are at least 3 methods that have been found thus far, according to my friend. Teachers often send in their own solutions and some even get included with the official solutions. My feeling is that questions like these are vital for our students intending to pursue higher mathematics, not just for those who happen to participate in contests. I also believe that variations on great questions like this will have an eternal life...

THE QUESTION

There's exactly one real number a for which ax² + (a+3)x + (a-3) = 0 has two positive integer solutions for x. What are these values of x?

Have fun with this!
For more information about the Math League, visit here.

Scroll down to see the solution I sent to the directors of the Math League...











First of all, I'll state the answer I obtained before giving the details:

A real number a which produces two positive integer solutions is a = -3/7 which does lead to the solutions x = 4 and x = 2. Proving that this is the only real value of a is far more challenging and I suspect that there's a much easier solution than the one I found. Anyway, here's my approach:

In order to simplify the original equation, divide both sides by a to make the leading coefficient equal to 1:

x^2 + (1+(3/a))x + (1 - (3/a)) = 0
To simplify further, let b = 3/a:
x^2 + (1+b)x + (1-b) = 0
From a well-known rule about the roots of a quadratic equation, the sum of the roots in this case equals -1-b and the product of the roots is 1-b. Since the roots must be positive integers, it follows that b is an integer less than -1. Remember for later:
b must be negative!

From the quadratic formula, we obtain the roots to be:
x = ((-1-b) +- SQRT(b^2+6b-3))/2.
In order for the roots to be integers, the expression b^2+6b-3 must be a perfect square.
I will rewrite this expression by completing the square:
(b+3)^2 - 12 and this must equal a perfect square , call it N^2.
Let b+3 = M, then we must have M^2 - 12 = N^2 or
M^2 - N^2 = 12. It's easy to guess a difference of perfect squares equal to 12, namely 16-4, but I will now prove this is the only solution:

Factoring, we have (M+N)(M-N) = 12 where M,N are integers.
The only pairs of factors to consider are: 12,1; 6,2; 4,3 as well as their negatives.
In order for M,N to be integers, the factors of 12 must BOTH be EVEN so the only possibilities are 6 and 2 or -6 and -2..
If M+N = 6, M-N = 2, then, by adding, M=4 and N=2. From b+3 = M, we would have b = 1, but this doesn't work since b must be negative (see above!)
Therefore M+N = -6 and M-N = -2. From this we have M=-4 and N=-2.
Then b = M-3 = -4 - 3 = -7.
b =-7 leads to the quadratic equation: x^2 + (1+(-7))x = (1-(-7)) = 0 or
x^2 - 6x + 8 = 0. Factoring we obtain the roots to be x = 4 and x = 2.

Since b = 3/a, it follows that -7 = 3/a or a = -3/7.

M^2 - N^2 = 12...Prove there is only one solution in positive integers and much more

Update:
(1) See the visualization for the difference of squares posted on 1-9-08.
(2) Read the comments in this post for considerable clarification and instructor guidelines and suggestions. Mathmom's and Eric's comments are particularly insightful.

This post can be developed into an activity for prealgebra through first-year algebra students (or even 2nd year algebra). The last part is more challenging.
The focus here is on developing a method/strategy that can be used to solve similar Diophantine equations. The other objective is to introduce the ideas and methods of proof. This problem may later be used to solve a recent math contest problem for which I obtained permission to discuss on this blog. I am fully aware that many students will 'solve' these equations by Guess-Test methods, but they need to go further.

STUDENT/READER PROBLEMS/ACTIVITIES

(a) Prove there is only one solution in positive integers for the equation:

M² - N² = 12

Note: If we omit the word positive, what would the solution(s) be?

(b) Determine all positive integer solutions:

M² - N² = 15

(c) Determine all positive integer solutions:

M² - N² = 36

(d) Let's investigate for what positive integer values of P, M² - N² = P has NO solutions in positive integers.

(i) Determine at least 5 positive integer values of P for which the above equation has no positive integer solutions.

(ii) (More challenging) Describe all values of P for which the above equation has no solutions. Justify your result.

Note: All students should have success with (i), although some may struggle to find 5 values. Part(ii) should challenge the student who has finished the other parts in rapid order and sits there complacently!

Additional Comment: If P is itself a perfect square, our equation is obviously related to the most famous equation in geometry. Thus, if P = 9 or P = 16, for example, students should recognize something! For this reason you may want to have students consider these values when doing this investigation. More to come...

Another View of sin(A-B) for a Special Case: An Investigation

[As always, don't forget to give proper attribution when using this in the classroom or elsewhere as indicated in the sidebar]

In a standard trig unit, students learn those wonderful formulas for the sin and cos of the sum and difference of angles. Many creative methods have been developed to derive these formulas and, depending on the ability of the group and teacher preference, these are demonstrated or not. Students are typically shown various mnemonics for recalling them on the big test, but, in this investigation, students will derive sin 15° using only 30-60-90 triangle ratios and the Pythagorean Theorem. We will then compare the result to that obtained by the traditional formulas for sin(45°-30°) or sin(60°-45°) and show equivalence by algebraic methods using radicals. This is not an attempt to develop a general approach to deriving sum/difference formulas, although readers are invited to try a generalization. You may recall other posts on this blog of a similar nature.

THE PROBLEM/INVESTIGATION
Refer to the triangle above. If the print is too small, click on the image to magnify.
∠A = 75° and ∠B = 15°

(a) In the triangle above, locate point D on side BC such that ∠CAD = 60° . Express the lengths of the sides of triangle CAD in terms of a.
[Note: We could avoid the variable a altogether and assign a value of 1 since this is a ratio problem.]
(b) Show that CB = a √3 +2a.
(c) Use the Pythagorean Theorem to show that AB = a √(8+4 √ 3)
(d) Verify the identity (√ 6 + √ 2)² = 8+4 √ 3. Use this to rewrite AB.
(e) Use above results to obtain an expression for sin 15°.
(f) Use the standard trig formula for sin(45°-30°) to obtain an expression for sin 15°.
(g) Show your results in (e) and (f) are equivalent.

An Instructional Aside
When introducing the formula for sin(A+B), for example, teachers sometimes motivate it effectively using numerical values or considering the special case A=B. Here's an alternative:
Consider the special case A+B = 90°
Ask students to verify the formula for sin(A+B) in this special case. Simple, but at least it's something slightly different to pique their curiosity.

Standard Disclaimer:
This investigation is not copied from some other source. As it is original and has not been edited by others, there's always the possibility of error. Please feel free to suggest corrections/edits/extensions...
You know I welcome your comments!

And vs. Or: Developing Deeper Understanding in Algebra

While some are awaiting the remaining chapters in Alec Klein's interview, I know the rest of my readers are wondering why I've gone into early hibernation. I'm actually doing some independent mathematical research apart from anything on this blog but I will not bore you with the details at this time. That may change (i.e., I may bore you later!).

In the meantime, since the central theme of this blog has always been developing student conceptual understanding, here are a couple of problems for you to consider giving to your Algebra 2 students (or beyond). Conjunction vs. Disjunction is often misunderstood by students and these ideas appear in so many contexts in mathematics, from absolute values to inequalities and beyond. Consider giving these as warm-ups, for review, practice for SAT's, etc. I'm not suggesting these are difficult or challenging problems. Their purpose is to promote deeper reflection on the part of the student. Students who have strong background and understanding will simply solve these quickly and not see why anyone would make a big deal over them. However, you may find other students who don't grasp the ideas as readily or have forgotten. Comparing/contrasting is a powerful heuristic when trying to develop a more profound understanding of mathematics...

1. If (a -4)² + (b+4)² = 0, what is the least possible value of a² + b²?

2. If (a-4)(b+4) = 0 , what is the least possible value of a² + b²?
(A) 0 (B) 8 (C) 16 (D) 32 (E) 64

Ask students to explain to each other, why the word 'least' is irrelevant in Question 1 but not in Question 2. Also, how does question 1 relate to circles?