[AM_20151226EIQR] Looking for a Pea among Quadratic Roots?

Question

Introduction

     This question is about finding the parameter  p, and not about solving for the “unknown”  x.  It is heavy on algebra, one has to be patient, careful and meticulous.  Please refer to this article for a recapitulation of (Vieta’s) theory of Quadratic Roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots
* any learner who is interested

[S1_20151226NPSW] Finding the General Term of a Sequence (2)

Problem

Introduction

     The above was discussed in this previous article.  The earlier parts of the problem are easy.  The major sticking point is finding the formula for  S_n.  We solved that using factorisation and observation, which I feel is the best way.  But what if you cannot do that and you are desperate (for example, in an exam or test)?

     This article introduces Newton’s Method, which can be used as a back-up method, even though it is not in the regular syllabus.

Solution (Newton’s Method)

 

Remark

     Note that number sequences in “IQ tests” (with no problem contexts) have been debunked.  In our case here, the numbers do have a certain regularity arising from the pattern of dots.  In fact this is an arithmetic progression.  What we are calculating is the sum of an arithmetic progression.  However, Newton’s Method extends beyond arithmetic progressions.

Suitable Levels

* Lower School Mathematics

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* GCE ‘A’ Levels H2 Mathematics (revision)

* other syllabuses that involve number patterns and sequences

* any precocious or independent learner who is interested

[S2_20151226EFQF] Factorisation without Trial and Error?

Problem

 

Introduction

     This problem was posed by a student going on to Secondary 1 (~ grade 7) next year.  This sort of problem is usually done at Secondary 2 or 3 (about grade 8 or 9).  This reminds me of my personal story.

     I accidentally discovered quadratic equations when I was in Primary 4.  I imagined a rectangle whose length is  2 cm  longer than the breadth.  If the breadth is  4 cm, the length is  6 cm and the area is obviously  24 cm².  But if I pretended that I knew the area but did not know the dimensions, I did not know how to solve it with the knowledge that I had at that time.  This started me on a quest to find out the answer.  I read secondary school guidebooks, asked my friend’s brothers and sisters, and even asked my Chinese teacher (who, after exams, offered to answer any question we had)!  Basically, I was offered two choices: (1) trial and error factorisation  and  (2) the quadratic formula.  I did not like guess and check (or hit and run?), and the quadratic formula looked formidable to me.

     So I started a quest to find a method of factorisation that did not require trial-and-error.  By secondary 1, after fiddling around with algebra, I managed to do it.  I reconstruct my derivation below.  And then I use my method to solve the above factorisation problem.

Derivation

Solution

Remark

     This looks like a Pyrrhic victory.  But like they say, it’s the journey and not the destination that matters.  Doing my own explorations prepared me for future learning and made me understand better.

     Nowadays, the new models of calculators give solutions to the associated equations and you can work backwards to get the factorisation.  Unfortunately, many students just blindly use this and forget to work backwards, giving the wrong factorisation.  If calculator gives 9 and ^-248/₂₉, and you write your factorisation as (x – 9)(x + ²⁴⁸/₂₉), your answer is wrong. Moral of the story: you still need to use your brain.

[S1_20151226NPGT] Finding the General Term of a Sequence (1)

Problem

Introduction

     This is a typical Secondary 1 type of problem involving number patterns.  Students are usually able to see the link between successive terms, but the general formula seems to be a challenge for most.

Strategy

     In case this is not obvious, every time you go to the next diagram, you add four dots on the outside.  So you can fill in the table very easily.  For diagram 5, there would be  19  dots and the total number of dots up to diagram  5  would be  55.

     What is the number of dots for diagram  1 000  or any number  n  for that matter?  Now, some students may have a problem predicting beyond the first few numbers.  What we need is a expression or formula that predicts the number of dots given the diagram number  n.  You know that the sequence  3,  7,  11,  15,  ...  follow a pattern where you keep adding  4.  Have you encountered a sequence in which  4  is added each time?  Yes!  It is the 4 times table.  Suppose we have the 4 times table.  [H08]   Let us do a comparison between that and  D_n.

diagram #	1	2	3	4	5	...	n
4 times table	4	8	12	16	20	...	4n
Dn	3	7	11	15	19	...	?

The numbers in  D_n  are always one less than those in the  4  times table.  So  D_n = 4n – 1.

Solution

n	number of dots for the  nth  diagrams Dn	Sum of number of dots for the first  n  diagrams Sn
1	3	3
2	7	10
3	11	21
4	15	36
5	19	55

     D_n = 4n – 1

     S_n  = n(2n + 1)   ©

Commentary

     How can we get the formula for  S_n?  We can do so by trying to factorise the numbers [H09], and then look for pattern.  [H04, H05]

           3 = 1×3   = 1×(2×1+1)

         10 = 2×5   = 2×(2×2+1)

         21 = 3×7   = 3×(2×3+1)

         36 = 4×9   = 4×(2×4+1)

         55 = 5×11 = 5×(2×5+1)

         ...

                      S_n = n(2n + 1)   ©  bingo!

But what if you have poor observational powers and if you are desperate?  There is a secret weapon to handle this!  Please refer to this article.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels

* Primary School Mathematics (challenge)

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve number patterns and algebra

* any precocious or independent learner who loves number patterns

[S1_Expository] Recurring Decimals and Rational Numbers

Problem

 

Introduction

     A student asked the above question on Facebook.  This article explains recurring decimals, which is part of the topic on real numbers in the Singapore Secondary 1 Mathematics syllabus. 

     By the way, I do not believe that Asians are inherently better at mathematics.   A few students are working over Christmas to prepare for the next years’ work.  Here we go!

Notation

     Personally, I prefer the horizontal bar notation which is what I learned during my time as a student, but nowadays in Singapore schools, we tend to use the dot notation.  There is no right or wrong about this, but it is just a matter of convention.  When in Rome, do as Romans do.

Coversion to a fraction

 

Concluding Remarks

     A rational number is a number that can be expressed as a ratio or  fraction  ^p/_q  where  p  and  q  are integers with  q ¹ 0.  The fraction can be proper or improper.  Since any recurring decimal can be converted to a fraction,

every (infinitely) recurring decimal is a rational number.

Now,

every finitely terminating decimal is also a rational number.

For example, 0.171 = ¹⁷¹/₁₀₀₀.  The numbers that cannot be converted to fractions are called irrational numbers.  How do these numbers look like?

The irrational numbers are exactly the numbers with non-terminating (infinite) and non-recurring decimal expansions.

Some examples of irrational numbers are
       p = 3.141 592 653 589 793 238 462 643 383 279 502 884 197 ¼
       e = 2.718 281 828 459 045 235 360 287 471 352 662 497 757 ¼

     Ö2 = 1.414 213 562 373 095 048 801 688 724 209 698 078 570 ¼

The decimal digits of irrational numbers never end, but they do not have any repeating pattern.  There are actually much “more” irrational numbers than rational numbers, but this is a fact that is technically profound, way beyond the secondary syllabus.  The interested reader can refer to this article.

Suitable Levels

* Lower Secondary Mathematics (Sec 1 ~ Grade 7)

* revision for GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve recurring decimals

* any independent learner who is interested

[S1_20151225ABEX] Apples and Cherries on Christmas?

Problem

The ratio of the mass of an apple to the mass of two cherries is  9 : 1.  The mass of the apple is  150 g.  What is the number of cherries that can be found in  y  kg?

Solution                

Remarks

     To obtain the answer, we made a simplifying assumption that all the apples and cherries are identical in mass.  The answer is an algebraic expression and it can be obtained by following the same procedure one would solve the problem if it were in concrete numbers.  Learning algebra is like learning a new but more powerful language.  It takes some time getting used to.  Since we do not know the value of  y,  we leave the answer in terms of  y.  But if we knew the value of  y,  we would know that the answer is  120 times that.  For example, with  3 kg,  we get (about)  360 cherries.

    

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H08. Make suppositions

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Primary 6 Mathematics (challenge)

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve ratios and algebra

Happy Holidays to all my readers!