[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!

[Pri_20151224RAMN] A Coin Problem with Constant Difference

Problem

Danny saved some  50-cent coins and  $1-coins in his coin box.  The total value of the  50-cent coins to the total value of the  $1-coins he had was in the ratio  2 : 5.  After  $14  worth of  50-cent coins and an equal value of  $1-coins were added to the coin box, the ratio of the total value of  50-cent coins to the total value of  $1-coins became  5 : 9.  How many coins of each type did Danny have in the end?

Introduction

     Here is a “Singapore math” coin problem that can be befuddling for kids and even for adults.  To rub salt to the wound (or pour oil to the fire?), the value of a collection of coins is different than its number.  Whilst a $1-coin obviously has a value of one dollar, you would need two 50-cent coins to make up a dollar.

Strategy

     Notice that after adding  $14  worth of coins to both types of coins,  the difference in the total value of the two types of coins remains the same.  Some people call this a “constant difference” problem.  But how do we exploit this constant difference, when the type of ratio units used in  2 : 5  are most likely not the same as those used in  5 : 9?  Well, we need to bring them to a common unit! [H09]   How?  Read on!

Solution   [H02, H06]

Ans: Danny had  54  $1-coins and  60  50¢-coins in the end.

Commentary

     I am using Distinguished Ratio Units in my presentation.  This makes it clear that the ratio units are of different types.  In the the “before” stage [H06], the difference in the value of the two sets of coins is  3  circle units.  In the the “after” stage, the difference in the value of the two sets of coins is  4  square units.  But we know these two differences refer to the same numerical number.  The Lowest Common Multiple of  3  and  4  is  12.  So both of them must me equal to  12  common units (which I envelop with triangles).  We multiply the numbers inside the circle units by  4  and we multiply the numbers in square units by  3.  I put these multiplications in quotation marks because we are not really changing the numbers of coins.  We are merely changing the type of units used.  I am saying that each square unit is the same as  3  triangle units and each circle unit is the same as  4  triangle units.

     Once we bring everything to common units (triangle units), we can see the  $14  added corresponds to  7  triangle units.  Henceforth the whole problem unravels easily.  [H11, H05]

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any independent learner who is interested

[AM_20151224QERI] Quadratic Roots and Use of Identities

Problem

Introduction

     Here is a fairly standard question on roots of quadratic equations, except that part (iii) is slightly more challenging.  To solve this question, one must know the square of sum identity well.

Recapitulation

     Please refer to this previous article  and  this article  for the theory on quadratic roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* other syllabuses that involve roots of quadratic equations

* any learner who is willing to learn

[AM_20151223DATP] A Horizontal Tangent and a Faux Asymptote

Problem / Question

 

Solution 1

Solution 2  (not using differentiation)

Remarks

     This problem just happened to be put as an exercise in a textbook under the applications of differentiation.  But who says one must use differentiation?  Once again, there are at least two ways to solve a mathematical problem.  In this instance, it happened that the equation of the curve can be put into a quadratic form that is amenable to analysis by the discriminant.  Mathematics is about mental flexibility and creativity, actually.

     An asymptote is a straight line that the curve goes near to (but does not touch), as  x  gets large or gets very negative.  The book’s use of the phrase  “tends towards the line  l” may be wrong or imprecise.  Technically, the line  l  is not an  asymptote, because if you analyse or plot the graph, the gap between the curve and the line does not really get closer and closer.  However, the ratio of  y  over  x  gets nearer and nearer to  -1  and the gradient of the curve also gets nearer and nearer to  -1.  What really happens is: as  x  increases, eventually the curve becomes almost parallel to the line, but does not go near it.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics,  “IP Mathematics”

* revision for  GCE ‘A’ Levels H2 Mathematics

* revision for IB Mathematics HL & SL

* revision for  Advanced Placement (AP) Calculus AB & BC

* other syllabuses that involve differentiation and/or quadratic functions

* any precocious or independent learner who wants to learn

[S1_20151221ABEX] The Table as the Years go by ...

Problem

Simon is  15q  years old.  He is now  5  times as old as his son.  How old will he be when his son is  28  years old?

Introduction

     This is an introductory algebra problem, good for getting used to the language of algebra.  As we have seen in this previous article, tabulation is a good way to help organise our information.  Although no one will penalise you for not using tables, once you start using tables, you wonder how you could ever survive without them.

     I present two approaches.  One way is to consider the number of years passed by.  A second way is to observe that the age difference always remains the same as time goes by.

Solution 1

	Now	future
Simon	15q	?
Son	3q	28

The number of years passed is  28 – 3q.

Simon’s future age = 15q + (28 – 3q) = 12q + 28

Ans: When the son is  28  years old, Simon will be  (12q + 28)  years old.

Solution 2     (Refer to table as above)

Note that the age gap always remains the same.

Age difference = 15q  – 3q = 12q.

Simon’s future age = 28 + 12q.

Ans: When the son is  28  years old, Simon will be  (12q + 28)  years old.

Remark

     The answer required is an algebraic expression, in terms of  q.  Since we do not know the value of  q,  do not try to evaluate the expression, but just leave it as it is.  When learning algebra, one needs to get comfortable working with unknowns.

     Also remember to be mentally flexible.  There may be more than one way to “skin the cat”.

Heuristics Used

H02. Use a diagram / model  [tabulation]

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

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

* other syllabuses that involve ratios and algebra

[Pri_20151223WNSV] Unravelling Four Whole Numbers

Problem

If  A,  B,  C  and  D  are whole numbers such that  A × B = 8,  B × C = 28, C × D = 63,  B × D = 36,  find the values of    A,  B,  C  and  D.

Introduction

     This question seems to be taken from a secondary school textbook from a chapter on linear equations.  However, I think a good  primary school pupil could attempt this.

Strategy

     The key to solving the above problem is to make observations.  When you multiply up the first two equations, you get an  A,  a  C  and two  Bs  in the product.  Hmmm ... This doesn’t look promising ...  Ah!  But when you multiply the second and the third equations together, you get an  B,  a  D  and two  Cs  in the product.  This can cancel (via division) with the fourth equation which has one B  and one  D  in the product.

Solution

Remark

     Always cancel as much as possible, to avoid large numbers and reduce chances of making careless mistakes.

     By the way, a whole number is a non-negative (zero or positive) integer that does not contain any fractional part.   As such, the set of whole numbers is {0, 1, 2, 3, 4, ...}.  Thus we do not need to consider the negative square roots.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics (Challenge)

* Lower Secondary School Mathematics (Challenge)

* other syllabuses that involve whole numbers

* anyone game for a challenge

[H2_VJC2015PromoQ10_IAXS] Volume of a Doughnut

Problem

Introduction

     This is a problem involving the calculation of the volume of solid of revolution of an enclosed region.   The junior colleges (or senior high schools) like to set this type of question. 

Technique

     If the axis of revolution is the  y-axis, the basic formula is   ò px² dy   with the appropriate lower and upper limits.  Notice that this only works for the region between one curve and the axis and when rotated, this will generate a solid with no hollow parts.  An enclosed region, however, consists of two curves.  In our case, when we make  x  the subject, we find that we have two choices.  One of them leads to a curve that is further away from the axis of rotation.  I call that the outer curve.  The other curve is the inner curve, and this is nearer the axis of rotation.  We need to subtract the volume generated by the inner curve from that generated by the outer curve.

Solution

Remarks

     In this example, the curve on the right happens to be the outer curve.  If the equation were

(x + 93)² + y² = 15²,  the circular region would be on the left of the  y-axis and the outer curve would be on the left.

     For your information, the above solid of revolution is a torus.  This is the shape of a doughnut, (or hoopla-hoop, circular tube, or Polo mint perhaps?).  It is the inner curve that gives the hole in the “doughnut”.

     You can imagine in your mind’s eye that as the circular disk revolves around the  y-axis, its centre traces out a circular path of  93 units.  By the Second Centroid Theorem of Pappus,

     volume = length of path of centroid × area of cross section = 2p(93) × p(15)² =  41850p²

In general, the volume of a torus with major radius  R  and minor radius  r  is

     volume = 2pR × pr² =  2Rr²p²

If you know this fact, you use it to check your calculations.  Although this is not in the H2 Syllabus, but it is something interesting to explore.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards                                [e.g. making  x  the subject]

H09. Restate the problem in another way  [symmetry: volume is twice of upper half]

H10. Simplify the problem                         [integration by substitution]

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* IB Mathematics HL (Applications of Integration)

* Advanced Placement (AP) Calculus AB & BC

* University / College Calculus

* other syllabuses that involve Applications of Integration

* any precocious or independent learner who loves to learn