[AM_20160125DAHT] Horizontal Tangents via Quadratic Discriminants

Problem

Introduction

     This is a Additional Mathematics textbook problem.  This question is of an intermediate level of difficulty.  The general method is by differentiation.  The equation of the curve happens to be capable of being put into a quadratic equation in  x.  Hence we can also use the theory of quadratic discriminants.  I present both methods of solution.

Method 1 (Using differential calculus)

Method 2 (Using quadratic discriminants)

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve differentiation or quadratic discriminants

* any independent learner who is interested

[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

[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.

[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

[H2_20151123APGP] Factor Theorem with Arithmetic and Geometric Progression

Question

Introduction

     This question tests students on their knowledge of arithmetic and geometric series.  They should also be familiar with Factor Theorem and methods of dealing with polynomials.  Once parts (i) and (ii) are solved, part (iii) is quite straightforward, provided that the student remembers how to deal with surds.

Review of Important Facts

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB HL Mathematics

* other syllabuses that series and Factor Theorem

[Pri20151116DNGC] Quadratic Plays Second Fiddle in Product-Difference Riddle

Question

What two numbers give a product of  21.5  and a difference of  6.1?

Introduction

     This question reminded me of a question that I set myself when I was in Primary 4 (» grade 4).  I imagined a rectangle with breadth 4  and  length 2 units longer then the breadth (i.e. 6) giving a area (product) of  24.  Then I pretended that I did not know the breadth and let it be  x.  This led to a quadratic equation  x(x + 2) which I did not know how to solve (if I did not know the answer).  So I accidentally discovered quadratic equations when I was in Primary 4.  This led me to a quest to learn the method of factorisation (by “trial and error” or “guess and check”) and the quadratic formula.  I never liked trial and error.  So I continued in my quest to invent a method of factorisation that did not require “guess and check”.  I finally succeeded doing that in secondary 1 (» grade 7).  This turned out to be a Pyrrhic victory.  The method I invented was quite similar to the quadratic formula.

     There is a place for “guess and check” in mathematics.  I present a simple solution to the above problem using just that.

Solution

smaller #	larger #	product
2	8.1	16.2	û
3	9.1	27.3	û
2.5	8.6	21.5	ü

Solved! J

Heuristics Used

H02. Use a diagram / model    (table)

H05. Work backwards             (if the smaller number is this, what is the bigger number?)

H07. Use guess and check

H09. Restate the problem in another way      (area = product)

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve decimal numbers

* anyone who loves to exercise their minds

[AM_20151105QFER] New Quadratic Equation satisfied from New Roots

Question

The roots of the quadratic equation  2x2 – 3x + 6 = 0  are  a  and  b. (i)   Without finding the value of  a,  show that  8a4 = 18 – 45a. (ii)  Find the quadratic equation whose roots are  (a2 + 1)  and  (b 2 + 1).

Introduction

     Do you know what a “root” is?  Is it like radish or ginseng?  Do you know what “satisfied” means?  Is it that nice feeling you get when you eat carrots?  Read on!

     The featured problem above is modified from an original question that contained an error.  The modified part is shown in red.  I present two solutions.  The first solution uses pretty much standard theory, and I use notations  a’  and  b’  to denote the new roots  (a ² + 1)  and  (b ² + 1)  respectively.  For the second solution, I present an alternative working part (i), and one using the method of substitution for obtaining new equations (not usually taught in schools at the secondary level) for part (ii).  But before that let me first explain what “root” and “satisfied” means.

Recapitulation of Standard Theory

Solution 1 – Using Standard Theory

Solution 2 – Using the Method of Substitution for part (ii)

 Remarks

     Once again we can see that there are many ways to skin the cat, as it were.  Mathematics is not about following a fixed procedure.  There are various truths, notions and rules that are inviolable.  But other than that, you can have as much creativity as you want!

HeuristicsUsed

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 ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots

* whoever loves roots and enjoy being satisfied by conquering mathematical challenges J

[U_20151029ITCX] Roger Cotes’ Integral of the Reciprocal of x^n – 1

Question

 

Introduction

     The book VisualComplex Analysis by Tristan Needham recounts the story of RogerCotes who considered the above problem.  Without ostensibly using complex numbers, Cotes discovered a geometrical principle that helped to factorise the denominator  xⁿ – 1,  and hence decompose the above integral.

     In this article, I am going to “cheat” by using complex numbers to split up the denominator.  The fact that the denominator splits completely into a product of simple linear factors makes it easy to decompose the integrand into partial fractions.  Once this is done, I can single out the one or two fractions with purely real linear denominators.  Then I can pair up the conjugate fractions to get fractions with real quadratic denominators.  In other words, I apply a divide-and-conquer strategy, splitting up a big problem into smaller problems (Heuristics!).  Then I collect all the partial answers together to form my final answer.

Solution

Remarks

     Note that I have only used real integration, not complex integration.  Complex numbers are used only to derive the various algebraic fractions.

HeuristicsUsed

H02. Use a diagram / model   (mentally: imagine roots of unity in a circle)

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* University / college level calculus

* high school students very strong and interested in calculus and complex numbers

* anybody who loves a challenging calculus problem and complex numbers

[AM_20151027DAX] Quadratic quagmire?

Question

Introduction

     This question was most probably taken from an Integrated Programme (IP) school in Singapore.  For your information, students in Integrated Programme schools do not take the GCE ‘O’ Levels, and each IP school is free to design its own individual curriculum.  In practice, they incorporate the mainstream GCE ‘O’ Level topics, as well as additional topics, and/or teach topics in advance, and may call their syllabuses by different names.  They tend to set more challenging questions than the mainstream schools, which are already targetting their internal examination standards above the ‘O’ Levels.  In other words, they tend to cram in more, but never less.  Part (d) tests approximate change, which had been taken out of the mainstream syllabus at this time of writing.
     Note that the IP schools tend to be schools that traditionally attract the academically best students from each cohort.  Even before the IP programme was introduced, these schools were already setting harder questions.  Anyway, this blog welcomes everybody from all around the world who is willing to learn, regardless of the type of school they are from, even home-schoolers and independent learners!  Let us see how we can employ re-usable tactics to tackle this question.

Get to the root of the matter, fast!

     The question is on the applications of differential calculus on a quadratic curve (a parabola).  Observe that the equation of the curve is given in completed square form.  From the equation, can you spot the line of symmetry (centre line) and the  y-intercept immediately (like within 5 seconds)?  [You need to know all the basic facts at your finger tips and make observations.]

The line of symmetry always passes through the maximum or (in this case) minimum point.  We know that the minimum is when the squared term  (x – 3)²  is zero.  So

                              the line of symmetry is  x = 3.

To get the  y-intercept, put  x = 0.  This gives  y = (-3)² = 9.  So 

                              C = (0, 9)  and equation of  CD is  y = 9

Since  PQ = 2k,  distance from  P  to the centre line = k.  All the above are basic observations that should be carried out mentally within one minute and you should be able to mark the diagram with pencil notes (shown above in blue).  Once this is done, let us get on to the real business.

Solution

Final Remarks

     For part (c), they have already told you it’s maximum, so you do not need to prove that it is maximum.  However, intuitively it is obvious there is a maximum: imagine if  k = 0  or  k = 3, then we get very thin rectangles with area zero.  As  k  increases from  0, the area gets bigger and after that shrinks towards zero again. 

     Actually, was this problem really so difficult?  What did we do to solve it?  Let’s review

· Know all basic facts and skills thoroughly at immediate recall (e.g. extremum of parabola lies on line of symmetry, how to spot that from completed square form, how to find y-intercept)

· Use heuristics: e.g. make observations.  Use simple facts you already know (e.g. subtract lengths, substitute values of  x  to find  y  which is “height” have x-axis etc,).  Practice positive psychology: instead of worrying, write down everything you can deduce.  Then try to find connections.

· Apply the formulas

· Use your intuition to see whether your answer makes sense.

· For part (d), if it is not in your syllabus, do not worry about it.  But if you are curious or feel the itch to learn more, it is also not too difficult.  See the boxed formulas in the solution above.  Just remember that the ratio of small changes  ^DA/_Dk  is approximately equal to the derivative  ^dA/_dk.  You can detach the  Dk,  bring it to the other side of the equation, and that allows you to approximate  DA.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

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

· other syllabuses that involve applications of differentiation

· anyone who is interested in calculus!

[H2_20150512PCD] Quadratic Discriminant for a Parametric Curve

Question

Introduction

     To begin with, do you notice that there are many letters (in italics) in the above?  There seems to be a confusing mix of variables and constants.  There are only three variables:  x,  y,  and  t.  The constants are  a,  s,  and  p.  There is another letter ‘l’,  which is the name/label for a straight line.  It is good to highlight or mentally mark these different things as different.

     Since the topic is on parametric differentiation, it seems that you need to use differentiation to solve this question.  Notice that the equations involved are at worst quadratic?  No square roots, cosines, logarithms, cubes, exponentials ... etc.  Whilst it is not wrong to use differentiation, there is a slick way – using quadratic discriminants.  In fact, this is the first thing you should think of if you see that the equations involved link to a quadratic equation.

Solution

Remarks

     We should always make it a habit to check and justify division by zero.  It is dangerous to divide an equation throughout by a variable or constant if you do not know what it is, or whether it is zero.  Make sure it is not zero before dividing.

     The quadratic discriminant method cannot be used unless you have things that reduce to quadratic equations.  But when it can be used, it is very powerful and it gives a direct answer.  Note that here we are not solving for the variable  t.  We are solving for the constant  s  in the first part, and for the constant  p  in the second part. 

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels, H2 Mathematics 

* International Baccalaureate Mathematics 

* other syllabuses that involve quadratic discriminants

[IBHL_SOTA201304_1B10c] Quadratic Discriminants

Question

Important Reminders

Solution

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics

[IBHL_SOTA201304_1B07] Quadratic Equations and Roots

Question

Important Reminders

Solution

     Actually we could have multiplied by  -4  or any multiple of  4  for that matter, but this is the set of integer solutions for which  a  is the least positive.

     For part (b), if we can solve the first equation easily, then the roots of the second equation can be obtained by just squaring your answers.  However, the LHS of the first equation cannot be factorised nicely, so we might as well use the quadratic formula to solve the second equation directly.

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics