[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

[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

[S1_20150516AELI] Apples and Pears on a Table?

Question

Michael has enough money to buy either 12 pears or 36 apples. if he intends to buy equal number of pears and apple, how many of each fruit can he buy with the money

Introduction

     This is an interesting algebra problem meant for secondary 1 students (» grade 7) that has the potential to lead to a system of complicated simultaneous equations in two variables.  Fortunately, there are some interesting and simpler approaches.  I present three of them below.

Solution 1 (using a Table and Unit Costs)

Solution 2 (Insight from proportion)

     This is perhaps the intended algebraic approach.  Observe that since  12 pears are as expensive as  36  apples,  1  pear is “equivalent” to  3  apples.  Michael’s budget is  36  apples-worth of money.  If there are  n  pears and  n  apples,  the  n  pears can be exchanged for  3n  apples.  We arrive at an equation as in solution  1.

  

Solution 3 (Acting it Out)

     You can role-play this with your friend using toy-fruits.  Your friend is the fruit-seller and you are the buyer.  No toy-fruits?  Well, use some counters, bottle caps, Lego bricks, ... whatever to represent the apples and pears.  If you are a lonely person, or if your friend is too busy, or your mother has thrown away all your toys since you are (sort of) more grown up already, maybe just do a thought experiment.  Imagine, at first, you took  36  apples to the check-out counter.  Then you saw some pears, and you grabbed  12 of them, ditching the apples.  You figure out that  1  pear is equivalent to  3  apples.  Just then, you change your mind.  You decide that you want an equal number of apple and pears.  OK, so you replace  1  pear with  3 apples.  You get  11  pears,  3  apples.  Keep on swapping pears for apples.   You get  10  pears,  6  apples.  Then  9  pears,  9 apples.  Bingo!

Final Remarks

     There is no magic potion for mathematics (although heuristics is a start).  There is also no one fixed method for you to memorise to solve problems.  As they say, you have more than one way to skin the cat.

Heuristics Used

H01. Act it out

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

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

* other syllabuses that involve ratios or algebra

JCCDQBHWH_FN021(b) Range of Composite Functions

[original source unknown]

Introduction

     This question is from the usual book which did not credit the source.  It comes from some unknown junior college in an unknown year.  It is a challenging question because most students are poor at finding the range of composite functions.  Furthermore, this question has a little twist: you are given the range of the composite function, but you are required to solve for something.  So you need to, in a way, work backwards and/or use inequalities (another weak point for many students).

     Students are reminded of the right-to-left convention for functions in JC as well as GCE ‘A’ level examinations.  This means in ‘fg’ the  g  is done first before the  f.  There are some university professors who use a left-to-right convention, but here we do not.  So take note.

     Again metacogntion and heuristics are very important and I will illustrate their use.

Stage 1:  Understanding the Problem

What is this (part of a) question about?

range of composite functions, solving for unknown

What is given in the question?

The range of  fg.

What is the question asking for?

The value of  k  that leads to the given range.

Stage 2:  Planning the strategy

What heuristics do you think can be used for this question?

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

Can you recall the definition of the range of a function?  The range of a composite function?

Stage 3:  Execution

Any observations that can make your job simpler?

     Yes.  Observe that  g(x)  is a quadratic with positive  x²  coefficient.  So this is a parabola that looks like a happy smile.  To locate the minimum point, we can complete the square (a technique learnt in secondary school).

     g(x)  =  x² + 2x – 1  =  x² + 2x + 1² – 1² – 1  =  (x + 1) ² – 2

when  x = -1,  g(x) = -2.  The minimum point is (-1,-2).  So the range of  g  is all the

numbers from  -2  upwards.  i.e.  R_g = [-2,` \oo `).

So what now?

     With the two-stage method, suppose now we have

                  x ` \in ` R_g

That means?

                  x > -2

That means?

                  x + k + 1 > -2 + k + 1

Why do you do that?

     I want to slowly manipulate the LHS to get  ln(x + k + 1)  which is  f(x).  Continuing,

                  ln(x + k + 1) > ln(k – 1)

                                f(x) > ln(k – 1)

i.e.                            R_fg = [ln(k – 1), ` \oo `).

Why is there no switching in the inequality sign?

     The slope of the graph of  the natural logarithm is always positive (albeit getting less steep for increasing  x).  So applying  ‘ln’  on both sides does not change the inequality.

What is the clue again?

     We are told that  R_fg = [ln 3, ` \oo `).  Aha!  *epiphany*  *light bulbs flashing*

ln(k – 1) must be equal to ln 3!!!  This can be solved easily!

figure 1 – working forward and backwards

Stage 4:  Evaluation

Is the answer correct?

     Substituting  k = 4,  we see that    ln(x + k + 1) =  ln(x + 5)  and with  x > -2,  this will be  > ln 3  as given in the clue.

And why  x > -2?

     This is because g(x) > -2, which we knew  from completing the square.  We treat the  ‘g(x)’  as the  ‘x’  when applying  f,  because this is what  fg(x)  really means.

Stage 5:  Reflection

What did we learn from solving this question?

     We used metacognition to do self-monitor and self-questioning during the 5 stage problem-solving process.

     We used the following heuristics.

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

     We learned to apply the definition of the range of fg.  There are two possible methods: the one-stage method and the two-stage method.  The latter is usually better.

     In the two-stage method, the range of  fg  is found by first finding the range of  g  and then applying the function  f  to it.  After the first step of finding the inequality for  g(x),  we can simply use  x  in the formula for  f.  How?  We set  x  to be in the range of  g(x) from the previous step,  then slowly manipulate the inequality until the expression for  f(x)  appears.  This will give us the range of  fg.

     From the formula for the range of  fg,  we learned how to make use of the given clue to work backwards to find the unknown k.

     Difficult questions can be tackled by thinking systematically and logically, and using heuristics and metacognition.  Mathematics is hard, but it is fun after you have learned it.  If you have really learned it, you become more powerful because you can use the same technique to solve all kinds of problems in future.

Any of your own reflections?  Please post in the comments below.