[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

[S1_AFMLCM_20150412] Conquering Algebraic Fractions

Question

Introduction

     This is an equation involving algebraic fractions, usually for secondary 1 (approximately grade 7) pupils in Singapore.  Many students (and teachers?) like to use the “cross-multiplying” method, as shown in Solution 1.  A usually more efficient method is to multiply every term by the Lowest Common Multiple (LCM) of all the denominators appearing in the equation, as shown in Solution 2.

Discussion

     Note that division by zero is not allowed.  Furthermore, in algebra, it is dangerous to cancel or divide by an unknown quantity, because there is a possibility that you are dividing by zero.  So any division or cancellation by an unknown quantity must be justified beforehand.  Mathematics is not a game of blind senseless manipulations.  If you look at the second solution, which is short and sweet (only 4 steps), multiplying through by the LCM of denominators not only avoids this awkwardness, but it clears all the fractions in one fell swoop.  The solution takes only  4  steps, and it is in fact the recommended method.  All students, whether “good” or “poor” in maths, should use the second method.  Teachers who refuse to use/teach the LCM method (out of habit, or because their own teachers taught them otherwise, or because this makes them or their pupils “uncomfortable”) are really doing the weaker students a huge disservice.  You are widening the achievement gap.  The better students are better, precisely because they use better methods.  The longer one’s working is, the higher the chances of making mistakes and the more time is wasted.  If the “weaker” pupils have to jump through lots of hoops to achieve a certain standard before they are allowed to learn this “advanced” method (actually it’s just the normal method), they will have to unlearn their old method and may get confused as they learn this method.  A triple whammy!  All learners need to practice anyway, so one might as well practice the correct thing right from the beginning and learn good habits (striving for efficient, effective, elegant solutions).  So please, please, please everyone: use the LCM method!

Suitable Levels

* Secondary 1 Mathematics

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

* other syllabuses that involve algebraic Fractions

* precocious children who want to learn algebra

[S2_20150402RLD] Fifty balls left behind

Question

This problem can be solved with Primary School knowledge using ratios.  The famous Singapore bar diagramming method can be used to model the situation, but I prefer my own Distinguished Ratio Units.  The former method is good for visualisation for beginners, while the latter is faster if you want to solve it quickly without fussing around drawing the perfect diagram.  My DRU method is also visual in another way, and it works with big numbers as well as small numbers.  Alternatively, this can be solved using algebra via simultaneous equations.

Solution 1 (Using my Distinguished Ratio Units method) [H02]

Explanation: Since the number of white balls is a multiple of  3,  I let “triangle” 3 represent the number of white balls.  I let “heart” 1 represent the number of red balls.  There are 50 more white balls than red balls.  [H04]  When the white balls are removed three at a time, the number of groups of three would be one-third of the number of white balls, i.e. 1 triangle unit.  [H04]  This number is less than the number of red balls (1 heart unit) by 50.  So 1 triangle unit plus 50 gives 1 heart unit.  [H04]  Following on from the heart to the “triangle” 3, one realises that 2 “triangle” units is the same as  100.  [H05]  So one triangle unit is  50.  [H11]  From here we can solve the rest of the problem.

Solution 2 (using Algebra)  [H13, H05]

                     w = k + 50 = 3h             –––––––––– [1]

                     r  = k         =   h + 50     –––––––––– [2]

for some unknown  k  and  h.  And then [1] – [2] gives  [H10]

                     w – r = 50 = 2h – 50

so                      2h = 50 + 50

                            h = 50

This quickly leads to

                           w = 150

and                       r = 100.

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

* Primary School Mathematics (“Ratio”)

* Lower Secondary School (“Simultaneous Linear Equations”)

* other syllabuses that involve ratio or algebra