[S2_20151027ACS] Completing the Square

Question

Introduction

     Many students in Singapore are taught to the method of “completing the square” in a rote fashion.  For example, to handle a quadratic expression like  x² – 6x + 8,  take half of the coefficient of the  x term, put it with  x, square that resulting binomial, and then also subtract the square of that same number, like this: 

                                        x² – 6x + 8 = (x – 3)² – (3)² + 8 = (x – 3)² –1

Note that the sign inside the squared binomial always follows the sign of the  x  term.  Just follow the procedure!  It works!

     Does the student understand why this works?  And, by the way, does this always work?  What if the coefficient of  x²  is not  1?  Well, we need to pull out that coefficient.  Many teachers teach pulling out that coefficient from all three terms  e.g. 

     3x² + 12x + 5 = 3[x² + 4x + ⁵/₃] = 3[(x + 2)² – (2)² + ⁵/₃] = 3[(x + 2)² – ⁷/₃]  = 3(x + 2)² – 7

A better way to do this is to just pull it out from the first two terms: 

     3x² + 12x + 5 = 3[x² + 4x] + 5  = 3[(x + 2)² – (2)²] + 5  = 3(x + 2)² – 7

Now it seems that as the questions get harder and harder, students need to memorise more and more procedures by rote.  And what if they encounter a question such as the one featured above?

What now?

     Let us go back to basics.  The square-of-sum identity is                                        

and the square-of-difference identity is                                       

Instead of memorising procedures (not that these are wrong in themselves), why not use the above identities and think backwards (which is a type of heuristic)?  You can even make it a game of “filling in the blanks”, as shown below.

Solution

     According to the identities, the two green patches must be the same and likewise the orange patches must be equal.  Working from the right end, we figure out that the orange patch must be  2  since  2² = 4  or  Ö4 = 2 .  Once we know this, we try to figure out the green space by matching the middle  y  terms:  2( ? y)(2)  =  32y.  So the unknown number (?)  must be  32 ¸ 4 = 8.  So the green spaces must be filled with  8y.  Since  (8y)² = 64y², we deduce that  k = 64.  Bingo!

Learning Points

     · You definitely need to know formulas and procedures, but ...

     · memorising procedures is not the holy grail of mathematics.

     · There is no holy grail of mathematics, but ...

     · heuristics (e.g. thinking backwards, pattern matching) are powerful problem solving tactics.

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

· Lower Secondary Mathematics

· other syllabuses that involve algebraic identities and completing the square

· anyone who loves a challenge!