**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

*x*^{2}– 6*x*+ 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:*x*

^{2}

**–**

**6**

*x*+ 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

*x*^{2}is not 1? Well, we need to pull out that coefficient. Many teachers teach pulling out that coefficient from all three terms e.g.
3

*x*^{2}+ 12*x*+ 5 = 3[*x*^{2}+ 4*x*+^{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:

3

*x*^{2}**+**12*x*+ 5 = 3[*x*^{2}**+****4***x*] + 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 2

^{2}= 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) = 32*y*. So the unknown number (?) must be 32 ¸ 4 = 8. So the green spaces must be filled with 8*y*. Since (8*y*)^{2}= 64*y*^{2}, we deduce that*k*= 64. Bingo!**Learning Points**

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

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

· heuristics (e.g.

*thinking backwards*,*pattern matching*) are powerful problem solving tactics.
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!

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