g(t) = 2t - 3t²

①To complete the square, you must first rewrite it in the form a² + 2ab + b², or a² - 2ab + b²

We have a -3t² here so we know that the completed square has the form -3(t + constant)² + ___ or -3(t - constant)² + ____

What you want to do first is to factor out -3.

g(t) = 2t - 3t² = -3(t² - ⅔t)


② Now, we can try to rewrite the term inside the brackets in the form a² - 2ab + b² since we have a -⅔t.


-⅔t can be rewritten as -2t(⅓)

So we can say a = t, b = ⅓

g(t) = 2t - 3t² = -3(t² - ⅔t) = -3(t² - 2t(⅓))


③ Rewrite the equation such that we have the b², which in this case is (⅓)²


g(t) = -3(t² - 2t(⅓)) = -3(t² - 2t(⅓) + (⅓)² - (⅓)²)

Then, simplify the -(⅓)² out from the brackets.

g(t) = -3(t² - 2t(⅓) + (⅓)²) - 3(-(⅓)²)

= -3(t - ⅓)² - 3(-1/9)

= -3(t - ⅓)² + ⅓

( Or ⅓ - 3(t - ⅓)² )

Putting it all together,


g(t) = 2t - 3t²

= -3(t² - ⅔t)

= -3(t² - 2t(⅓) + (⅓)² - (⅓)²)

= -3((t - ⅓)² - 1/9)

= -3(t - ⅓)² - 3(-1/9)

= -3(t - ⅓)² + ⅓

And yet some of my students still believed that there is always the “ = 0 “ behind when the question asks to complete the square.

They were surprised when I told them it was an expression. They were able to recognise the completing the square procedure when the other side has = 0. But when presented in the format that Zwen gave (without the = 0 phrase), my students would assume the = 0 exists and proceed to do as such.

Which is obviously flawed of course, but they still did that anyway even during the days leading up to the O Level paper.

They'll need to revisit the concept of quadratic expressions and parabolas, and completing the square to find the turning points

i.e vertex form

a(x - h)² + k where the turning point has coordinates (h,k)

Thank you all.