Complete the Square

Back in IGCSE, students should be familiar with this type of expansion
\[ (x+2)^2 = (x+2)\times(x+2) = x^2 + 4x + 4\]
As you can see from the expansion itself, there are three terms to complete a perfect square a.k.a \((x+2)^2\). Now, if we take a look at similar ones, we can spot the general pattern of this very particular type of expansion
\[
\begin{eqnarray*}
(x+3)^2 &=& x^2 + 6x + 9 \\
(x+4)^2 &=& x^2 + 8x + 16 \\
(x+6)^2 &=& x^2 + 12x + 36 \\
(x-7)^2 &=& x^2 -14x + 49 \\
\end{eqnarray*}
\]

That is all the fun with expanding. Now, the real question is: "How can we do the opposite or technically, complete the square?". The answer is simple, you need 3 terms to complete a square. Let's take a look at the following example
\[
x^2 + 6x + 11 = (x^2 + 6x + 9) + 2 = (x+3)^2 + 2
\]

And it is that simple. Of course, more practice is needed to master this skill. Why is this an important skill? Let's say \( y = x^2 + 6x + 11 \) is the graph of a parabola, by completing the square \(y = (x+3)^2 + 2\), we can answer a few questions here

1. Coordinates of the vertex \( (-3,2)\)
2. Axis of Symmetry \(x = -3\)
3. Minimum Value \( y = 2 \) and it happens at the vertex when \( x = -3 \)

You have seen all the good things about Completing the Square, it is important to work on the skill. Other example of the skill involving dealing with fractions such as 

\[
\begin{eqnarray*}
x^2 + x + 1 &=& \underline{x^2 + 2\times x \times \left(\frac{1}{2}\right) +  \left(\frac{1}{2}\right)^2}- \left(\frac{1}{2}\right)^2 +1 \\
x^2 + x + 1 &=& \left( x + \frac{1}{2}\right)^2 + \frac{3}{4}

\end{eqnarray*}
\]

 And one more with the negative, again, it is important to leave the constant (-7) alone as we can deal with it later 

\[
\begin{eqnarray*}
-2x^2 + 12 x - 7 &=& -2(x^2 - 6x) - 7 \\
&=& -2(\underline{x^2 - 6x + 9} - 9) - 7 \\\
& = & -2((x-3)^2 - 9) - 7 \\
& = & -2(x-3)^2 + 18 - 7 \\
&=& -2(x-3)^2 + 11

\end{eqnarray*}
\]