National 5 Maths
Completing the Square

Look at the coefficient of \(x\). It is 10. We halve it to get 5 and then square that. So 5² = 25.

We are going to add this 25 into the expression but then immediately subtract it out again. That might seem like a strange thing to do, but it will give us the perfect square that we need.

The coefficient of \(x\) is 6. We halve this to get 3 and then square that: 3² = 9.

We add this 9 into the expression but then immediately subtract it out again. The first three terms will be our perfect square.

The coefficient of \(x\) is -2. We halve this to get -1 and then square that: (-1)² = 1.

We add this 1 into the expression but then immediately subtract it out again.

The coefficient of \(x\) is -4. We halve this to get -2 and then square that: (-2)² = 4.

We add this 4 into the expression but then immediately subtract it out again.

This example is trickier because the coefficient of \(x\) is odd. One like this hasn't yet appeared in a National 5 exam paper, but the course specification says that the coefficient can be any integer, so be prepared!

We halve the \(5\) to get \(\frac{5}{2}\) and then square that: \( (\frac{5}{2})^2=\frac{25}{4}. \)

So we add the \(\frac{25}{4}\) into the expression and then immediately subtract it out again.

We halve the \(-9\) to get \(-\frac{9}{2}\) and then square that: \( (-\frac{9}{2})^2=\frac{81}{4}. \)

So we add the \(\frac{81}{4}\) into the expression and then immediately subtract it out again.

The coefficient of \(x\) is \(-8\) so we first calculate \((\frac{-8}{2})^2=(-4)^2=16\).

Then we complete the square:

Now it is easy to state the coordinates of the turning point, because perfect squares like \((x-4)^2\) can never be negative. The smallest they can ever be is \(0\). In this example, \((x-4)^2=0\) when \(x=4\).

Finally, when \(x=4\), we can just read off from the formula that \(y=-11\). So the turning point is \((4,-11)\).

(a)  This part is just a straightforward 2-mark completing the square question. $$ \begin{eqnarray} &\ & x^2+8x+15 \\[6pt] &=&\ x^2+8x+16-16+15 \\[6pt] &=&\ (x^2+8x+16)-16+15 \\[6pt] &=&\ (x+4)^2-1 \end{eqnarray} $$

(b)  This part requires us to use the fact that the turning point of the graph of \(y=\left(x+a\right)^2+b\) is \(\left(-a,b\right).\) So the turning point is \(\left(-4,-1\right).\)