A My guess is that these students have a weakness in algebra. It is difficult to know which parts in particular so I am going to tell you how I approach this. I begin by reminding students of the quadratic formula that arises from the general quadratic equation (ax2 + bx + c = 0) that they will have used to solve quadratic equations in their GCSE maths: x = b + C (b2 - 4ac) 2a

The process of completing the square helps us arrive at this formula.

I don’t then dive into the algebraic manipulation to find the formula, but go through examples demonstrating the technique of completing the square.

To be able to find the solution to the equation I try to create “a complete square = a complete square”. The following are examples of complete squares: 9, x2, (x + 4)2, (x - 3)2. To understand the completing of the square it is important that students are familiar with the general forms of the last two expressions - (x + p)2 = x2 + 2px +p2 and (x - p)2 = x2 - 2px + p2.

A key point is understanding that the value of the x-term coefficient in the expanded expression is twice the value of the constant term before expansion. Another point not to be missed is that the constant in the x2 term must be 1.

The best way to explain the approach to the method is through an example, such as “Complete the square for the expression 4x2 - 6x + 1 = 0”. In this case, the coefficient in the x2 term is 4. To make this coefficient equal to 1 we divide each term by 4, then bracket the whole expression and multiply by 4. Thus we have really multiplied the whole expression by 44, which is of course equal to 1 and so leaves the original expression unchanged. Do not assume that students know this.

The expression becomes 4(4x24 - 6x4 + 14) which can be simplified to 4(x2 -3x2 + 14). Encourage them to leave the terms in fractional form as this makes the process much easier.

Easy so far? The next part is where students get confused and need to be reminded of the general term, in this case (x - p)2 = x2 - 2px + p2.

We begin by looking at the first two terms of our simplified expression: x2 -3x2. From the general case, we have 2p = 32, so p = 34. As you can see, the value for p is half the x-term coefficient. This will always be the case, so it is worth reminding students that to halve a fraction we double the denominator.

We now write (x - 34)2. Although this is a complete square, when it is expanded this expression is not yet the same as our original expression.

Get the students to discover this by asking them to expand this new expression, which becomes x2 - 3x4 - 3x4 + 94. This is then simplified to become x2 - 3x2 + 94.

The next part is key to their understanding of the process, that of the “bit extra”. Compare the original expression (from within the bracket): x2 - 3x2 + 14 with the newly expanded expression: x2 - 3x2 + 94. Notice the difference in the constant term and ask them how to make the first expression from the newly expanded expression: we need to subtract 84 from the new expression: 94 -84 = 14.

Implementing this, we have x2 - 3x2 + 94 - 84, which is (x - 34)2 - 84 ; our original expression, in full, then becomes 4(4x24 - 6x4 + 14) = 4((x - 34)2 -84).

We can go back and look at the steps we have taken: * Make the coefficient of the x2 term equal to 1.

* Write the bracket as (x q one-half of the x-term coefficient)2 + the constant term - the x2-term coefficient.

Now we can see how the technique helps with solving another quadratic equation. In this case the coefficient of the x2-term is already is already 1.

x2 + 6x - 1= 0 First, we complete the square: (x + 3)2 - 32 - 1= 0 (x + 3)2 - 9 - 1 = 0 (x + 3)2 - 10 = 0 Adding 10 to both sides, we have “a complete square” = “a complete square”:(x + 3)2 = (C10)2 Taking the square root of both sides, we then have: x + 3 = +C10, x = - 3 + C10 or, for the negative root, x + 3 = - C10, x = - 3 - C10.

I have found two sites worth a visit. www.mathcentre.ac.uk has material for revision and is useful for students who have missed topics.

I also found an American site that has some great examples of how you can extend pupils’ mathematical thought. For example there are students discussing the mathematics of bungee jumping: www.learner.orgresourcesseries34.html?=yesvodid=102308pid=933NoNo

Wendy Fortescue-Hubbard is a teacher and game investor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses. www.nesta.org.uk

Email your questions to Mathagony Aunt at teacher@tes.co.uk Or write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX