Why Doesn't Completing the Square Always Work?

I was meant to be going to bed but then I saw that someone had googled the above question and come to me.

The honest answer is "it does always work" but I imagine the problem my putative reader has is that some quadratics are really inconsiderate and don't have real roots.


Well imaginary and complex numbers are Further Pure nowadays but I will hazard an answer.

A quadratic is lovely because it has its famous formula: -b +/- (sqt b²-4ac)/2a

The b² - 4ac bit is called the discriminant and determines the kinds of solutions the equation can have.

I can't really draw a graph tonight but we see it in terms of the points the quadratic crosses the x-axis of the graph.

It goes a little bit like this:

b² - 4ac = a square number (36, say) - quadratic factorises (ie can be reduced to (x+1)(x-2) say).

b² - 4ac = +ve numbers, including fractions - quadratic cuts x-axis twice but can't factorise.

b² - 4ac = 0 - quadratic hits x-axis once (x-axis is a tangent to quadratic)

b² - 4ac = -ve number, quadratic does not touch x-axis at all.


If b² - 4ac is less than 0 it means we need a square root of a negative number. At c1 we can't do that at all. Later on one learns that numbers which square to give -ve numbers are called imaginary numbers and have all kinds of key roles. For now we just don't really want them,

so for our A/S Level purposes the quadratic does not have real roots (ie the values would be complex numbers - an imaginary with a real number) and we simply cannot calculate it at all.

You just end up with a complex number statement like (-2+sqrt -1)/2 or something. As I say, probably great for FP1, no use for C1.

However, that does not mean that completing the square doesn't work - it can ALWAYS be used to solve a quadratic. It just means that it can produce imaginary numbers.