An amazing site which explains completing the square much better than I ever could:
http://www.mathsisfun.com/algebra/completing-square.html
Showing posts with label Completing the Square. Show all posts
Showing posts with label Completing the Square. Show all posts
Sunday, 1 May 2011
Sunday, 17 August 2008
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.
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.
Politics Break & Completing The Square
well with the A Level results this week the annual debating on dumbing down continues. All I will say is that I can do A Level maths now and I couldn't in 1994: also that I can tell by the textbooks I have bought, borrowed and scrounged, that material has been moved, generally into higher modules over that time (complex numbers, trigonometry, series, and algebraic manipulation).
Having said that, you need, according to AQA's website, around 59/75 on C1 to get an A. This doesn't sound much but you can drop marks very easily and you don't have a lot of leeway.
I'll be signing up next month for C1 and S1 and possibly C2 in January so I can't really afford to make sarky comments about dumbing down to be honest. I don't fancy being bitten on the arse by an angry polynomial or a not very normal distribution.
By the way, for the benefit of one reader who googled completing the square "when it doesn't divide by two", it goes something like this.
3x2 + 2x -1 = 0
3[x2 + 2/3x] -1 = 0
3[(x+ 2/6)2 - 4/36] - 1 = 0
3(x + 1/3)2 -12/36 - 1 = 0
3(x + 1/3)2 = 4/3
Then you start to solve the completed quadratic:
(x + 1/3)2 = 4/9
x + 1/3 = +/- 2/3
x= 1/3
x= -1
You can check it alongside the quadratic equation if you like.
If you look at my fairly hasty working out you can see that whether the x2 term divides by anything isn't the point: you simply take out the co-efficient from the x2 term and divide the x term by it as well, as I've done here.
I will do more on completing the square soon but I wanted to see if I could answer that reader's query. I hope they have popped back to check!
Having said that, you need, according to AQA's website, around 59/75 on C1 to get an A. This doesn't sound much but you can drop marks very easily and you don't have a lot of leeway.
I'll be signing up next month for C1 and S1 and possibly C2 in January so I can't really afford to make sarky comments about dumbing down to be honest. I don't fancy being bitten on the arse by an angry polynomial or a not very normal distribution.
By the way, for the benefit of one reader who googled completing the square "when it doesn't divide by two", it goes something like this.
3x2 + 2x -1 = 0
3[x2 + 2/3x] -1 = 0
3[(x+ 2/6)2 - 4/36] - 1 = 0
3(x + 1/3)2 -12/36 - 1 = 0
3(x + 1/3)2 = 4/3
Then you start to solve the completed quadratic:
(x + 1/3)2 = 4/9
x + 1/3 = +/- 2/3
x= 1/3
x= -1
You can check it alongside the quadratic equation if you like.
If you look at my fairly hasty working out you can see that whether the x2 term divides by anything isn't the point: you simply take out the co-efficient from the x2 term and divide the x term by it as well, as I've done here.
I will do more on completing the square soon but I wanted to see if I could answer that reader's query. I hope they have popped back to check!
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