A nice collection of Problem solving resources arranged by topics on this site by @LeanneShawAHS
There are 3 standard ways of solving quadratic equations once they are in the form:
ax² + bx + c = 0
I think I generally teach them in that order probably without much thought as to why. I guess the formula needs to be derived by using completing the square and factorising seems to follow on from multiplying out double brackets, which comes before all of this. The I question that I sometimes get from students is “what’s the point in learning factorising if the two other methods always work?”. Well, it’s quicker and you can do it on a non-calc exam is probably a standard response.
But have you tried using the formula without a calculator to solve a quadratic that you know will factorise? Have a go. Plug this:
x² - 3x - 28 = 0
…and solve without a calculator.
It’s a surprisingly satisfying experience, one that I would not want to deny my students.
You’ll need to know your square numbers because b² – 4ac will always give you a square number for quadratics that factorise. But the arithmetic is perfectly reasonable and is likely to be so for most quadratics that can be factorised.
When I did this recently, I had a great question from one of my students, as I was aching my brain trying to make up a quadratic that I knew would factorise. “If you just picked one randomly, what are the chances that you would be able to factorise it?”
I’ve since had chance to investigate this further. It’s a great question and there is a lesson in here, or at least an extension question to explore once the fundamentals of using the formula are secure.
I started approaching it by using the formula and focussing on b² – 4ac and what values of a, b and c would yield square numbers. To simplify the problem, I started with a=1, so I was looking for when b² – 4c = 1,4,9,16,25, etc.
I then looked at it from the other end, i.e. starting with e.g. (x+1)(x+n) what values of a, b, and c are yielded. Then vary further to look at (x+m)(x+n). I just started with a few values of n and m to see if I could spot patterns. I won’t spoil the fun by revealing those patterns, but this is very open-ended and could provide some intrigue for the right learners.
The red equilateral triangle side length 4cm sits inside the larger pink equilateral triangle such that the “border” is 1cm wide.
What is the ratio of the height of the red triangle to the height of the pink triangle?
Can you solve using trigonometry or only using Pythagoras?
The border is now 2cm, whilst the side length of the red triangle remains 4cm. What is the ratio of heights now?
Explore what happens for other border widths. Can you generalise for any border width w?
Geogebra file here.
I got a few solutions posted on Twitter for this, but the most elegant (so far…) has to be this from @mathforge. As he said: no Trig, no Pythagoras, just ratio.
First up, this post is NOT going to be about this:
In fact, we were going the other way: converting fractions to recurring decimals, specifically looking at fractions where the denominator is prime. It was a fascinating session, great for deepening subject knowledge. This blog post is my attempt to reflect on what I learned and my thoughts about how I might use this in the classroom, probably Year 10 or 11, but really any group that is confident with bus stop division could investigate this.
I was already aware of some pretty cool things that happen with sevenths, mainly from Don Steward’s blog:
Why should this happen? Why do we only see these six digits with sevenths? The process of bus stop division helps us see why and this is where I feel I would start with a class. This is good practice of a technique that should be secure but often isn’t in Year 9 / 10. It helps learn the seven times table and I don’t think it is too tedious to ask students to perform these six calculations manually:
Some learners might want to find 2/7 by doubling 1/7. And then maybe find 3/7 by adding 1/7 to 2/7 and so on. Even if they stick with the bus stop division, they will find that they are essentially doing the same six calculations:
10 ÷ 7 = 1 rem 3 20 ÷ 7 = 2 rem 6 30 ÷ 7 = 4 rem 2 40 ÷ 7 = 5 rem 5 50 ÷ 7 = 7 rem 1 60 ÷ 7 = 8 rem 4
The ones digit is always zero and the tens digit must be less than 7 so there are only these 6 options. Can we extend this rule to other fractions with denominator less than 10? Of course, all others except 3, 6 and 9 will terminate – why is this?
There are also some interesting things to notice about the 1/7 “wheel” before moving onto higher primes. I won’t spoil it for you, but suggest you include the fractions around the outside of the wheel to spot some of the patterns.
We then moved onto looking at 13ths. 11th are interesting too for different reasons, so I can see why we went onto 13th because there are some surprising relationships between 7th and 13ths. At this point, we started using calculators and I would do the same with a class. Or even better, open a spreadsheet, which I am always keen to do!
I’m not sure of the value of kids typing 12 calculations into their calculators, so I might give this image above as a print out for them to write on. Hopefully they will soon spot that there are two sets of recurring digits, i.e. 076923 in 1/13 + 5 others and 153846 in 2/13 + 5 others and which, this time can be written as two wheels:
Ideally at this point you’ll have the class hooked and they would be asking all sorts of questions. Well, maybe enough of them to get everyone else thinking. I would try really hard to encourage the students to come up with these questions. This is what I hope they will ask, but I hope they will also ask questions I hadn’t thought of, something which is a really special moment in any lesson.
On the image above, I have shown that as you move clockwise around the wheel, you are effectively multiplying by 10. This is obvious going from 1/13 to 10/13 but actually occurs for all other hops if we ignore the whole number part. i.e.
So if go 3 hops we have multiplied by 1000. I think this goes some way to explaining why the fractions that are opposite each other must sum to 1.
The next prime fraction: 17ths recur after its maximum of 16 digits, so effectively we have one, rather large wheel. Unfortunately, Excel gives up after 16 decimal places (please let me know if there is a way around this) but you can still see some patterns in here:
Beyond that, we found that the following fractions recur after the maximum number of digits so have only one wheel: 3,7,17,19,23,43,59,61 – someone had a much better calculator than me that showed 32 digits!
Other fractions worth exploring are:
With the fractions that recur after a relatively few numbers of digits, we can find factors.
Because 1/41 is a decimal that recurs after only 5 digits, if follows that 41 must be a factor of 99999.
And in the larger wheels there are all sorts of patterns – pentagons and triangles in the wheel of 31ths apparently.
I’m not sure how far I’d go with a class, there could be several lessons worth in here. It would depend on how they responded, of course. Maybe 13ths would be sufficient unless…
This post captures only some of what we worked on in this session and highlights to me the depth of subject knowledge that can be gained by attending the ATM conference. Other sessions were more pedagogical in nature, but this one was pure fascinating mathematics and I was grateful to be surrounded by so many knowledgable and friendly people!