Sometimes it’s worth taking a risk and changing course of a lesson halfway through. And sometimes it pays off.
Today’s lesson was supposed to be about algebraic proof and we started with these nice questions from Don Steward. I thought they looked like good practice for multiplying out double brackets at the same time as introducing algebraic proof.
By the way question 3 is particularly tricky.
Following on from the previous lesson, lots of them started by trying values for n. Great to then have the discussion on what makes a proof vs. an example.
We got into a good discussion on Question 5 and I wanted to know if they were familiar with the difference of two squares. This is where the lesson changed course completely. None of them could tell me what it was called but I got the sense that they had seen it before. So next, I wrote these questions on the board:
Factorise:
- n² – 9
- 4n² – 25
- 81 – n²
- 100 – 81
They raced through these because they had all spotted the short cut. So next, I put this up, again from Don Steward:
Which followed on nicely from the final of the 4 questions I had put on the board. Many of them initially struggled to see the link immediately or see the pattern in the numbers. But with a bit of time and just the right amount of help (i.e. not much actual help, just encouragement!) they started to find others. I heard them forming statements like: “I need to number that multiply to give 2016, then I need another two numbers that add to give one of those number and subtract to give the other.”
About 5 minutes before it was time to pack away there was a great buzz in the room as 3 students found some other numbers and then the race was on. Lots of solutions started coming, but nobody got all 12 (including me!).
I went away and built a spreadsheet to investigate further, but I still didn’t find all 12. Can you help???
I’ve no idea how to leave square symbol in here, so below is a list of all the pairs on numbers you square and subtract;
505 503
79 65
171 165
45 3
65 47
254 250
50 22
90 78
46 10
130 122
54 30
71 55
You were on the right lines with finding the pairs, but didn’t take your spreadsheet far enough.
As your students did, I spotted that the difference between the numbers multiplied by their sum has to take us back to 2016.
I started by breaking 2016 down into it’s factors 2x2x2x2x2x3x3x7.
The difference can only ever be a factor of 2016 (and their sum also a factor). So it’s then a case of working through all the factors and seeing which ones give positive integer pairs.
There may be a more elegant method but I then resorted to a spreadsheet to list all the factors and work out which ones worked. I can send you the spreadsheet if you want me to (I can’t figure out if I can post it here).
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The trick with question 3 is to not expand the bracket but to take out a factor of (n+1) from each term and the factorise the second bracket;
n(n+1) + (n+1)(n+2)
= (n+1)(n + n + 2)
= (n+1)(2n + 2)
= (n+1)2(n+1)
=2(n+1)^2
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