## Forming and Solving Equations

While checking the work of a year 11 student on Friday I came across a question that could have been a great one for the higher GCSE students to practice their skills together and also their selection of which mathematics to use.

The question was to find the area of this triangle:

A great question. One that to you or I is straightforward but that would take GCSE level students and below a bit of thinking and let’s them hone their skills.

The way to tackle it is to use Pythagoras’s Theorem to form an equation, solve for x then find the area. I feel is beneficial as it combines Pythagoras’s Theorem with a decent amount of algebra then includes the find the area bit at the end.

In this case though, that wasn’t the question. There was more information on offer and the question was:

Which is still a fairly nice form and solve an equation problem.

*3x + 1 + 3x + x – 1 = 56*

*7x = 56*

*x = 8*

*A = 0.5×7×24 = 84*

There is a niceness to this question that goes beyond the question itself. It shows us a great way of differentiating within lessons. Just be leaving out a tiny portion of the information, in this case the perimeter, we can make the question much harder. This idea is something I’ve been working on in various places. M1 questions can be made much easier by providing a diagram, for example.

*Have you used questions in a similar way? If so I’d love to see them, please do get in touch.*

*Cross-posted to Betterqs here.*

## The root of the problem

Yesterday, while discussing various proofs and problems, my friend Steve posted this and said it was his favourite problem of the last few years.

He laid it down as a challenge for me and the others involved in the discussion to solve.

The question:Find all solutions in non negative integers a,b to (a)^1/2 + (b)^1/2 = (2009)^1/2I was out when I saw the question so couldn’t crack on with it straight away, but I did throw a few ideas around in my head. My first thought was “one equation, two unknowns, Hmmmm, I could square it, rearrange the original and sub it it to eliminate, actually no. I would just be proving a=a, b=b, and 2009=2009 then, so that’s a bit silly.” It always amuses me when I, or someone else, manages to prove something like this. My Year 13s still laugh about the time one of them started and finished with F = ma after two pages of working!

My next thought was to complete a prime factorisation on 2009. As it happens I didn’t need to my brother posted “I’ve simplified root 2009 to 7root41” which was what my prime factorisation was for. I realise now that I probably should have seen the answer and this point, but I didn’t.

I thought it would be a good idea to rearrange the equation to find a:

I replaced root 2009 with 7 root 41 and then it was clear: for a to be an integer, b has to be a multiple of 41. This is because any integer b would make the first and last terms integers, but only multiples of 41 will make the middle term an integer. The term is 14x(b)^1/2(41)^1/2 and you need a root 41 to rationalise the one that’s there. Moreover, b must be 41 multiplied by a square number, as any other factors of b would leave another surd.

From there it was easy to spot the answers.

The root a term plus the root b term must equal 7 root 41, so if root a = root 41 then root b = 6 root 41. Likewise if root a = 2 root 41 then root b = 5 root 41. In general if root a = n root 41 then root b = (7-n) root 41 (n is a natural number less than 7). Eliminate your roots and you get {a,b} = {41n^2, 41(7-n)^2} n € N, n < 7.

A truly lovely problem, with a thoroughly beautiful solution.

Nb I apologise for my inability to code the correct maths symbols etc.

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