# Ask Uncle Colin: A STEP in the right direction

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I'm struggling with a STEP question. Any ideas?

Given:

1. $q^2 - pr = -3k$
2. $r^2 - qp = -k$
3. $p^2 - rq = k$

Find p, q and r in terms of k.

- Simultaneous Triple Equation Problem

Hi, STEP, and thanks for your question!

This is an absolute biter that took me several attempts to get under control. The key was to spot that the difference between each successive pair of equations is $2k$, and that there's a common factor in the differences.

By which I mean:

• The difference between equations (1) and (2) is $r^2 - q^2 + p(r-q) = 2k$, which is $(r-q)(p+r+q) = 2k$ (*)
• The difference between equations (2) and (3) is $p^2 - r^2 + q(p-r) = 2k$, which is $(p-r)(p+r+q) = 2k$

Dividing both sides of both equations by $(p+r+q)$ gives $(r-q)=\frac{2k}{p + r + q} = (p-r)$, which means $2r = p+q$.

If we add (2) and (3), we get $p^2 + r^2 - q(p+r)=0$.

Replacing $p$ with $2r-q$ gives $5r^2 - 7rq + 2q^2 = 0$; that factorises as $(5r-2q)(r-q)=0$. That means either $5r-2q=0$ or $r-q=0$.

If $r=q$, then (*) says that $k=0$, and it follows from (3) that $p = \pm r$, and from (2) that all three variables are the same - which is a very dull solution.

The other option is that $5r=2q$. Knowing that $p+q=2r$, we can deduce $r = -2p$ and $q = -5p$.

Going back to equation (1) gives $27p^2 = -3k$, so $p=\frac{1}{3}\sqrt{-k}$, $q = -\frac{5}{3}\sqrt{-k}$ and $r = -\frac{2}{3}\sqrt{-k}$.

To check it works, suppose $k = -9$, so that $p=1$, $q=-5$ and $r=-2$. Substituting those into (1), (2) and (3) gives the answers 27, 9 and -9, as desired.

Hope that helps!

- Uncle Colin

* Thanks to @ajk_44 for suggesting this might interest me.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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##### Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.