Dear Uncle Colin,

I'm told that \(x\sqrt{x} - 5\sqrt{x} = 2\) and I have to find \(x - 2\sqrt{x}\). Everything I try seems to make it worse! Any ideas?

Mastering A Cubic - Help Is Needed

Hi, MACHIN, and thanks for your message! At first glance, that's a strange one. We can solve it, but I'm not quite sure a) what technique the question would *like* you to use, or b) whether my approach is the best one. So, with that in mind, let's have at it!

### What's ugly?

I don't like the look of those square roots, so let's let \(y = \sqrt{x}\) and turn it all into a cubic: \(y^3 - 5y = 2\). That doesn't have any immediately obvious solutions, but it certainly looks nicer.

It also means that we're looking for the value of \(y^2 - 2y\).

### Can we make that nicer?

Of course we can! Let's play completing the square and write \(y^2 - 2y\) as \((y-1)^2 - 1\). If we let \(z = y-1\), we can rewrite the cubic again, this time as \((z+1)^3 - 5(z+1) = 2\).

Expanding that gives \(z^3 + 3z^2 - 2z - 6 = 0\), which is simple to factorise: it's \((z+3)(z^2 - 2)\), which has solutions when \(z = -3\) or \(z = \pm \sqrt{2}\).

### Naively...

We might then say "we want \(z^2 - 1\), which is therefore either 8 or 1" - however, we need to be careful: not all of the \(z\) solutions are valid for \(x\) - because \(y=\sqrt{x}\), \(y\) is non-negative, which means \(z\) is at least -1. Only \(z=\sqrt{2}\) satisfies that - which means that \(z^2 - 1\) can only take the value of 1.

Hope that helps!

- Uncle Colin

* Updated 2018-11-28 to correct an $x$ that should have been a $z$. Thanks, @snapey1979!

## 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.