Ask Uncle Colin: A Cubic That Won’t Come Good

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.

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