\(x^2 - 16\sqrt{x} = 12\)

What is \(x - 2\sqrt{x}\)?

The above problem showed up on my feed and my first thought was that doesn't look too hard it's either a factoring problem or you need to complete the square. That's the same reaction my son had too when I showed it to him.

But a little substitution (\(z = x^2)\) shows that its actual a quartic equation in disguise:

$$z^4 - 16z - 12 = 0$$

The wording strongly suggests that \(z^2 - 2z\) or some variant is a factor which is a useful shortcut but that led me down the following path on how to generally factor a quartic. The good news here is that the equation is already in depressed format with no cubic terms.

Some links for the procedure:

http://www.maa.org/sites/default/files/Brookfield2007-103574.pdf

A little easier to read:

http://www.sosmath.com/algebra/factor/fac12/fac12.html

How it works:

1. First we need to find the resolvent cubic polynomial for \(z^4 - 16z - 12 = 0\).

That works out to \(R(y) = y^3 + 48y - 256\).

2. Using the rational roots test we only have to look at \(\pm2^0\) ... \(2^8\) for possible roots but since we only can use roots that are square we only have to test \(\pm2^0, \pm2^2, \pm2^4, \pm2^6\) and \(\pm2^8\). Plugging them in we find \(2^2=4\) is indeed a root. So there is a rational coefficient factorization for our original quartic.

3. Now we can use the square root of the resolvent root i.e. 2 and its inverse to get the following factorization (they are the coefficients of the z term): $$(z^2 - 2z - 2)(z^2 + 2z + 6) = z^4 - 16z - 12 = 0$$

4. At this point we could factor the 2 quadratics and plug the solutions back in to find \(x - 2\sqrt{x}\) which in terms of z is \(z^2 - 2z\). But we can shortcut slightly for one of the solutions since the if the first factor is the root then \(z^2 - 2z - 2 = 0\) which implies \(z^2 - 2z = 2\)

5. Interestingly for \(z^2 + 2z + 6 = 0\) we have the two roots \(1 \pm i\sqrt{5}\) Plugging either

one into \(z^2 - 2z\) and you get -6 anyway!

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