Dear Uncle Colin,

I have a quartic expression and I want to know whether it can be expressed as a perfect square. How would you do it?

Quite Ugly Algebra – Rooting Turns Into Clean Squares?

Hi, QUARTIC, and thanks for your message!

Like quadratics, quartics have a discriminant. Unfortunately, the discriminant is too unwieldy to be much use here, and doesn’t actually answer your question – so we won’t be looking at it.

I can think of two methods for this: an algebraic one (matching coefficients) and a calculus one (finding turning points).

For the purposes of this post, I’ll consider two expressions:

  • 9x4−30x3+37x2−20x+4, which is a square
  • 9x4−30x3+37x2−20x+9, which is not a square

Matching coefficients

If the quartic is a perfect square, its square root is an arbitrary quadratic that can be expressed as ax2+bx+c. So the quartic is (ax2+bx+c)2, or a2x4+2abx3+(b2+2ac)x2+2bcx+c2.

That’s a bit of a mess, but it does allow us to pattern match quite easily.

If the quartic is a square, then its x4 coefficient must be a square (a2) and its unit term must be a square c2. Both of our candidates satisfy this test. Without loss of generality, we can say that a is positive.

  • For the first expression, a=3 and c=±2
  • For the second, a=3 and c=±3

We can work out a candidate b from the x3 coefficient, which must equal 2ab. In both cases, that gives b=−5.

Then we can check whether it works using the x coefficient, which must equal 2bc.

In both cases, the x coefficient is −20, so c=2 is the only possible value – which means the first expression is (3x2−5x+2)2 and the other is not a square.

Calculus

If the quartic is a perfect square, all of its zeros – including the complex ones – must also be turning points.

Let’s look at the derivative of (ax2+bx+c)2, which is (2ax+b)(ax2+bx+c).

That has (generally) three solutions: one at x=−b2a and two more at x=−b±b2−4ac2a.

That is to say, the two ‘outer’ turning points are evenly spaced around the inner one – the curve is symmetric about its inner turning point. (The turning points might involve complex numbers. That’s ok.) This is a necessary but not sufficient condition for the quartic to be a square – as we’ll see immediately: in both of the expressions, we have the same cubic derivative: 36x3−90x2+74x−20.

If the expression has a line of symmetry, then it’s midway between the points of inflexion! So we can differentiate again: 108x2−180x+74; we don’t even need to solve it, the zeros are evenly spaced around 1802×108, which is 56.

This tells us that 6x−5 is a factor of 36x3−90x2+74x−20. We can do the division and get 6x2−10x+4 back out; this factorises as 2(3x−2)(x−1).

The outer turning points are therefore at x=23 and x=1, which are indeed equally spaced about x=56.

But are they zeros? You have two options. You may plonk them back into the original expressions and see whether you get 0; or you may work out (3x−2)2(x−1)2 and see that you match exactly one of the options.

Do you know of a better way? Let me know!

- Uncle Colin