For example, a quadratic equation in the form $$ax^2+bx+c=0$$ cannot be solved by simply "moving" the terms around; we wouldn't be able to reduce $x$ to just one term without introducing new terms.

Is there a way to describe this? Say I'm trying to teach the derivation of the quadratic formula. How would I explain that we must introduce new terms, construct a perfect square trinomial, factor it, etc. rather than simply attempt to solve it via basic algebraic manipulation?

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    $\begingroup$ I usually say that we rewrite the LHS. $\endgroup$ – Dasherman Feb 26 at 16:21
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    $\begingroup$ Completing the square is a basic algebraic manipulation. $\endgroup$ – NoNames Feb 26 at 16:21
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    $\begingroup$ @NoNames it's basic once you saw it... $\endgroup$ – Michelle Feb 26 at 16:25
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    $\begingroup$ I'd say you can't do it with the four basic arithmetic operations (addition, subtraction, multiplication, or division) like you can with a linear equation. You need to use a square root, but since square roots don't distribute over addition ($\sqrt{a^2 + b^2} \neq a+b$), you have to rewrite $ax^2 + bx+c$ as a perfect square first. That gets you to completing the square and the quadratic formula. $\endgroup$ – Amaan M Feb 26 at 16:28
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    $\begingroup$ I say to my students that there is not a way to properly undo the operations performed onto the variable so that it ends up isolated, noting the exponent and fundamentally, order of operations. $\endgroup$ – Andrew Chin Feb 26 at 16:39

The following older (mostly before 1900) usage may be of interest. Equations of the form $ax^2 + c = 0$ such that $a \neq 0$ used to be called pure quadratic equations (sometimes incomplete quadratic equations) and equations of the form $ax^2 + bx + c = 0$ such that $a \neq 0$ and $b \neq 0$ used to be called affected quadratic equations (sometimes complete quadratic equations). See this google books search.

Footnote * on p. 280 of College Algebra by Edward A. Bowser:

1888 1st edition at google books and 1893 reprint of 1st edition at internet archive

The term adfected, or affected, was introduced by Vieta, about the year 1600, to distinguish equations which involve, or are affected with, different powers of the unknown quantity from those which contain one power only.

  • $\begingroup$ This is interesting to know, but I honestly don't see how it answers the question? $\endgroup$ – Torsten Schoeneberg Feb 27 at 16:13
  • $\begingroup$ @Torsten Schoeneberg: $ax^2 + c = 0$ can be solved by "moving terms around" (isolate $x^2$ and extract the root) whereas $ax^2 + bx + c = 0$ (or any polynomial equation containing two or more different powers of $x,$ such as a trinomial equation) cannot be solved in this manner—some form of factoring (e.g. complete the square, factor out $x$ when $c=0,$ etc.) or other essentially algebraic method is needed. The older terms I gave seem to be exactly what the OP is asking for, and besides these older terms, I strongly suspect any other such terms are extremely obscure and/or locally used. $\endgroup$ – Dave L. Renfro Feb 27 at 16:49
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    $\begingroup$ (continuing) However, if you mean that the phrases I gave do not actually describe the process of solving by "moving terms around", but instead are simply names for polynomials in which the process can or cannot be successfully carried out, then I agree. However, I don't think there is a reasonably well known phrase for that process. If I had to name this process in a paper or something, I might call it "by arithmetical methods". @Sean Xie (who might be interested in these comments) $\endgroup$ – Dave L. Renfro Feb 27 at 16:54
  • $\begingroup$ I guess my point is that to me, "extracting the root" is actually the non-trivial step which goes beyond "moving terms around". And that one happens either way. I would say, completing the square is a clever way of "moving terms around" which reduces any quadratic equation to one where you now do the non-trivial step of extracting the root. So in a way, the distinction here is just one between those equations where you do the non-trivial step right away, and one where you first use smart (but basic) algebra and then perform the non-trivial step. $\endgroup$ – Torsten Schoeneberg Mar 2 at 21:40
  • $\begingroup$ I find it quite important to teach students that "solving" $x^2=a$ by writing $x=(\pm)\sqrt{a}$ is actually sort of handwaving magic (which to be made proper would use calculus to prove existence of roots of positive real numbers, and whose poor understanding commonly leads to trouble later when one introduces complex numbers, let alone other fields). Whereas completing the square is an (ingenious but, once seen,) straightforward trick using only basic algebra i.e. "moving terms around". $\endgroup$ – Torsten Schoeneberg Mar 2 at 21:48

If you were teaching students about this, then perhaps you could mention strategies that ultimately don't work. If we rewrite $$ ax^2+bx+c=0 $$ as $$ x=-\frac{ax^2+c}{b} $$ then $x$ is still written in terms of $x^2$, meaning that we can't get anywhere. Similarly, if we make $x^2$ the subject of the equation, then we don't get anywhere. The crux of the derivation of the quadratic formula is realising that \begin{align} ax^2+bx+c &= a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right) \\[4pt] &= a\left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right) \, . \end{align} From here, there is only a single term containing $x$, and the rest that follows is the basic algebraic manipulation you mentioned earlier. You might want to make things look cleaner by writing the equation $$ x^2+\frac{b}{a}x+\frac{c}{a}=0 $$ as $$ x^2 + 2Bx + C = 0 \, , $$ with $2B=b/a$ and $C=c/a$. Then, the factorisation is $$ (x+B)^2-B^2+C=0 \, , $$ and here it is abundantly clear why completing the square is such a powerful method.


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