Terminology to describe that, eg, $ax^2+bx+c=0$ cannot be solved by simply moving terms around to isolate $x$ 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?
 A: 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.

A: 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.
