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I'm trying to teach some secondary school students on how to complete the square. The goal is to rewrite: $$y = ax^2 + bx + c \ \ \Rightarrow \ \ y = a(x-h)^2 + k$$ The first thing I did was to ask them to confirm, via FOIL: $$\left( x + \tfrac{b}{2} \right) \left( x + \tfrac{b}{2} \right) \ \ = \ \ x^2 + bx + \tfrac{b^2}{4}$$ With this as a guide, they were able to rewrite something like: $$y = x^2 + 6x - 10 \ \ \Rightarrow \ \ y = (x + 3)^2 -19$$ Problems occur when the leading coefficient does not equal one. So far, I am not able to explain why I need to reduce the leading coefficient to $1$ before completing the square. (Moreover, some of them are having problems with distributions.) For example: \begin{align} y \ \ & = \ \ 2x^2 + 8x - 9 \\ & = \ \ 2(x^2 + 4x - 4.5) \\ & = \ \ 2(x^2 + 4x + 4 - 8.5) \\ & = \ \ 2(x^2 + 4x + 4) - 17 \ \ = \ \ 2(x + 2)^2 - 17 \end{align}

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    $\begingroup$ My experience with covering this topic is that it is a good one for uncovering the issue discussed in this question about equals signs. $\endgroup$ – Chris Cunningham Aug 8 '17 at 2:02
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    $\begingroup$ I find it easier for them to group the two non-constant terms before factoring out the leading coefficient. It doesn't eliminate all difficulty but it's one less thing to worry about. $\endgroup$ – rnrstopstraffic Aug 8 '17 at 3:27
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    $\begingroup$ If they're not sure why you factor out the leading coefficient, why don't you ask your students what they'd do instead? Either they come up with a method that also works, or their failure to do so demonstrates the reason why you taught it that way. $\endgroup$ – Daniel Hast Aug 8 '17 at 3:31
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    $\begingroup$ I would just refer back to the usual joke about an engineer and mathematician trying to make tea with a kettle, wood stove and water faucet. $\endgroup$ – DRF Aug 9 '17 at 11:53
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    $\begingroup$ A good way to explain the need to factor out the lead coefficient is to work backwards from the goal. Order of operations has us squaring the binomial $(x-h)$ before multiplying the number in front of it. Thus, working in the other direction, that number has to be factored out before the binomial can be "un-squared". $\endgroup$ – G Tony Jacobs Aug 9 '17 at 13:38
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This may not really answer your question, and, it may be inappropriate to the level of your class, but, I find the idea of looking at an algebraic manipulation algebraically is at times helpful. The whole point of doing algebra is to engage in the kind of thinking I give below so I think there is something to gain from taking this sort of "abstract" approach. The idea here is to discover how arbitrary $h$ and $k$ and indeed $A$ must relate to $a,b,c$ in order that $$ y = ax^2+bx+c = A(x-h)^2+k. $$ This is itself and algebra problem. Do algebra to do algebra.

Consider $$ ax^2+bx+c = A(x^2-2xh+h^2)+k $$ or $$ ax^2+bx+c = Ax^2-2hAx+Ah^2+k $$ requires that these be the same polynomial. Hence, equating coefficients, $$ \begin{array}{cc} x^2: & a=A \\ x: & b=-2Ah \\ 1: & c = Ah^2+k \end{array} $$ So, the $h$ and $k$ we seek are given by: $$ A = a, \ \ h = -\frac{b}{2a}, \ \ k = c - \frac{b^2}{4a}. $$ Therefore, $$ ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a} $$ Then, give numeric examples galore until it sinks in.

It's likely as or more important to explain why it is important to complete the square. Of course the danger of my method, and it is a real danger is that students may black-box this solution. Heaven forbid teachers just say well, you set $h = b/2a$ and $k = c-\frac{b^2}{4a}$ and that's it. The universe in which I present this to a class is the same universe in which the students are also held responsible for reproducing the abstract result in addition to the expectation they do numerically specific examples with ease. Ease born of practice.

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    $\begingroup$ I like this approach; I have used a somewhat similar one as outlined here. $\endgroup$ – Benjamin Dickman Aug 13 '17 at 4:41
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The goals of my approach are:

  • Emphasize that the purpose of completing the square is a process of writing something that is equal to the original expression.
  • Emphasize that there are choices to make, differentiating between a choice being useful versus it being true (avoiding the words "correct" or "right" which conflate useful and true).
  • Avoid unnecessary algebra and quantifiers; my students have a hard enough time with $x$; adding in $b$ and $c$ obscures my previous two goals.

Complete the square: $x^2 + 8x + 19$

First ask yourself: "is it a perfect square?" If it was, you would be able to find two numbers that add up to 8, multiply to 19, and are the same number.

If not, ask yourself what the two numbers would have to be to add to 8 and be the same number. Then recognize that the constant 19 is the wrong constant. Ask yourself: What is the constant you want it to be? Then don't let your dreams be dreams, just write it:

$\begin{align*} & x^2 + 8x + 19 \\ = & x^2 + 8x + 16 \end{align*}$

Notice that this is useful (it factors now) but not true (they aren't equal). Fix what you've written so that it is also true.

$\begin{align*} & x^2 + 8x + 19 \\ = & x^2 + 8x + 16 & + 3\end{align*}$

Then factor the part of the result that you have engineered to work out.

$\begin{align*} & x^2 + 8x + 19 & \\ = & x^2 + 8x + 16 & + 3 \\ = &(x + 4)^2 & + 3\end{align*}$


I have tried a variety of approaches, and I like the abstract skills that this approach allows students to practice. For example, when a student does this:

$\begin{align*} & x^2 + 8x + 19 \\ = & x^2 + 8x + 4 + 15 \end{align*}$

I can ask students if that was a good idea and tease out the distinction that there is a difference between something being true and useful. Since one of the primary goals of algebra courses is convincing students that they have the freedom to do whatever they want when it is true (sure, multiply both sides by 10 if you want man; worst case you can undo it later), this synergizes well with my approach to the rest of algebra.

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I don't know if the following will be of much use to your specific case (our audiences are quite different---I teach this to college freshmen), here is my usual approach:

First, we typically start the quarter by getting a good handle on geometric transformations of graphs of functions. Given some function $f:\mathbb{R}\to\mathbb{R}$ with a known graph, the graph of the function defined by $$ x \mapsto B f\left( \frac{x-h}{A} \right) + k $$ is the same as the graph of $f$

  1. shifted right $h$ units (where $h<0$ is a shift to the negative right, or left),
  2. scaled horizontally by a factor of $A$,
  3. reflected across the $y$-axis if $A < 0$,
  4. scaled vertically by a factor of $B$,
  5. reflected across the $x$-axis if $B < 0$, and
  6. shifted up $k$ units (again, down is negative up).

We can then consider the humble parabola defined by $x \mapsto x^2$. This lovely function has a zero at zero, goes through the points $(1,1)$ and $(-1,1)$, and has inverses on the domains $[0,\infty)$ and $(-\infty,0]$ given by the positive and negative square roots, respectively.

We might first note that if we transform the graph of $x\mapsto x^2$, the two multiplicative constants can be combined: \begin{equation} B\left( \frac{x-h}{A} \right)^2 + k = \left( \frac{B}{A^2} \right)(x-h)^2 + k, \end{equation} so we may fairly conclude that any transformation of the graph will be given by a function of the form \begin{equation*} x \mapsto C(x-h)^2 + k, \end{equation*} where $C$ represents some kind of scaling (we may as well understand it as a vertical scaling, but it combines both the horizontal and vertical scalings into one gooey mess), and $h$ and $k$ are translations, as usual. In particular, the vertex of this graph is the point $(h,k)$, and the vertex will represent a minimum or maximum for the function, depending on whether $C > 0$ or $C < 0$.

We can also expand this mess in order to obtain \begin{align} C\left( x-h \right)^2 + k &= C (x^2 - 2hx + h^2) + k \\ &= \left(C\right)x^2 + \left( -2hC \right)x + \left( h^2C + k\right) \\ &= ax^2 + bx + c, \end{align} which is a more familiar expression to most of my students, as they have mostly seen the material before in high school. The next "obvious" (for certain values of "obvious") question is: can we go the other way?

That is, if we are told that $$ f(x) = ax^2 + bx + c,$$ can we understand the graph of $f$ as the graph of our basic parabola subject to some elementary transformations? That is, given arbitrary $a,b,c$ (with $a\ne0$), are there $C,h,k$ such that $$ f(x) = ax^2 + bx + c = C(x-h)+k?$$ The answer is "Yes!", with the details as provided by James S. Cook's answer to this question.

In this exposition, the idea is that the geometry motivates exploration of the problem. We want to know how to "complete the square" so that we can find the vertex of the transformed parabola, or so that we can determine an inverse on some domain (i.e. solve a quadratic equation).

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    $\begingroup$ Excellent answer as to the "why" we care. Of course, computationally there are so many other reasons once we go on to integration theory, Laplace transforms and pretty much any application which involves a quadratic model symbolically, but, surely graphing is the first and most important application for students to appreciate and master. $\endgroup$ – James S. Cook Aug 8 '17 at 18:39
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I present completing the square as a recipe.

Given $x^2 + bx$, add $\big(\frac{b}{2}\big)^2$ to both sides. You obtain a different expression, but your new expression can be written neatly as a perfect square:

\begin{align*} x^2 + bx &\xrightarrow{\text{add }\left(\frac{b}{2}\right)^2} x^2 + bx + \big(\frac{b}{2}\big)^2 = \big(x + \frac{b}{2}\big)^2 \end{align*}

At some point, I emphasize how important it is that we started with $x^2$ without a (non-one) coefficient: the recipe produces a new expression of the form $(x + \text{something})^2$, which, when expanded, will never produce anything whose $x^2$ term has a non-one coefficient.

So, maybe you can really play up that it's a "faithful, but simple" method that only works for expressions of the form $x^2 + bx$: The recipe deals exclusively with quadratics whose leading term is $1$, end of story.

Note: You face similar problems getting students to factor $2x^2 - 3x + 5$ using the "ac method" or whatever you prefer to call it. Students are used to the process find a pair of numbers that multiply to this, and add to that from the "easy" quadratic case. But, unlike the "easy" case, here we don't just get to jump right to a factorization. Why? Because $(x + \text{thing1})(x + \text{thing2})$ will never produce quadratics whose leading coefficient isn't $1$, and we finally need that. So, we have to add another step or two to the old method.

So I will make a bit of a big deal about the jump from expressions like $x^2 + 6x + 1$ to $2x^2 + 6x + 1$: Our version of completing the square requires us to have just $x^2$. Is all hope lost; do we need a brand new version of completing the square?

This is where I take the approach suggested by DRF's comment: We reduce a superficially new problem to a known, already solved problem. By adding the minor modification of factoring out the leading coefficient, we get the much vaunted $x^2 + [\text{something}]x$ on which our recipe relies.


I'll add that there is a super slick, more general version of completing the square that can handle non-one leading coefficients without factoring/division, given here by André Nicolas (and I think the linked answer is the first place I saw it). It's more appropriate for solving equations by completing the square (instead of rewriting quadratic expressions), but it's so nice that I think it deserves to be more widely-known.

The main idea is that $(2ax + b)^2 = 4a^2x^2 + 4abx + b^2$, so given an expression like $ax^2 + bx$, you'll multiply by $4a$ then add $b^2$, at which point you can write your new expression as a perfect square.

I have yet to mention this for a remedial algebra / precalculus class, but occasionally I'll get a math ed student and show them this method in addition to the usual approach.

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My Approach: \begin{align} y \ \ & = \ \ 2x^2 + 8x - 9 \\ & y+9 = \ \ 2(x^2 + 4x + \underline{\quad}\ ) \\ & y+9+8 = \ \ 2(x^2 + 4x+4) \\ & y+17 = \ \ 2(x^2 + 4x + 4) \\ & y+17 = \ \ 2(x + 2)^2 \\ \end{align}

Notes - Moving the constant to the left in the first step avoids the errors some students might make dividing this number. Now, when you add the '4' to complete the square, you ask "What did we add to the right side?" "4" "don't forget the number to multiply, here, '2', so we added 8 to the right, and we'll add 8 to the left."

Last, the format of that last line is the one I prefer. It's the "Vertex Form" and students can quickly see (-2,-17) is the vertex.

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I think it is easier to practice doing this without the y (with the equation set to zero). Of course it is same thing but for someone just learning, doing it as just a single equation is easier and concentrates the mind on roots.

I would also try to do it very mechanically, writing out each step (like when you are in pre-algebra and first learn how to sort things out with first order equations in X only.)

Just some thoughts.

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    $\begingroup$ I would discourage you from setting the expression equal to zero. Students in algebra classes have enough trouble with the impulse to randomly set things equal to zero as it is, and you calling the expression an "equation" (it has no equals sign unless you set it equal to zero for no reason) is also damaging to their use of precise language. $\endgroup$ – Chris Cunningham Aug 9 '17 at 19:07
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    $\begingroup$ The point was to practice it for a while in a simpler situation. Not giving them the problem as is and then a method of removing the y. But actually learning it as is with just the X's. Obviously there are many situations where that formula is exactly all you have. For example the characteristic equation of a 2nd order diffyQ. Or most chemical equilibrium problems. Or just deriving the quadratic formula. Once the student is well practiced at just doing these problems all the time as is, they can execute fine with the y thrown in. $\endgroup$ – guest Aug 11 '17 at 3:36

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