## Thursday, November 25, 2010

### Deriving the Quadratic Formula: James Tanton's Twist

I've always enjoyed showing students how to derive the quadratic formula. I don't test them on it, so the stress level is lower. And it's late in the term, so they appreciate a break from the pressure, and most really do try to get it. I get a few making those appreciative sounds that happen when the lightbulb goes on, and that makes it especially fun.

But it's hard slogging through some of the weird steps. Here's the standard derivation, if you haven't done it in a while. Check it out, and imagine trying to explain it to people who are pretty fragile around math.

So the math education gods were smiling on me last week, and the day before I brought this topic to my students, I interviewed James Tanton, who (out of the blue) showed us his twist on this. (Thanks, James, for helping me with my lesson plans!)

If you just can't find the time to watch the video, it goes something like this:
ax2+bx+c=0
We want a perfect square in the first term,
so we multiply both sides by a:                  a2x2+abx+ac=0
We want the second term to have a factor of 2, and to keep the first term a perfect square,
so we multiply both sides by 4:            4a2x2+4abx+4ac=0
We almost have what we see in the box above, but we want b2 and not 4ac,
so we do a little adding and subtracting:   4a2x2+4abx+b2 = b2-4ac
Now factor the left side:
(2ax+b)2 = b2-4ac
Taking the square root of both sides steals away a solution, so we include a plus or minus:
`2ax+b = ±√ b2-4ac  `
Subtract b from both sides and divide both sides by 2a, and you've got it. Much prettier than the standard derivation, I think.

I used this in class last Thursday, and I think it went much more smoothly than the standard version. I always do it twice, so I did it again on Monday. Students said they got it, and some liked it. I haven't spent enough time on completing the square (our days together are numbered, at this point in the term), so I don't expect their understanding goes very deep, but it's a start.

I'll do this again next semester in my intermediate algebra course, with a better grounding in completing the square. I look forward to it. Maybe all our conics problems will be easier, with James Tanton's brilliant help.

1. Wow!! Thank you for sharing!

2. That is very nice! Never seen that before.

3. I wonder if anyone has ever done this as a math circle, and gotten the students to do the derivation?

4. I tend to dislike these two methods (there are some other variants too), because although they are algebraically pretty simple,

1) it's hard to imagine how someone thought of them, and

2) they don't connect to other things students are doing, and

3) they are never useful again.

There are a couple of other methods (which you probably know) which seem to me to not have these disadvantages.

A) If you have done, or will do, horizontal shifts of functions, do a shift: let x = t-h, so you are trying to solve

a(t-h)^2 + b(t-h) + c = 0

which simplifies to

at^2 + (b-2ah)t + (c-bh+ah^2) = 0.

Algebraically, you want rid of that annoying "t" term; graphically, you want to move the vertex to the origin. So let h = (b/2a), and crank through the algebra to get

t^2 = (b^2-4ac)/(4a^2).

Putting this back in for x = t-h gives the answer. The algebra is a bit more involved than your versions, but the key idea is easy to remember, it connects to the topic of shifts of functions, and it is something which is useful over and over again (write things in a variable in which they look nicer).

B. Probably at some point students have factored quadratics which have integral roots. They know that, given

x^2 + (b/a)x + (c/a) = 0,

you are looking for two numbers r and s such that

r + s = -(b/a), and
rs = (c/a).

Such "sum and product" problems go back to ancient Babylonia. If you try some problems with specific numbers (and students will have done so when factoring), it doesn't take long to hit on a general strategy: for example, if you want

r + s = 20, and
rs = 51,

you know that the numbers have to average to 10: one is 10 plus something, the other is 10 minus something. More formally, you can say

r = 10 + t
s = 10 - t

and then the second equation you are trying to solve is (10+t)(10-t)=51 - a difference of squares! It is easy to find t^2=49, t=7 and hence r,s are 17 and 3.

So do the same in letters:

r = -(b/2a) + t
s = -(b/2a) - t

Then r+s= -(b/a) automatically, so we just have to solve

(-(b/2a)+t)(-(b/2a)-t) = (c/a)

which gives a difference of squares,

t^2 = (b^2-4ac)/(4a^2),

and the formula comes out. Again, the algebra is more involved, but the idea is easy to remember and familiar: just do what you were doing to factor quadratics, but do it in general with letters. (Also a nice historical interlude - you can show problems that students were solving on clay tablets in 1900 BC...) There is also a general principle here - they have factored quadratics ad nauseam by this point. Rather than do the same thing over and over again, do it once with letters.

5. 6. I absolutely agree with Andrew that solving problems via isolated techniques is not at all respectful to the learning process and the general pursuit of knowlegde. But one's judgment on this depends on context. For example, the derivation I presented in that interview, for the majority of my classes, comes at the end of a general tale of connecting complex problems to simpler ones E.g. x^2 = 9 is easy to solve, and so is 4x^2 + 12x + 10 =1 if you can make it look just as nice, hence the whole completing the square gig (which also explains how you come to think of the approach I present). This is what the "translation" method you present does.

Also the completing the square method connects geometric thinking to algebra which is a mighty powerful general technique - useful all over the place in what I do with kids.

All approaches are good if they come couched in a bigger picture story. Teachers should always have at the forefront of their mind "what is this doing for my students beyond the immediate technique at hand?" If there is no clear answer to it, then I say, don't do it!

It is fun, when students are up for it, to present multiple ways to prove the same formula. The idea that different paths of thinking lead to different insights is in and of itself a great lesson.

We both might be saying that we want to teach mastery of thinking, not really the mastery of doing (though the doing is, of course, expected!)

7. James, I didn't mean to be as critical as it may have sounded. I only meant to suggest other options. I'll sometimes use the completing the square argument as well - I just think it has been cemented in many classrooms as the *only* argument when in fact it is one among many, and probably not the best.

By the way, is the geometrical completion of the square common knowledge? I can't draw it here of course... but think of the equation

x^2 + px = q

(take p and q to be positive for now) as being

one square of side length x
plus p rectangles of length x and width 1
gives a certain known area q.

Now glue (p/2) of the rectangles to one side of the square of side length x, and glue the other (p/2) of the rectangles to the top of the square. Now we have an equation that says

square with a corner bit missing
has a certain known area q.

Now literally complete the square! Glue in a square of side length (p/2) in the top corner - so you get, on the left side, a completed square of side length (x + (p/2)), and on the right side, the value (q+(p/2)). So now you have a square with known area (q+(p/2)), so you can solve for the side length (x+(p/2)), and consequently for x.

I find this is most valuable when you do a few examples where everything works out in integers.

(If the equation isn't in the form x^2+px=q for positive p and q, you can do a variant on the argument - just move the terms so all the coefficients are positive.)

My apologies if everyone knows that one.

8. Oops, typo: for every place I typed "(q+(p/2))" above, read instead "(q+(p/2)^2)".

9. Yep ... Completing the square is indeed, literally, completing a square! (Hence the name "quadratic" for these degree two equations - they are the equations solved by completing quadrangles.) I am not sure if the geometric interpretation is well known. After all, many teachers are surprised when I point out that x^2 is called "square" for a specific geometric reason. (Just as x^3 is called x cubed and just as we don't have a name for x^4!)

I don't mean to self-promote, but I go through how I approach quadratics to kids in a beginning algebra class in a pamphlet on my website: www.jamestanton.com. Just click on the "Everything you wanted to know about quadratics" piece on the right of the screen. It doesn't get into the other approaches you suggested, only because they come later in a different course. (BTW... You will see I do translations in graphs too, so I am setting the stage.) 