Teaching the quadratic equation and the roots

[Astronuc]

https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/

An interesting article about solving ax² + bx + c = 0 = (x-R)(x-S), where R and S are the roots.

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

In my classes, we were never 'spoon fed' any formula, but rather, we worked through a derivation.

 

[PeroK]

https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/

An interesting article about solving ax² + bx + c = 0 = (x-R)(x-S), where R and S are the roots.

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

In my classes, we were never 'spoon fed' any formula, but rather, we worked through a derivation.

If we are really worried about a student's ability to solve a quadtratic equation unless it is made easier, then how are they going to cope with higher mathematics? Deriving trig identities, for example, requires a variety of techniques and cannot be boiled down to one simple rule.

And yesterday, there was a tricky homework problem here about the interior area of a triangle. Not everything can be made easy.

 

[vanhees71]

Isn't this more complicated than the standard way to "complete the square"?

First we can simplify the task by deviding by $A$:
$$x^2+p x + q =0, \quad p=\frac{b}{a}, \quad q = \frac{c}{a}.$$
Then "complete the square"
$$\left (x+\frac{p}{2} \right)^2 + q-\frac{p^2}{4}=0.$$
Then you get
$$x_{1/2}=-\frac{p}{2} \pm \sqrt{\frac{p^2}{4}-q}=-\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}.$$
You can of course also write it in the way as written on the website
$$x_{1/2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

 

[malawi_glenn]

I teach both ways, factorization and completing the square. However, Swedish school insists in using formula sheets during national exams...

Calling it a new way is perhaps a big lie. I have been doing it for almost 10 years

 

[vela]

Then "complete the square"

I think this step mystifies a lot of students. It relies on the student knowing that $(a+b)^2 = a^2+2ab+b^2$, which, for some reason, confuses a lot of students. I often find it frustrating when I see a student painfully multiplying everything out and simplifying instead of just writing the damn answer down. For a student like that, adding and subtracting $(p/2)^2$ to complete the square seems like something pulled out of the air.

Loh's approach just avoids the "trick" and breaks the problem down into steps that are easier to understand for most students. The advantage is a student should be able to solve for roots even if they can't remember the quadratic formula. The drawback is that STEM students do need to learn how to complete the square eventually, so you're just delaying the inevitable.

Of course, it's not an either-or situation. You could teach Loh's approach and then use it to derive the quadratic equation. Then hopefully, more students won't see it as a mysterious formula.

 

[malawi_glenn]

It relies on the student knowing that (a+b)2=a2+2ab+b2, which, for some reason, confuses a lot of students

Students should learn this, and realize why it is true.

 

[cormsby]

In one of my classes lonnng ago, a student asked why the process is called "completing the square", so I showed the class its geometric origin. From that, I showed the derivation of the quadratic formula. Many students told me this demonstration made it much more clear, so I presented it thus ever since.

 

[MidgetDwarf]

I teach both ways, factorization and completing the square. However, Swedish school insists in using formula sheets during national exams...

Calling it a new way is perhaps a big lie. I have been doing it for almost 10 years

I went against the order of the book when I was teaching Algebra to 7th and 8th graders. Ie., they give them the quadratic equation first, then many chapters later introduce completing the square. I found this approach very odd. So instead, I went over factoring, completing the square, graphing, and somewhere in the middle of graphing, we did the derivation for completing the square.

Im sure I pissed off a few other faculty by doing this. But my reasoning is that, when they are asked to graph quadratic equations. Completing the square tells them the whole story. Plus it makes them work with fractions again lol.

 

[MidgetDwarf]

Moreover, I am a bit skeptical of the derivation in the paper being a new one. Since I have seen it used by an old teacher whose class I observed (long deceased).

 

[malawi_glenn]

Moreover, I am a bit skeptical of the derivation in the paper being a new one. Since I have seen it used by an old teacher whose class I observed (long deceased).

It is a special case of Vieta's formulas https://en.wikipedia.org/wiki/Vieta's_formulas

 

[vanhees71]

I think this step mystifies a lot of students. It relies on the student knowing that $(a+b)^2 = a^2+2ab+b^2$, which, for some reason, confuses a lot of students. I often find it frustrating when I see a student painfully multiplying everything out and simplifying instead of just writing the damn answer down. For a student like that, adding and subtracting $(p/2)^2$ to complete the square seems like something pulled out of the air.

But precisely these "binomial formulae" were hammered into us like crazy. It was so boring! I never understood, what's the fuzz about it. You just multiply the product out once in your life and then remember the formula. It's no deep thing at all. I think, in the recent decades the didactics of math and also the natural sciences became worse and worse, though I see some tendency in the physics didactics to swing back to a more useful approach.

Poh's approach just avoids the "trick" and breaks the problem down into steps that are easier to understand for most students. The advantage is a student should be able to solve for roots even if they can't remember the quadratic formula. The drawback is that STEM students do need to learn how to complete the square eventually, so you're just delaying the inevitable.

Well, I found this approach much more complicated than to complete the square.

Of course, it's not an either-or situation. You could teach Poh's approach and then use it to derive the quadratic equation. Then hopefully, more students won't see it as a mysterious formula.

Sure, it's good to have alternative approaches at hand to help students individually, but I think the most simple derivation is the best.

 

[Mark44]

Well, I found this approach much more complicated than to complete the square.

As did I.

Of course, it's not an either-or situation. You could teach Poh's approach and then use it to derive the quadratic equation. Then hopefully, more students won't see it as a mysterious formula.

Minor note. It's Loh's approach (Po-Shen Loh).

Sure, it's good to have alternative approaches at hand to help students individually, but I think the most simple derivation is the best.

Agree.

 

[Mark44]

In one of my classes lonnng ago, a student asked why the process is called "completing the square", so I showed the class its geometric origin.

IMO, presenting something both in algebraic form and in geometric form makes for better understanding. A long time ago I was told that text and images are processed on different sides of the brain, so if this is still held to be the case, these two approaches make it more likely that the student will retain the idea.

 

[vela]

Minor note. It's Loh's approach (Po-Shen Loh).

Fixed.

 

[robphy]

Possibly useful:
https://www.cmu.edu/math/people/faculty/loh.html
https://arxiv.org/abs/1910.06709 "A Simple Proof of the Quadratic Formula" (Po-Shen Loh)
https://www.poshenloh.com/quadraticdetail/

 

[symbolipoint]

In my classes, we were never 'spoon fed' any formula, but rather, we worked through a derivation.

Or at least, teacher worked through the derivation, and showed it as classroom instruction. That was how it was done and it worked. Very small mystery, solved, and understandable.

 

[symbolipoint]

I went against the order of the book when I was teaching Algebra to 7th and 8th graders. Ie., they give them the quadratic equation first, then many chapters later introduce completing the square. I found this approach very odd. So instead, I went over factoring, completing the square, graphing, and somewhere in the middle of graphing, we did the derivation for completing the square.

Im sure I pissed off a few other faculty by doing this. But my reasoning is that, when they are asked to graph quadratic equations. Completing the square tells them the whole story. Plus it makes them work with fractions again lol.

? thinking;
? thining;
Excellent! You understand the order by which parts of this can be taught for better understanding and you did what is logical for this instruction.

 

[MidgetDwarf]

It is a special case of Vieta's formulas https://en.wikipedia.org/wiki/Vieta's_formulas

Ahh. Makes sense, I finally had time to read and work through the wiki. I must confess, I am a bit weak in Ring Theory.

 

[symbolipoint]

https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/

An interesting article about solving ax² + bx + c = 0 = (x-R)(x-S), where R and S are the roots.

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

In my classes, we were never 'spoon fed' any formula, but rather, we worked through a derivation.

I just began to take a look at the referenced article. The equation you display in post #1 looks like what Algebra 1 students learn to do or use. (Maybe really for Algebra 2). Algebra 1 students learn quickly how to multiply two such simple linear binomials. After simplifying, student or whoever, can simply equate corresponding parts and solve - usually with ordinarily known numbers not including the variable x.

 

[symbolipoint]

The geometric, rectangular interpretation is great! Some people may not be aware of it until some years after graduating with a degree in something. That kind of technique for CompletingTheSquare gives a person a neat way to derive the solution.

 

[Swamp Thing]

Here is a nice idea that's relevant to this thread, I hope: https://math.hmc.edu/funfacts/complex-roots-made-visible/

 

[Muu9]

Here's a nice playlist building towards a derivation of the quadratic formula based on the geometric approach to completing the square:

 

[symbolipoint]

That video lesson in post#22 also shows how to understand multidigit multiplication.

Further on in the video, that way to show negative multiply by negative be positive is wild.

 

[mathwonk]

exactly this "new" method of solving the quadratic, which I rather like, is described on pages 47-48 of Lectures on Elementary mathematics, by LaGrange, where he attributes it to Diophantus, writing over 1700 years ago.
https://www.gutenberg.org/files/36640/36640-pdf.pdf

reading the article again, it is clear that Mr. Loh himself expressed only skepticism that the method might be new, and that implication was due mainly to the journalist.

 

[Vanadium 50]

t is a special case of Vieta's formula

1. Unfortunately, at least in the US, this seems to be an obscure and lost piece of lore. It's very existence, much less its proof by induction.

2. You can memorize it by singing it to Row Row Row Your Boat. (The same reason I remember Albania's chief export is chrome)

4. I learned it, like millions of other students, by completing the square. It's not clear to me that "assume a solution of the form" might not be as good a starting point.

 

[mathwonk]

I apologize for summarizing the discussion for my own clarification.

As you all know, solving a quadratic equation is showing how to find two numbers, r,s if one knows their sum r+s and their product rs. I.e. the solutions, r,s, of x^2 - bx + c = 0, satisfy r+s = b, and rs = c. The key idea is that if one could also find their difference, one would be done. i.e. (r+s) + (r-s) = 2r, and (r+s)-(r-s( = 2s.

All solutions are then based on these two formulas, which show that the square of the sum, and the square of the difference, are equivalent up to a multiple of the product:

(r+s)^2 = r^2 + 2rs + s^2;
(r-s)^2 = r^2 -2rs + s^2.

Consequently, (r-s)^2 = (r+s)^2 - 4rs.

Thus, if one knows the sum of two numbers, and their product, hence the right hand side of this equation, one can find the square of their difference, i.e. the left hand side. Then by taking a square root one finds also their difference.
I.e.square the sum, and subtract 4 times the product, and one has the square of the difference,

Since knowing both the sum and the difference of two numbers gives both numbers, it follows that one can find two numbers if we know their sum and product.

Thus to solve x^2 - bx + c = 0, for roots r,s, we know r+s = b and rs = c. Hence (r-s)^2 = b^2-4c.

Hence (r-s) = sqrt(b^2-4c), so 2r = (r+s) + (r-s) = b + sqrt(b^2-4c).

Similarly, 2s = b - sqrt(b^2-4c).

Thus r,s = (1/2)(b±sqrt(b^2-4c)).

This is the way I personally like to understand the solution of quadratic equations, essentially the method of Diophantus.

The reason I like this is I can understand the reason for every step. In high school when I read the derivation of the final formula, by completing the square, it seemed to me entirely unmotivated, and looked clunky and ugly. Over 50 years later, when I finally encountered Lagrange's clear explanation, I at last felt satisfied about it.

I had memorized the formula in high shool, and always had success using it, but never felt that I understood how someone would have thought of it. That is always my goal. So this is to me another argument for reading the classics, i.e. the masters.

 

[vanhees71]

Well, just "completing the square" is much more straight forward
$$x^2+p x+ q=0 \; \Leftrightarrow \; (x+p/2)^2-p^2/4+q=0 \; \Leftrightarrow \; (x+p/2)^2=p^2/4-q \; \Leftrightarrow \; x_{1/2}=-p/2 \pm \sqrt{p^2/4-q}.$$
The Vieta formula for quadratic equations simply follows from
$$x^2+px + q=(x-x_1)(x-x_2)=x^2 -(x_1+x_2)+x_1 x_2 \; \Leftrightarrow \; x_1+x_2=-p, \quad x_1 x_2=q.$$

 

[MidgetDwarf]

I apologize for summarizing the discussion for my own clarification.

As you all know, solving a quadratic equation is showing how to find two numbers, r,s if one knows their sum r+s and their product rs. I.e. the solutions, r,s, of x^2 - bx + c = 0, satisfy r+s = b, and rs = c. The key idea is that if one could also find their difference, one would be done. i.e. (r+s) + (r-s) = 2r, and (r+s)-(r-s( = 2s.

All solutions are then based on these two formulas, which show that the square of the sum, and the square of the difference, are equivalent up to a multiple of the product:

(r+s)^2 = r^2 + 2rs + s^2;
(r-s)^2 = r^2 -2rs + s^2.

Consequently, (r-s)^2 = (r+s)^2 - 4rs.

Thus, if one knows the sum of two numbers, and their product, hence the right hand side of this equation, one can find the square of their difference, i.e. the left hand side. Then by taking a square root one finds also their difference.
I.e.square the sum, and subtract 4 times the product, and one has the square of the difference,

Since knowing both the sum and the difference of two numbers gives both numbers, it follows that one can find two numbers if we know their sum and product.

Thus to solve x^2 - bx + c = 0, for roots r,s, we know r+s = b and rs = c. Hence (r-s)^2 = b^2-4c.

Hence (r-s) = sqrt(b^2-4c), so 2r = (r+s) + (r-s) = b + sqrt(b^2-4c).

Similarly, 2s = b - sqrt(b^2-4c).

Thus r,s = (1/2)(b±sqrt(b^2-4c)).

This is the way I personally like to understand the solution of quadratic equations, essentially the method of Diophantus.

The reason I like this is I can understand the reason for every step. In high school when I read the derivation of the final formula, by completing the square, it seemed to me entirely unmotivated, and looked clunky and ugly. Over 50 years later, when I finally encountered Lagrange's clear explanation, I at last felt satisfied about it.

I had memorized the formula in high shool, and always had success using it, but never felt that I understood how someone would have thought of it. That is always my goal. So this is to me another argument for reading the classics, i.e. the masters.

I found rat Lagrange book you linked to in this thread very readable and insightful. Have something similar for Analysis?

 

[mathwonk]

I cannot offer a similarly clear and authoritative explanation of analysis, possibly because analysis, as we know it, is only a few hundred years old, as compared to the thousands of years old nature of elementary algebra. (Also, I myself do not thoroughly understand analysis.) Of course Archimedes gave arguments of an analytical nature, but Riemann only defined his integral less than 200 years ago. So perhaps the best sources now for expositions of elementary analysis are in relatively modern books, such as Courant's or Apostol's Calculus. More advanced, and more abstract, sources include Berberian's Fundamentals of real analysis, and Dieudonne's Foundations of modern analysis. (Arnol'd liked the course in analysis by Goursat, and I own that, but hesitate to recommend it as an elementary treatment.)

But I can offer an authoritative explanation of an aspect of analysis, namely infinite series, and its use in expanding elementary transcendental functions, by a great classical master, Euler. The difficulty is in understanding what Euler means, as he uses freely the notion of infinitely large and small numbers, in place of precise arguments with limits. Still, since it is Euler, it may well be interesting and illuminating. (Even after thinning out my collection for a big move, I still have all the books I mention on my shelf, including Archimedes and Riemann, and have greatly benefited from perusing both of those.)

http://www.17centurymaths.com/contents/introductiontoanalysisvol1.htmThere is a modern analysis book, not a historical one, apparently quite clear and useful, and free of charge, that is linked in the first post of the following thread:
https://www.physicsforums.com/threads/what-are-infinitesimals.1052290/#post-6882767

 

[mathwonk]

actually, perhaps the beginning of writing about analysis occurs in Euclid, Book V, definition 5, which says in essence, that two pairs of magnitudes A,B and C,D, are "in the same ratio", i.e. the real numbers A/B and C/D are equal, if and only if, for every (positive) rational number r, we have rA≥B, if and only if rC≥d, (and rA≤B if nd only if rC≤D). Thus the real numbers A/B and C/D are equal if and only if they partition the rational numbers into the same two families, (later they define the same "Dedekind cuts").
Euclid uses this "limiting" criterion to prove the basic triangle similarity theorem, Prop. 2, Book V, using his theory of area.
So to me, Euclid contains the beginnings of analysis. especially as Euclid was made more precise by Hilbert. I.e. every point on a line divides the line into two "sides", and every subdivision of the line into two sides, determines a point of that line. This is the origin of the notion of "completeness" of the real number line, the basic notion of analysis.

 

[mathwonk]

Here is an example of how euler, without differential or integral calculus, derives the correct formula for the infinite series expansion of an exponential function, by imprecise but amazingly intuitive approximations.

we know that a^0 = 1, hence, since small changes in the argument yield small changes in the output which are approximately linear, (the fundamental principle of the differential calculus), if w is very small, “infinitely small”, then a^w is essentially equal to 1 + kw, where k the derivative of the exponential function at 0. I will simplify by taking the base of the exponential function a, to be so that derivative is 1, (I am taking a = e), hence for w infinitely small we have a^w = 1+w.

Now let z be a finite quantity, and n an infinitely large integer, so that w = z/n is infinitely small.

Then a^z = a^[(z/n)(n)] = [a^w]^n = (1+w)^n = (1 + (nw/n))^n = (1+z/n)^n

= (by the binomial theorem) 1 + z + (z^2/2!)[n(n-1)/n^2] + (z^3/3!)[n(n-1)(n-2)/n^3] + ………

But since n is infinitely large, n/(n-1) = n/(n-2) = 1, and this gives:

a^z = 1+ z + z^2/2! + z^3/3! + ……….

In particular for the base, a = a^1 = 1+1 + 1/2! + 1/3! + ……….One might find it interesting to try, by modern limiting arguments, to justify this derivation.

Pardon me, I realize I have deviated from the purpose of this thread, so I will not post further in this different vein.

 

[mathwonk]

indeed, re post #27, one man's "straightforward", is another man's "unmotivated, clunky, and ugly". all of which makes the world go round, cheers!

 

[vanhees71]

I'd say, it's not a good idea to learn analysis from such old textbooks. There was a big crisis in the 19th century, and reading any book before the works of Weierstraß, Cauchy et al one has to be aware that it's outdated.

Also Euclid is missing the importance of the completeness assumption, which has been explicated first by Hilbert in his famous axiomatic foundation of the subject.

 

[mathwonk]

Here is a comment along the lines of the original article linked by Astronuc on solving the quadratic after translating by the average of the roots. (Of course you may be well aware of this extension also.)

Suppose we try this for cubics! Here we have x^3 + bx^2 + cx + d = 0, and the sum of the roots this time is again -b, but now there are 3 of them, with average -b/3. Note this is also the geometric "center" of the cubic in some sense, since that is where the unique inflexion point of the cubic is found.

Now substituting x = y -b/3 into our equation does not give us a "pure" cube, since there is still a term in y, but it does eliminate the y^2 term! So we have not "completed the cube", but we have simplified things a bit.

Now it turns out one can solve a cubic of this "reduced" form, which we will write as x^3 = px + q. Indeed the key is an extension of the idea rediscovered by Mr. Loh, of writing our solution x as a sum of two other numbers. (This is the famous method of Scipione del Ferro, published by Cardano in 1545.)

I.e. suppose X = a+b, so that X^3 = (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = 3ab(a+b) + (a^3+b^3)
=3abX + (a^3+b^3).

Thus if we have a cubic of the special form X^3 = (3ab)X + (a^3+b^3), then X = a+b will give a solution.

Now let us look at a general reduced cubic X^3 = pX + q. If only we had numbers a,b such that p = 3ab, and q = a^3+b^3, we would have solved our cubic.

But note that this says we want a and b such that 27a^3b^3 = p^3, or a^3b^3 = p^3/27, and a^3+b^3 = q. That says we know both the sum and the product of the numbers a^3 and b^3. Hence we can find both a^3 and b^3 by solving a quadratic!

I.e. a^3 is a solution of the quadratic T^2 -qT + (p^3/27) = 0. Then we let a be any cube root of a solution of this quadratic, and take b so that 3ab = p. That gives the 3 solutions of our cubic!

example: to solve X^3 = 9X + 28, we want 3ab = 9, and a^3+b^3 = 28, which suggests a = 1 and b = 3, and indeed X = 1+3 = 4 works!

This discussion is modeled on that in the great algebra book of Euler, chap XII, p. 240 of the following link:
https://archive.org/details/ElementsOfAlgebraLeonhardEuler2015/page/240/mode/2up

 

[mathwonk]

vanhees71 makes a good point that one should certainly not limit oneself to what is available in very old books. My suggestion is that, especially with later insights, neither should one ignore them, as one can learn something from anything written by a great master.

 

[mathwonk]

It has finally dawned on me how my high school textbook might have made completing the square look more motivated to me. My book had pretty much the derivation in post #27, which made me ask "and how did you think of that?" after the first step. Probably it was my weakness with fractions that was the problem.

I did not readily see that
X^2 + bX + c = (X+b/2)^2 -b^2/4 + c, but I knew very well that (X+b)^2 = X^2 + 2bX + b^2.

So if they had only written it as follows, I would perhaps have found it clearer:

X^2 + 2bX + c = (X^2 + 2bX + b^2) -b^2 + c = (X+b)^2 -b^2 +c. Hence

X^2 + 2bX + c = 0 iff (X+b)^2 = b^2-c iff X+b = ≠ sqrt(b^2-c) iff X = -b ± sqrt(b^2-c).

I.e. X^2 + 2bX almost cries out to be completed by adding b^2.

In the same way, when using the method emphasizing that the cofficients tell us what the sum and product of the roots r,s is, I find it clearer to write X^2 -bX + c, so that b = r+s and c = rs.

I don't know if any of my students would have been helped by this, as although I am still struggling to see how to make things clear, I have long been retired from the classroom. Maybe it would not have helped those for whom the problem was seeing that (X+b)^2 = X^2 + 2bX + b^2, but that could have been explained by the many nice geometric illustrations which have been given here, (going back to Euclid, Book II, Prop. 4)
http://aleph0.clarku.edu/~djoyce/elements/bookII/propII4.htmlWell I just spent a good hour + watching the videos linked in post 22 by Muu9, and in my opinion that guy, James Tanton, has to be the greatest (algebra) teacher in the world. (I was slightly reassured that he is human, when he did the computation -5 -8 = -11 in his head.) Thank you Muu9!

 

[haushofer]

I recently became aware of the trick one uses to solve for the roots of 3d order polynomials: the Tschirnhaus transformation, x --> y=x+p , to eliminate the term quadratic in x. For 2nd order polynomials you can do the same thing to remove the linear term linear, solve for y and transform back.