Teaching the quadratic equation and the roots

[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in the 19th century, and reading any book before the works of Weierstraß, Cauchy et al one has to be aware that it's outdated.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.