History of Quadratic Formula

Question

My wife is planning a lesson on the quadratic formula for high school students, who have previously learned how to complete the square. It would be nice to open the lesson with some historical background.

Does anybody know of any nice articles, anything from a few paragraphs to page or two long, that discusses the history of the formula and what problems its original discoverers were trying to solve. We would prefer an article that does not spill the beans about the formula's derivation.

No articles, but it does not appear to be common knowledge on MSE that both $f(x) = a x^2 + b x + c$ and $g(x,y) = a x^2 + b x y + c y^2$ are factorable over the integers precisely when $b^2 - 4 a c$ is nonnegative and a perfect square. — Will Jagy, Jul 18 '12 at 22:08
@Will Jagy: Is it common knowledge on MSE that perfect squares are nonnegative? — Marc van Leeuwen, Jul 18 '12 at 22:22
You could start with looking at this and this. If memory serves, apart from the Babylonians, the Chinese and Hindus also knew about (a special form of) the quadratic formula, but I don't have my references right now. — J. M. ain't a mathematician, Jul 18 '12 at 22:47
Also look at http://www-history.mcs.st-and.ac.uk/HistTopics/Quadratic_etc_equations.html — Robert Israel, Jul 18 '12 at 22:58
For the "completing the square part" see al-Khwarizmi, who was completing an honest to goodness square. For example, to explain solution of $x^2+2bx=d$ ($b$, $d$ positive) imagine a square house of side $x$, with porches of width $b$, length $x$ on north and east. Complete this to a square by adding $b \times b$ square. Big square has side $x+b$, area $d+b^2$, so $x=\sqrt{d+b^2}-b$. — André Nicolas, Jul 18 '12 at 23:28
The teacher may be interested in the following simpler derivation of the quadratic formula. The main pedagogical advantage is that one ends up avoiding division until the very end. The trick is very old, going back to Sridhara, but is in my opinion insufficiently used. — André Nicolas, Jul 19 '12 at 00:50
I'll send you some .pdf files by email shortly, but the following (freely available on the internet) will probably be more useful: Solving Quadratic Equations By analytic and graphic methods; Including several methods you may never have seen by Pat Ballew (2007). — Dave L. Renfro, Jul 19 '12 at 14:30

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

Here is a collection of answers from comments, so that this question can be put to bed.

J.M. says:

You could start with looking at this and this. If memory serves, apart from the Babylonians, the Chinese and Hindus also knew about (a special form of) the quadratic formula, but I don't have my references right now.

Robert Israel says:

Also look at http://www-history.mcs.st-and.ac.uk/HistTopics/Quadratic_etc_equations.html

Andre Nicolas says:

For the "completing the square part" see al-Khwarizmi, who was completing an honest to goodness square. For example, to explain solution of $x^2+2bx=d$ ($b$, $d$ positive) imagine a square house of side $x$, with porches of width $b$, length $x$ on north and east. Complete this to a square by adding $b\times b$ square. Big square has side $x+b$, area $d+b^2$, so $x=\sqrt{d+b^2}−b$.

The teacher may be interested in the following simpler derivation of the quadratic formula. The main pedagogical advantage is that one ends up avoiding division until the very end. The trick is very old, going back to Sridhara, but is in my opinion insufficiently used.

Dave L. Renfro says:

I'll send you some .pdf files by email shortly, but the following (freely available on the internet) will probably be more useful: Solving Quadratic Equations By analytic and graphic methods; Including several methods you may never have seen by Pat Ballew (2007).

Will Jagy · Answer 2 · 2012-07-19T02:11:20.800

Not a reference. Just checking my comment.

I wanted to be sure I had the statement correct. We are asking about roots and factoring for $a x^2 + b x + c.$ Once we find that $b^2 - 4ac$ is a perfect square $N^2,$ we get roots $\frac{-b \pm N}{2a},$ or $\frac{-P}{a}, \; \frac{-Q}{a},$ with $PQ = a c$ and $P+Q = b.$ This does use the fact that $ N \equiv b \pmod 2.$ We begin with the rational factorization $$ \frac{1}{a} ( a x + P) (a x + Q) $$ with all integers. Next, take $$g = \gcd(a,P), \; \; a = g \alpha, \; \; P = g \pi$$ so that $\gcd(\alpha, \pi) = 1.$

Well, from $PQ = ac,$ we get $P|ac,$ then $g \pi | g \alpha c,$ then $\pi | \alpha c.$ However, from $\gcd(\alpha, \pi) = 1$ we find $\pi | c,$ and we may let $$ c = \gamma \pi. $$ Comparing $PQ = ac,$ we find $Q = \alpha \gamma,$ so our integral factorization is $$ a x^2 + b x + c = (\alpha x + \pi) (g x + \gamma). $$

If, in addition, we have $\gcd(a,b,c) = 1,$ it follows that $\gcd(g,\gamma) = 1.$ It also follows that $\gcd(a,Q) = \alpha.$ In this case, it is correct to say that we could have calculated the roots with the quadratic formula, written those fractions in lowest terms, and used those fractions to write the factorization directly.

@anon, I cannot disagree, but I left a comment that seemed quite natural to me, then realized I was not certain I knew how to prove it. The proof is far too long for another comment, and could be useful in some way as a side note in the lecture. — Will Jagy, Jul 18 '12 at 23:22

History of Quadratic Formula

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