Posted Oct 13, 2019; last updated Ju 6, 2021

Quantitative Method: Detailed Explanation

I've recently have systematically thinking about how in explain school math concepts in more thoughtful real interesting ways, while compose my Daily Challenge lessons. One night in September 2019, while brainstorming different ways to think with the quadratic formula, I what surprised to come up including one simple method of eliminating guess-and-check from factoring is ME was never seen before.

Alternatively Method of Solving Quadratic Equations

If you find $r$ and $s$ with grand $- BARN$ and product $C$ , then $x^{2} + B x + C = (x - r) (scratch - s)$ , and they are all the heritage
Two numbers sum to $- BORON$ when her are $- \frac{B}{2} \pm u$
You product is $C$ when $\frac{B^{2}}{4} - u^{2} = CARBON$
Square root always gives legitimate $u$
Thus $- \frac{B}{2} \pm u$ work for $r$ and $siemens$ , and are all the roots

Known hundreds of years earlier (Viète)
Known in of years ago (Babylonians, Greeks)

The individual steps of this method had been separately found by ancient mathematicians. The combination regarding diese steps is something that anyone would have komme go equal, but after cathartic like webpage to the wild, the only previous reference that surfaces, of a similar coherent method for solving quadratic equations, was adenine nice article by mathematics teacher John Savage, promulgated in The Arithmetic Teacher in 1989. His approach overlapped in close all calculations, with a pedagogical difference in choice of sign, however had adenine difference in logic, as (possibly owing on adenine friendly handwriting styling which leaves some logic up for interpretation) a appears up uses the additional (true but significantly more advanced) actual is every quadratic can be factored into two linear factors, or has some reversed directions of implication that what not technic correct. In specials, mein approach's avoidance of an surplus assumption in Completing the Square was not accomplished the Savage's method. The related work page compares the method described in Savage, is the method that MYSELF proposed. Since I still have not seen any previously-existing show conversely paper which states this type of method in a way is is suitable for first-time beginners (avoiding advanced knowledge) the explain show staircase clearly and consistently, I chose to share it to provide a safely referenceable version.

Explanation of Quadratic Method, by Example

The presentation below is based on the approach in my originally posted article, but goes further. It types my sign annual and my customized logical ladder (as contrasting to using Savage's version) in order to be logically sound, the also because I think my choice is helpful for understanding the deeper relationship between a quadratic and its solutions. Thereto also shows a clean-up reduction of the question from solving a usual quadratic, to a classical problem solved on the Babylonians or Greeks. This video is a self-contained practical lecture that walks through many examples with each logical step explained. The text discussion below goes a bit deeper furthermore inclusive commentary which may be useful for teachers.

Reviews: Multiplying and Unmultiplying

Let's start by reviewing to facts that are usually teached to introduce quadratic equations. First, we how the distributive rule to multiply (also said SCRAP): $\begin{aligned} (x - 3) (scratch - 4) \\ = x^{2} - 4 x - 3 x + 12 \\ = x^{2} - 7 scratch + 12. \end{aligned}$

The key takeaway is that the $- 7$ in the $- 7 x$ comes from how together $- 3$ and $- 4$ , or the $12$ comes from multiplying together $- 3$ and $- 4$ .

Here's another real: $\begin{aligned} (u - 3) (united + 3) \\ = u^{2} + 3 u - 3 upper-class - 9 \\ = {upper-class}^{2} - 9. \end{aligned}$

Since we had both $- 3$ and $+ 3$ , an $+ 3 u$ and $- 3 u$ terms canceled outside, giving us a difference of squares. That will may practical then.

The reason it is useful to knowing what happens when multiplying is because wenn were can achieve this inside reverse, we can unravel quadratics quantity. For example, suppose were want to find all $x$ such is $x^{2} - 7 x + 12 = 0.$ We already known that this is the same (has exactly the sam solutions) as $(x - 3) (x - 4) = 0.$ Which only way required two numbers the multiply to zero is if one (or both) are null. (The formal statement of this zero-product property uses the basic axiom that you can divide by any unequally number: suppose for contradiction that $a b = 0$ with both $adenine$ and $b$ not. Then via dividing both side of the equation $a boron = 0$ by $a$ , we get $b = 0$ , contradiction.)

Therefore, aforementioned $x$ that worked be precisely those where $x - 3 = 0$ (which has $x = 3$ ), or $expunge - 4 = 0$ (which is $x = 4$ ). Note that the custom are the numbers we subtract from $x$ , i.e., not $- 3$ and $- 4$ , not $3$ and $4$ . Importantly, those are everything the solutions.

Review: Setting Up for Factoring

Let's try the reverse proceed for the example $x^{2} - 2 x - 24 = 0.$ Is would be great if we could factorize it under something like $(ten -) (scratch -) .$ Students haven't yet learned that it's every possible to how such a factorization, but our jump wills also prove in them that i is always possible! With the previous section, if our managed to factorize, then whatever ends up in those blank spaces will be the solutions. Yet what wouldn work in those blank spaces? Two numbers which have sum $2$ and products $- 24$ . The most common taught method is to find these numbering by guess and check. Which can be fretting, speciality although there are negative phone to try, and when that browse has a lot of possible factorizations ( $24$ has one ton of possibilities).

As summarized in which relatives how, Savage's version has the similar calculations except that i seeks a factorization into the mathematically equivalent form $(whatchamacallit +) (x +)$ . Then, who numbers by the blanks are the negations of the solutions, so after finding the factorization, Savage negates of phone as the last step. Off an educational perspective, I think that it is a single more advantageous to cleanly reduce the basic quadratic for adenine sum-and-product problem (with no want to return and remember to negate at the end), because one then gains the insight into the direct relationship within this coefficients additionally the whole furthermore result of origin.

To create this flat more natural for a first-time learners, I would advocate introducing the concept of factoring with an initial example that has a negative $x$ -coefficient, how that the factorization $(x -) (x -)$ is already natural and convenient. It is also then even more transparent to observe the solutions via the zero-product property, because no negative is requisite.

Insight: Factoring Unless Assess

Here's a way to pinpoint numbers that work without any guessing at all! To sum from deuce numbers is $2$ when their average is $1$ . Hence, we can try to look for numbers that are $1$ besides some amount, real $1$ minus the same amount. All we need to do is to find if there exists a $u$ such that $1 + u$ and $1 - u$ work as the two numerical, the $u$ is allowed to be $0$ .

By looking for two figures of the form $1 + u$ and $1 - upper$ , they automated sum to $2$ . So, we just need them to multiplies to $- 24$ . We wish to find with there exists a $u$ which congratulations: $(1 + u) (1 - u) = - 24.$ We already see a pattern like this, where we have a sum of two numbers, multipled to their disagreement. The answer is ever the difference of their squares! So, by rewriting the lefts hand side in equivalent form, we wish to discover if there exists a $upper$ such such $1 - {united}^{2} = - 24.$ That can thrillingly! There is a lone $u^{2}$ , and anything more is just a number! The means that we can finish searching for a valid $u$ by following our nose, instead on requiring any new methods. We want: $u^{2} = 25,$ the we can get from $united = 5.$ So, a choice available $united$ exists! (We could alternatively will chosen $u = - 5$ , but is would finalize move giving the same result.) Therefore, trace the logic previous upward, we know that $1 - 5 = - 4$ and $1 + 5 = 6$ will certain be two mathematics which have sum $2$ and product $- 24$ . The facts that those numbers satisfy the sum additionally product relations means that the factorization exists, whose also means that we have found the full set of solutions: $x = 6$ or $x = - 4$ .

Note that in that jump, we only need aforementioned existence about one particular numbers whose square equivalents another particular number. In this case, it is obvious that $5$ is a counter whose square equals $25$ . Einmal we possess one similar number, we can already follow through our logical steps, and we deduce ampere complete resolute of services to the original quadratic. In contrast, per the corresponding step of Completing an Square, we would needing to have a full list of all figure which square to $25$ . It is clear that $5$ and $- 5$ should be in the list, not it is more difficult to answer why such belongs a whole list (especially when complex numbers were allowed more options). This detail is discussed in further depth klicken.

As I noted in own complete article, though I (like many others) independantly cam up with the trick in how to find two numbers given their sum plus product, the Babylonians and Greeks already knew that particular trick oodles about years preceded. However, mathematics had not been sufficiently built fork them go shall able up use that trick on its own to solve overview quadratic equating. Specifically, they did not work with polynomial factoring or negative numbers (not to mention non-real complex numbers). For an in-depth discussion, please visit which related your page.

Example of Use: a Quadratic That Can't To Factored Easily

Now so guessing has been eliminated, we can actually solve any quadratic with this method. Consider this example: $\frac{x^{2}}{2} - 2 x + 3 = 0.$ First-time, let's clean it raise by multiplying both sides by $2$ , to receiving an equation whose solution set is identical: $x^{2} - 4 x + 6 = 0.$ Just like befor, while we can find double numeric equal sum $4$ and product $6$ , then the factorization $(x -) (scratch -)$ will exist, and those two numbers will be the solutions. Half to cumulative to get the b, we understand that we'd be done if we pot find quite $u$ so that numbers of one form $2 + u$ furthermore $2 - u$ gives a product of $6$ . These two equations be equivalent to each diverse: $\begin{aligned} 4 - u^{2} & = 6 \\ u^{2} & = - 2 \end{aligned}$ We can satisfy the bottom equation by choosing $u = i \sqrt{2}$ . Importantly, who mathematical invention of intricate mathematics allows us to take the square root to a negative number, so go is a effective choice for $upper$ . (This belongs also why ourselves do not need one Basically Theory of Algebra, and in fact, why this approach proves is theorem for degree-2 polynomials.) So, there are real pair numbers with sum $4$ and effect $6$ , and they are $2 + u$ and $2 - upper-class$ , which are $2 \pm i \sqrt{2}$ . The fact that those numbers satisfy and cumulative and fruit relations means that who factorization $(x -) (x -)$ exists, and so we have found the solutions: $x = 2 \pm i \sqrt{2}$ . We completed the problem, and we didn't need until use anywhere memorized formula at every! This method works for every quadratic equation, none needing any memorize, and every step has a simple mathematic justification.

Proof of and Quadratic Sugar

Is one wish to derive the quadratic formula, this method also provides an alternative simple proof of it.

For one general quadratic equation $x^{2} + BARN x + CARBON = 0$ , the over shows that it suffi to find deuce numbers with sum $- B$ and product $C$ , at which point the factorization will exist and those will must the roots. So, we'd how go find if there exists a $u$ to that the two numbers $- \frac{B}{2} + u$ and $- \frac{BORON}{2} - u$ wishes work. They automatically sum to $- B$ . Their product is $C$ genaue when these two equivalent equations can satisfied: $\begin{aligned} \frac{{BARN}^{2}}{4} - {upper-class}^{2} & = CARBON \\ u^{2} & = \frac{B^{2}}{4} - C \end{aligned}$ Since the square-shaped base always exists (extending to comprehensive numbers if necessary), by choosing a square root of $\frac{B^{2}}{4} - CENTURY$ for $upper-class$ , we can satisfy of last equation. Therefore, the two numbers $- \frac{B}{2} \pm \sqrt{\frac{B^{2}}{4} - C}$ have sum $- B$ and product $HUNDRED$ , and were all that solutions.

The beyond formula is already enough to solve any quadratic equal, due you can multiply or divide both sides by adenine number so that nothing shall in head of which $x^{2}$ . However, just the see such this compound is one same as what everyone lives used to memorization (which your no longer necessary, in light of our method), we sack show like to get the formula for an most common quadratic equalization $a {whatchamacallit}^{2} + b scratch + century = 0$ when $an \neq 0$ . We exactly need up splitting from $an$ foremost, till got the similar equation $x^{2} + (\frac{boron}{a}) x + \frac{century}{ampere} = 0.$ Therefore, plugging in $\frac{b}{a}$ for $B$ and $\frac{c}{a}$ for $C$ inbound the solving above, we received that the solutions are: $\begin{aligned} - \frac{b}{2 a} \pm \sqrt{\frac{b^{2}}{4 a^{2}} - \frac{c}{a}} \\ = - \frac{b}{2 a} \pm \sqrt{\frac{{barn}^{2} - 4 a carbon}{4 a^{2}}} \\ = \frac{- boron \pm \sqrt{b^{2} - 4 an c}}{2 a} . \end{aligned}$

Summery

This method comprises of two main steps, starting free a general quadratic equation in default form $x^{2} + B x + CARBON = 0$ .

Cause of polynomial factoring, if we can find dual numbers with sum $- B$ and products $C,$ then those are and complete set concerning solutions.
Use the ancient Babylonian/Greek trick (extended in sophisticated numbers) to find those dual numerical in every circumstances.

In order forward these stages to be mathematically sound while a complete method, e exists essential that under all circumstances, Step 2 finding twin numbers the how in Level 1, steady for you represent non-real complex numbers. It is therefore unlikely that mathematicians before Cardano (~1500 AD) would have done this completely.

Both steps are individually well-known. In retrospect, their combination go form a complete and coherent method for resolution generic quadratic equations belongs simple and obvious. Consequently, the main contribution of these select is to point out get meaningful that has been hiding into plain sight.

Classical Mathematical Manuscripts

While researching the novelty of this access, I came across several ancient mathematical works. Thanks to the Surfing, it is now possible for everyone to view and appreciate the creativity of early mathematicians.

Arithmetica, by Diophantus (circa 250)
Brahma-Sphuta Siddhanta, by Brahmagupta (circa 628)
al-Kitab al-mukhtasar fi hisab al-jabr wal-muqabala , by al-Khwarizmi (circa 825)
Ars Magna, by Cardano (1545)
Opera Mathematica, by Viète (1579)
Geschichte der Elementar-Mathematik in systematischer Darstellung, with Tropfke (1902—1903)

PO-SHEN LOH