Posted Oct 13, 2019; last updated Feb 13, 2020

Quadratic Method: Detailed Explanation

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

Alternative Method of Solving Quadratic Equations

  1. If you findrandswith sumBand productC, thenx2+Bx+C=(xr)(xs), and they are all the roots
  2. Two numbers sum toBwhen they areB2±u
  3. Their product isCwhenB24u2=C
  4. Square root always gives validu
  5. ThusB2±uwork asrands, and are all the roots

Known hundreds of years ago (Viète)
Known thousands of years ago (Babylonians, Greeks)

The individual steps of this method had been separately discovered by ancient mathematicians. The combination of these steps is something that anyone could have come up with, but after releasing this webpage to the wild, the only previous reference that surfaced, of a similar coherent method for solving quadratic equations, was a nice article by mathematics teacher John Savage, published in The Mathematics Teacher in 1989. His approach overlapped in almost all calculations, with a pedagogical difference in choice of sign, but had a difference in logic, as (possibly due to a friendly writing style which leaves some logic up for interpretation) it appears to use the additional (true but significantly more advanced) fact that every quadratic can be factored into two linear factors, or has some reversed directions of implication that are not technically correct. In particular, my approach's avoidance of an extra assumption in Completing the Square was not achieved by Savage's method. The related work page compares the method described by Savage, with the method that I proposed. Since I still have not seen any previously-existing book or paper which states this type of method in a way that is suitable for first-time learners (avoiding advanced knowledge) and justifies all steps 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 uses my sign convention and my own logical steps (as opposed to using Savage's version) in order to be logically sound, and also because I think my choice is helpful for understanding the deeper relationship between a quadratic and its solutions. It also shows a clean reduction of the problem from solving a standard quadratic, to a classical problem solved by the Babylonians and Greeks. This video is a self-contained practical lesson that walks through many examples with each logical step explained. The text discussion below goes a bit deeper and includes commentary which may be useful for teachers.

Review: Multiplying and Unmultiplying

Let's start by reviewing the facts that are usually taught to introduce quadratic equations. First, we use the distributive rule to multiply (also called FOIL):(x3)(x4)=x24x3x+12=x27x+12.

The key takeaway is that the7in the7xcomes from adding together3and4, and the12comes from multiplying together3and4.

Here's another example:(u3)(u+3)=u2+3u3u9=u29.

Since we had both3and+3, the+3uand3uterms canceled out, giving us a difference of squares. That will be useful later.

The reason it is useful to know what happens when multiplying is because if we can do this in reverse, we can solve quadratic equations. For example, suppose we want to find allxsuch thatx27x+12=0.We already know that this is the same (has exactly the same solutions) as(x3)(x4)=0.The only way for two numbers to multiply to zero is if one (or both) are zero. (The formal justification of this zero-product property uses the basic axiom that you can divide by any nonzero number: suppose for contradiction thatab=0with bothaandbnonzero. Then by dividing both sides of the equationab=0bya, we getb=0, contradiction.)

So, thexthat work are precisely those wherex3=0(which isx=3), orx4=0(which isx=4). Note that the solutions are the numbers we subtract fromx, i.e., not3and4, but3and4. Importantly, these are all the solutions.

Review: Setting Up for Factoring

Let's try the reverse process for the examplex22x24=0.It would be great if we could factorize it into something like(x)(x).Students haven't yet learned that it's always possible to find such a factorization, but our approach will also prove to them that it is always possible! By the previous section, if we managed to factorize, then whatever ends up in those blank spaces will be the solutions. But what would work in those blank spaces? Two numbers which have sum2and product24. The most commonly taught method is to find these numbers by guess and check. That can be frustrating, especially when there are negative numbers to try, and when the product has a lot of possible factorizations (24has a ton of possibilities).

As summarized in the related work, Savage's version has the similar calculations except that he seeks a factorization into the mathematically equivalent form(x+)(x+). Then, the numbers in the blanks are the negatives of the solutions, so after finding the factorization, Savage negates the numbers as the final step. From an educational perspective, I think that is a bit more advantageous to cleanly reduce the standard quadratic to a sum-and-product problem (with no need to return and remember to negate at the end), because one then gains the insight into the direct relationship between the coefficients and the sum and product of roots.

To make this even more natural for a first-time learner, I would advocate introducing the concept of factoring with an initial example that has a negativex-coefficient, so 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 negation is needed.

Insight: Factoring Without Guessing

Here's a way to pinpoint numbers that work without any guessing at all! The sum of two numbers is2when their average is1. So, we can try to look for numbers that are1plus some amount, and1minus the same amount. All we need to do is to find if there exists ausuch that1+uand1uwork as the two numbers, anduis allowed to be0.

By looking for two numbers of the form1+uand1u, they automatically sum to2. So, we just need them to multiply to24. We wish to find if there exists auwhich satisfies:(1+u)(1u)=24.We already saw a pattern like this, where we have a sum of two numbers, multiplied by their difference. The answer is always the difference of their squares! So, by rewriting the left hand side in equivalent form, we wish to find if there exists ausuch that1u2=24.This is exciting! There is a loneu2, and everything else is just a number! That means that we can finish searching for a validuby following our nose, instead of requiring any new methods. We want:u2=25,which we can get fromu=5.So, a choice foruexists! (We could alternatively have chosenu=5, but that would end up giving the same result.) Therefore, tracing the logic back upward, we know that15=4and1+5=6will definitely be two numbers which have sum2and product24. The fact that those numbers satisfy the sum and product relations means that the factorization exists, which also means that we have found the full set of solutions:x=6orx=4.

Note that in this approach, we only need the existence of one particular number whose square equals another particular number. In this example, it is obvious that5is a number whose square equals25. Once we have one such number, we can already follow through our logical steps, and we deduce a complete set of solutions to the original quadratic. In contrast, at the corresponding step of Completing the Square, we would need to have a full list of all numbers which square to25. It is clear that5and5should be in the list, but it is more difficult to answer why that is a complete list (especially when complex numbers are allowed as options). This detail is discussed in further depth here.

As I noted in my complete article, although I (like many others) independently came up with the trick of how to find two numbers given their sum and product, the Babylonians and Greeks already knew that particular trick thousands of years prior. However, mathematics had not been sufficiently developed for them to be able to use that trick on its own to solve general quadratic equations. 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 the related work page.

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

Now that guessing has been eliminated, we can actually solve any quadratic with this method. Consider this example:x222x+3=0.First, let's clean it up by multiplying both sides by2, to obtain an equation whose solution set is identical:x24x+6=0.Just like before, if we can find two numbers with sum4and product6, then the factorization(x)(x)will exist, and those two numbers will be the solutions. Halving the sum to get the average, we see that we'd be done if we can find someuso that numbers of the form2+uand2ugive a product of6. These two equations are equivalent to each other:4u2=6u2=2We can satisfy the bottom equation by choosingu=i2. Importantly, the mathematical invention of complex numbers allows us to take the square root of a negative number, so there is a valid choice foru. (This is also why we do not need the Fundamental Theorem of Algebra, and in fact, why this approach proves that theorem for degree-2 polynomials.) So, there are indeed two numbers with sum4and product6, and they are2+uand2u, which are2±i2. The fact that those numbers satisfy the sum and product relations means that the factorization(x)(x)exists, and so we have found the solutions:x=2±i2. We completed the problem, and we didn't need to use any memorized formula at all! This method works for every quadratic equation, without needing any memorization, and every step has a simple mathematical justification.

Proof of the Quadratic Formula

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

For a general quadratic equationx2+Bx+C=0, the above shows that it suffices to find two numbers with sumBand productC, at which point the factorization will exist and those will be the roots. So, we'd like to find if there exists auso that the two numbersB2+uandB2uwill work. They automatically sum toB. Their product isCprecisely when these two equivalent equations are satisfied:B24u2=Cu2=B24CSince the square root always exists (extending to complex numbers if necessary), by choosing a square root ofB24Cforu, we can satisfy the last equation. Therefore, the two numbersB2±B24Chave sumBand productC, and are all the solutions.

The above formula is already enough to solve any quadratic equation, because you can multiply or divide both sides by a number so that nothing is in front of thex2. However, just to see that this formula is the same as what everyone is used to memorizing (which is no longer necessary, in light of our method), we can show how to get the formula for the most general quadratic equationax2+bx+c=0whena0. We just need to divide byafirst, to get the equivalent equationx2+(ba)x+ca=0.Then, plugging inbaforBandcaforCin the solutions above, we get that the solutions are:b2a±b24a2ca=b2a±b24ac4a2=b±b24ac2a.

Summary

This method consists of two main steps, starting from a general quadratic equation in standard formx2+Bx+C=0.

  1. Because of polynomial factoring, if we can find two numbers with sumBand productC,then those are the complete set of solutions.
  2. Use the ancient Babylonian/Greek trick (extended to complex numbers) to find those two numbers in every circumstance.

In order for these steps to be mathematically sound as a complete method, it is essential that under all circumstances, Step 2 finds two numbers to use in Step 1, even if they are non-real complex numbers. It is therefore unlikely that mathematicians before Cardano (~1500 AD) could have done this completely.

Both steps are individually well-known. In retrospect, their combination to form a complete and coherent method for solving general quadratic equations is simple and obvious. Therefore, the main contribution of this method is to point out something useful that has been hiding in plain sight.

Historical Mathematical Manuscripts

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

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