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

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 find$r$and$s$with sum$-B$and product$C$, then${x}^{2}+Bx+C=\left(x-r\right)\left(x-s\right)$, and they are all the roots
2. Two numbers sum to$-B$when they are$-\frac{B}{2}±u$
3. Their product is$C$when$\frac{{B}^{2}}{4}-{u}^{2}=C$
4. Square root always gives valid$u$
5. Thus$-\frac{B}{2}±u$work as$r$and$s$, 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):$\begin{array}{rl}& \phantom{\rule{1em}{0ex}}\left(x-3\right)\left(x-4\right)\\ & ={x}^{2}-4x-3x+12\\ & ={x}^{2}-7x+12.\end{array}$

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

Here's another example:$\begin{array}{rl}& \left(u-3\right)\left(u+3\right)\\ & ={u}^{2}+3u-3u-9\\ & ={u}^{2}-9.\end{array}$

Since we had both$-3$and$+3$, the$+3u$and$-3u$terms 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 all$x$such that${x}^{2}-7x+12=0.$We already know that this is the same (has exactly the same solutions) as$\left(x-3\right)\left(x-4\right)=0.$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 that$ab=0$with both$a$and$b$nonzero. Then by dividing both sides of the equation$ab=0$by$a$, we get$b=0$, contradiction.)

So, the$x$that work are precisely those where$x-3=0$(which is$x=3$), or$x-4=0$(which is$x=4$). Note that the solutions are the numbers we subtract from$x$, i.e., not$-3$and$-4$, but$3$and$4$. Importantly, these are all the solutions.

### Review: Setting Up for Factoring

Let's try the reverse process for the example${x}^{2}-2x-24=0.$It would be great if we could factorize it into something like$\left(x-\phantom{\rule{1em}{0ex}}\right)\left(x-\phantom{\rule{1em}{0ex}}\right).$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 sum$2$and product$-24$. 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 ($24$has 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$\left(x+\phantom{\rule{1em}{0ex}}\right)\left(x+\phantom{\rule{1em}{0ex}}\right)$. 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 negative$x$-coefficient, so that the factorization$\left(x-\phantom{\rule{1em}{0ex}}\right)\left(x-\phantom{\rule{1em}{0ex}}\right)$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 is$2$when their average is$1$. So, we can try to look for numbers that are$1$plus some amount, and$1$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 numbers, and$u$is allowed to be$0$.

By looking for two numbers of the form$1+u$and$1-u$, they automatically sum to$2$. So, we just need them to multiply to$-24$. We wish to find if there exists a$u$which satisfies:$\left(1+u\right)\left(1-u\right)=-24.$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 a$u$such that$1-{u}^{2}=-24.$This is exciting! There is a lone${u}^{2}$, and everything else is just a number! That means that we can finish searching for a valid$u$by following our nose, instead of requiring any new methods. We want:${u}^{2}=25,$which we can get from$u=5.$So, a choice for$u$exists! (We could alternatively have chosen$u=-5$, but that would end up giving the same result.) Therefore, tracing the logic back upward, we know that$1-5=-4$and$1+5=6$will definitely be two numbers which have sum$2$and product$-24$. 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:$x=6$or$x=-4$.

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 that$5$is a number whose square equals$25$. 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 to$25$. It is clear that$5$and$-5$should 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:$\frac{{x}^{2}}{2}-2x+3=0.$First, let's clean it up by multiplying both sides by$2$, to obtain an equation whose solution set is identical:${x}^{2}-4x+6=0.$Just like before, if we can find two numbers with sum$4$and product$6$, then the factorization$\left(x-\phantom{\rule{1em}{0ex}}\right)\left(x-\phantom{\rule{1em}{0ex}}\right)$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 some$u$so that numbers of the form$2+u$and$2-u$give a product of$6$. These two equations are equivalent to each other:$\begin{array}{rl}4-{u}^{2}& =6\\ {u}^{2}& =-2\end{array}$We can satisfy the bottom equation by choosing$u=i\sqrt{2}$. Importantly, the mathematical invention of complex numbers allows us to take the square root of a negative number, so there is a valid choice for$u$. (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 sum$4$and product$6$, and they are$2+u$and$2-u$, which are$2±i\sqrt{2}$. The fact that those numbers satisfy the sum and product relations means that the factorization$\left(x-\phantom{\rule{1em}{0ex}}\right)\left(x-\phantom{\rule{1em}{0ex}}\right)$exists, and so we have found the solutions:$x=2±i\sqrt{2}$. 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 equation${x}^{2}+Bx+C=0$, the above shows that it suffices to find two numbers with sum$-B$and product$C$, at which point the factorization will exist and those will be the roots. So, we'd like to find if there exists a$u$so that the two numbers$-\frac{B}{2}+u$and$-\frac{B}{2}-u$will work. They automatically sum to$-B$. Their product is$C$precisely when these two equivalent equations are satisfied:$\begin{array}{rl}\frac{{B}^{2}}{4}-{u}^{2}& =C\\ {u}^{2}& =\frac{{B}^{2}}{4}-C\end{array}$Since the square root always exists (extending to complex numbers if necessary), by choosing a square root of$\frac{{B}^{2}}{4}-C$for$u$, we can satisfy the last equation. Therefore, the two numbers$-\frac{B}{2}±\sqrt{\frac{{B}^{2}}{4}-C}$have sum$-B$and product$C$, 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 the${x}^{2}$. 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 equation$a{x}^{2}+bx+c=0$when$a\ne 0$. We just need to divide by$a$first, to get the equivalent equation${x}^{2}+\left(\frac{b}{a}\right)x+\frac{c}{a}=0.$Then, plugging in$\frac{b}{a}$for$B$and$\frac{c}{a}$for$C$in the solutions above, we get that the solutions are:$\begin{array}{rl}& -\frac{b}{2a}±\sqrt{\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}}\\ & =-\frac{b}{2a}±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\\ & =\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.\end{array}$

## Summary

This method consists of two main steps, starting from a general quadratic equation in standard form${x}^{2}+Bx+C=0$.

1. Because of polynomial factoring, if we can find two numbers with sum$-B$and product$C,$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.