## A simple proof of the quadratic formula

If you’re reading this blog you have probably memorized (or used to have memorized) the quadratic formula, which can be used to solve quadratic equations of the form

$ax^2 + bx + c = 0.$

But do you know how to derive the formula? Usually the derivation is presented via completing the square and it involves some somewhat messy algebra (not to mention the idea of “completing the square” itself).

My colleague Gabe Ferrer recently brought to my attention a remarkable new paper by Po-Shen Loh, A Simple Proof of the Quadratic Formula. This paper is remarkable for several reasons: first of all, it’s remarkable that anyone could discover anything new about the quadratic formula; it’s also remarkable for a research mathematician to publish something about elementary mathematics. (But Po-Shen Loh is not your average research mathematician either; he does lots of really cool work making mathematics more accessible for all kinds of learners.) I’m going to explain the basic idea but I highly recommend actually reading the paper, which not only explains the ideas but also does a great job putting everything in proper historical context. Loh has also made a whole web page dedicated to explaining the ideas, with a video, worked examples, etc.; it’s definitely worth taking a look!

# The Setup

Suppose we have a quadratic equation we want to solve,

$x^2 + bx + c = 0.$

To make things simpler, we’ll assume that $x^2$ has a coefficient of $1$. (If we have a quadratic equation with some other coefficient $ax^2$, we can always divide everything by $a$ first.)

Now imagine we knew how to factor the quadratic. Then we could rewrite the equation into the form

$(x - r)(x - s) = 0$

which would imply that $x = r$ and $x = s$ are the two solutions. If we multiply out the above factorization (using, you know, “FOIL”), we get

$x^2 - (r+s)x + rs = 0$

which means we’re looking for values $r$ and $s$ whose product is $c$ and whose sum is $-b$.

So far, so good; everyone learns this much in high school algebra. The way one usually goes about factoring quadratic polynomials is to make informed guesses for values of $r$ and $s$ and check whether their sum and product give the right coefficients.

# The Insight

The key insight at this point, however, is that we don’t actually have to guess! Starting from $r + s = -b$, let’s divide both sides by $2$:

$\displaystyle \frac{r+s}{2} = -\frac{b}{2}$

The left-hand side is the average of $r$ and $s$, which lies halfway in between them on the number line. Let’s use $z$ to denote the distance from $r$ to $-b/2$. Since $-b/2$ is halfway in between $r$ and $s$, $z$ must also be the distance from $-b/2$ to $s$. So we can write $r$ and $s$ in the form

$r,s = -b/2 \pm z$

Now, we know their product has to be $c$, and multiplying them is particularly easy because we get a difference of squares:

$\displaystyle c = rs = \left(-\frac{b}{2} + z \right) \left(-\frac{b}{2} - z \right) = \left(-\frac{b}{2} \right)^2 - z^2$

Now solving for $z$ is easy; just move $z^2$ to one side of the equation by itself and take the square root:

$\displaystyle z = \pm \sqrt{\frac{b^2}{4} - c}$

That means the solutions are

$\displaystyle r,s = -\frac{b}{2} \pm z = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - c}.$

If you like, you can use the same method starting from $ax^2 + bx + c = 0$ to derive the usual quadratic formula including an arbitrary value of $a$, although the required algebra gets a bit messier.

# Using it in practice

One particularly nice thing about this derivation is that it corresponds to a simple algorithm for solving an arbitrary quadratic equation $x^2 + bx + c = 0$, so there’s no need to memorize a formula at all:

1. Note that the two solutions must add up to $-b$, so their average is half of $-b$, and hence they can be written as $-b/2 \pm z$.
2. Write down the equation $(-b/2 + z)(-b/2 - z) = b^2/4 - z^2 = c$, and solve for $z$.
3. The solutions are $-b/2 + z$ and $-b/2 - z$.

Of course if you need to solve something of the form $ax^2 + bx + c = 0$, you can add an extra step to divide through by $a$ first.

And that’s it! I really hope this new method will make its way into classrooms around the world; Loh makes the argument (and I agree) that it really is much easier for early algebra learners to grasp. And again, I really encourage you to go look at Loh’s web page to read more, especially about the historical context: at what point in human history could someone have come up with this idea? And why didn’t they? (Or if they did, why did we forget?) All this and more are in the original paper, which is a really fascinating and accessible read.

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 3 Responses to A simple proof of the quadratic formula

1. Fernando says:

Roger Cooke argues, in his history of mathematics textbook, that the Old Babylonian method for solving quadratics was based on understanding the relationship between the average $(a+b)/2$ and the semidifference $(a-b)/2$. If he is right, this method is over 3,000 years old. In any case it is certainly not new.

• Brent says:

Thanks for the reference! Indeed, Po-Shen Loh goes into much more detail on the history, and is careful to claim that it is almost certainly not new in the sense of no one ever having thought of it before, but only in the sense that it seems to have been all but forgotten, and in modern times no one presents the derivation in this way.