That new quadratic equation solving technique

A New York Times article describes a new technique for solving quadratic equations which emerged from an article by Poh-Shen Loh. At first I was skeptical, but I decided to work through it enough to get a real feel for it. Though this technique does appear to be old information in a new guise, it does have some practical advantages over some of the traditional techniques (e.g. factoring, completing the square, the Quadratic Formula). I plan to use some of my students as guinea pigs…

By the way, kudos to reporters Kenneth Chang and Jonathan Corum for getting real math into the New York Times.


Loh’s technique emerges from a few well-known facts. First, any quadratic equation can be written as \[ a x^2 + bx + c = 0 \] In this article, we will assume that the equation is monic (i.e. $a=1$) so the equation simplifies to \[ x^2 + bx + c = 0 \] (If the equation is not monic, just divide both sides by $a$ and adjust the coefficients.)

Second, in that equation, $-b$ is the sum of the roots and $c$ is the product of the roots, since if the roots are $r$ and $s$, the factored form will be \[ (x-r)(x-s) = 0 \] Multiply the left side out and you’ll see that \[ x^2 – (r+s)x + r s = 0\] so $-b=r+s$ and $c=r s$. Notice also that $- \frac{b}{2}$ is the average of the roots, which is also the $x$-coordinate of the line of symmetry.

Loh’s method

The new method focuses on $u$, which is the distance between each root and the average of the roots. Thus the roots are $-\frac{b}{2} \pm u$.

It’s easy to find $u$ by realizing that $c$ is the product of the roots. What’s cool here is that the product of the roots is a factored form of a difference of squares so it multiplies cleanly, leaving no middle term:

c & = & \left( – \frac{b}{2} – u \right) \left( – \frac{b}{2} + u \right)\newline
c & = & \frac{b^2}{4} – u^2\newline
u^2 & = & \frac{b^2}{4} – c\newline
u & = & \sqrt{\frac{b^2}{4} – c}

Thus the roots are

x & = & – \frac{b}{2} \pm u\newline
& = & – \frac{b}{2} \pm \sqrt{\frac{b^2}{4} – c}

This looks suspiciously like the Quadratic Formula…


Let’s solve $x^2-4x-12=0$ using this method.

First, $-\frac{b}{2} = -\frac{-4}{2} = 2$ is the average of the roots.

Now $c=-12$ is the product of the roots so we solve for $u$:

-12 & = & ( 2 – u ) ( 2 + u)\newline
-12 & = & 2^2 – u^2\newline
u^2 & = & 2^2 + 12\newline
u & = & \sqrt{2^2 + 12}= 4\newline

So the roots are $2-4 = -2$ and $2+4=6$

Relationship to the Quadratic Formula

The Quadratic Formula $x = \frac{- b \pm \sqrt{b^2 – 4 ac}}{2 a}$ gives the roots of a standard form quadratic equation of the form $a x^2+b x+c = 0$.

We are working with a monic quadratic (in which $a=1$) so that simplifies to

x & = & \frac{- b \pm \sqrt{b^2 – 4 c}}{2}\newline
& = & \frac{- b}{2} \pm \frac{\sqrt{b^2 – 4 c}}{2}\newline
& = & – \frac{b}{2} \pm \sqrt{\frac{b^2-4c}{4} }\newline
& = & – \frac{b}{2} \pm \sqrt{\frac{b^2}{4} – c}\newline
& = & – \frac{b}{2} \pm u

You’ll notice that this is exactly what we get in the end of the “Loh’s Method” section.

Is this new or useful?

It’s clear that Loh’s method is equivalent to the monic version of the Quadratic Formula. What I dislike about the Quadratic Formula, though, is that most students just memorize it without understanding, and even those of us who do understand its derivation still use it by rote. Loh’s method is simple enough to derive each time, summarized below, and the calculations seem a bit simpler:

  1. If necessary, make the equation monic by dividing by $a$
  2. Realize that $-\frac{b}{2}$ is the average of the roots
  3. Let $u$ represent the distance between $-\frac{b}{2}$ and each root, so the roots are $-\frac{b}{2} \pm u$
  4. Realize that $c$ is the product of the roots, i.e. $c = ( – \frac{b}{2} – u ) ( – \frac{b}{2} + u )$
  5. Solve that equation for $u$, giving the roots as $-\frac{b}{2} \pm u$

Furthermore, like the Quadratic Formula and completing the square, Loh’s method can solve any monic quadratic equation. Unlike traditional factoring, it requires no guess-and-check (what multiplies to $c$ and adds to $-b$?) It works either in the real or complex fields. In the real case, of course, some equations cannot be solved, e.g. $x^2+1=0$.