The problem above, from the October 2019 SAT, illustrates an interesting and powerful principle that almost always eludes students: a quadratic expression is a structure where the base element does not have to be a simple variable lacking a coefficient or an exponent.

Students all know that *x*² + 4*x* – 12 = 0 is a quadratic equation, and they understand that the quadratic expression here can be factored as (*x* + 6)(*x* – 2), which gives the solutions *x* = -6 and *x* = 2. They also understand that a quadratic equation might not be provided in standard form; the previous equation might be provided in a different form, such as *x*² – 12 = -4*x*, but that can be rewritten as a standard form equation, and they know how to work with that, so that situation rarely presents any difficulty.

However, when faced with finding the zeros of an expression such as our introductory example *x* – 2√*x* – 3, or (3*t*)² – 5(3*t*)- 14, or 4*x*² + 4*x* – 3, or *x*⁴ + 4*x*² – 12. or (*x* – 5)² + 4(*x* – 5) + 4, most students (and many prep tutoring novices) will not recognize (or even know) that these are also quadratic equations that are susceptible to precisely the same solving methods as is *x*² + 4*x* – 12 = 0, so they will instead attempt to apply a variety of other algebraic operations in search of a solution. This can be successful, but it’s not optimal.

It comes as a revelation when students and less-experienced tutors are shown that a quadratic expression can be “in” something other than a simple variable such as x. Here are a few examples of quadratic expressions demonstrating this phenomenon:

*x*² + 4*x*– 12 is a quadratic in x, which can be factored as (*x*+ 6)(*x*– 2)*x*– 2√*x*– 3 is a quadratic in √*x*; it can be written as (√*x*)² – 2√*x*– 3, which can be factored as (√*x*+ 1)(√*x*– 3) (see October 2019-NC-18)- (3
*t*)² – 5(3*t*)- 14 is a quadratic in 3*t*, which can be factored as (3*t*+ 2)(3*t*– 7) (see March 2019-NC-19) - 4
*x*² + 4*x*– 3 is a quadratic in 2*x*, as it can be written as (2*x*)² + 2(2*x*) – 3, which can be factored as (2*x*+ 3)(2*x*– 1) (see May 2018-NC-12) *x*⁴ + 4*x*² – 12 is a quadratic in*x*², and it can be re-written as (*x*²)² + 4*x*² – 12, which can be factored as (*x*² + 6)(*x*² – 2)- (
*x*– 5)² + 4(*x*– 5) + 4 is a quadratic in*x*– 5, which can be factored as [(*x*– 5) + 2][(*x*– 5) + 2) = (*x*– 3)² (see Test 1-C-36, May 2017-NC-5)

Problems incorporating expressions with this structure don’t appear very often on the SAT. Still, I think understanding this phenomenon can stimulate a deeper understanding and appreciation of the math, and that can have wider benefits that extend beyond the quite useful time savings that can be harvested on these problems.

We call problem types such as this that have been intentionally constructed to include streamlined solutions “wormhole” problems, because there’s a portal available that takes you quickly from problem to solution without traveling the longer, more general solving route. There are about two dozen classes of wormhole questions that have appeared on the SAT over the years, represented by a few per test on average, of which this quadratic pattern is just one. It behooves tutors and students to familiarize themselves with these templates and, more importantly, to embrace increased flexibility in their analysis and solving approaches. The rewards are there for the taking.

**Daniel Kirchheimer** is the Director of Learning Systems Development for 1600.io, which provides educational support resources for tutored students and self-studiers to advance the company’s goal of helping young people become well-prepared for college and life.