How to Solve *Any* Quadratic Equation by Factoring
Yes, you read that correctly.
Table of Contents
§ Two dilemmas
It felt like a typical day of mathematics tutoring. It turned out to be anything but.
A student who I used to work with regularly came in with algebra homework. He was learning about factoring and the quadratic formula. When he asked how I knew when we could factor a quadratic equation \(ax^2 + bx + c = 0, a \neq 0\), there were two cases that were apparent to me:
- \[a=1\]
- \[a \neq 1\]
The first case was easiest. I taught him that, when \(a=1\) (i.e. \(x^2 + bx + c = 0\)), you want two numbers with
\[\left\{ \begin{array}{c} \text{their sum} = -b \\ \text{their product} = c \end{array} \right.\](The second case was not much harder; I told him to just divide the whole equation by \(a\) and he’ll end up in a situation similar to the first case.)
This allows us to factor the quadratic expression into the form:
\[(x - \underline{\hspace{1em}})(x - \underline{\hspace{1em}})\]However, there was something not immediately obvious to me (without using the guess-and-check method for factoring most of us learned in school): how do we mathematically determine which two numbers solve the system?
The student was also overwhelmed by the idea of having to remember the quadratic formula for the cases where the quadratic expression “cannot be factored cleanly”.
While I told him something placating that day, deep down, I didn’t really know what to tell him. I left work that day deeply dissatisfied with the guess-and-check method and the quadratic formula – never having questioned either before.
§ Enter Po-Shen Loh
After digging around on YouTube for alternative methods of solving quadratic equations, I found a video by Po-Shen Loh, Carnegie Mellon mathematics professor and former coach of the US International Mathematics Olympiad team.
It’s ~40 minutes long, so here I go over solving a general quadratic equation using the method he goes over in the video.
§ Solving the general case (and recovering something familiar along the way)
Let’s solve
\[ax^2 + bx + c = 0, a \neq 0\]First, we divide everything by \(a\) (so that we can use the sum-product method I described above). We get
\[x^2 + \frac{b}{a}x + \frac{c}{a} = 0\]If we can find two numbers with
\[\left\{ \begin{array}{c} \text{their sum} = -\frac{b}{a} \\ \text{their product} = \frac{c}{a} \end{array} \right.\]we can factor the left-hand side of the equation like so:
\[(x - \underline{\hspace{1em}})(x - \underline{\hspace{1em}}) = 0\]and the two numbers which fill in the blanks will be the solutions of the equation.
Now, the two key ideas in Po-Shen Loh’s video are:
- The sum of two numbers is twice their average.
- The two aforementioned numbers must be an equal distance \(u\) away from their average.
Two numbers which add up to \(-\frac{b}{a}\) are \(-\frac{b}{2a} + u\) and \(-\frac{b}{2a} - u\). So, we need a \(u\) such that
\[(-\frac{b}{2a} + u)(-\frac{b}{2a} - u) = \frac{c}{a}\]Now we solve for \(u\).1
\[\begin{align*} \frac{b^2}{4a^2} - u^2 &= \frac{c}{a} \\ \frac{b^2}{4a^2} - \frac{c}{a} &= u^2 \\ \frac{b^2}{4a^2} - \frac{4ac}{4a^2} &= u^2 \\ \frac{b^2-4ac}{4a^2} &= u^2 \\ u &= \sqrt{\frac{b^2-4ac}{4a^2}} \\ u &= \frac{\sqrt{b^2-4ac}}{2a} \end{align*}\]We can substitute our expression for \(u\) back into the expressions for the two numbers we found for the sum:
\[\begin{align*} -\frac{b}{2a} + u &= -\frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a} \\ &= \boxed{\frac{-b+\sqrt{b^2-4ac}}{2a}} \end{align*}\]and
\[\begin{align*} -\frac{b}{2a} - u &= -\frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a} \\ &= \boxed{\frac{-b-\sqrt{b^2-4ac}}{2a}} \end{align*}\]These are the two solutions of \(ax^2 + bx + c = 0, a \neq 0\). So,
\[\boxed{x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]We recovered our foe friend, the quadratic formula!
§ Back to tutoring
Sadly, the next time I saw the same student, he seemed ready to move on to the new topics he was learning in school that week.
I, however, remain amazed.
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Using the negative solution for \(u\) yields the same results. Try it yourself (or watch the video)! ↩
