# Quadratic equations

I’m currently working on a post about continued fractions but it’s ended up taking a lot of time to complete, so while I work on that, here’s a post about a neat way to solve quadratic equations which you might not have seen before.

A quadratic equation is an equation of the form

$a x^2 + b x + c = 0,$

where $a$, $b$ and $c$ are real numbers (called the coefficients), $a \ne 0$ and $x$ is a variable. The solution is given by the well-known quadratic formula,

$x = \frac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}.$

The usual method of deriving this formula is a process called “completing the square”, which takes advantage of the identity

$a x^2 + b x = a \left( x^2 + \frac b {2 a} \right)^2 - \frac {b^2} {4 a^2},$

allowing the solution to be found by rearrangement. However, there is an alternative method which is interesting because it can be generalised to solve polynomial equations of greater degree. This method takes advantage of the fact that for every pair of real numbers $\alpha$ and $\beta$,

$a (x - \alpha) (x - \beta) = a x^2 + b x + c$

if and only if $\alpha$ and $\beta$ are the solutions. The left-hand side expands out to $a x^2 - a (\alpha + \beta) x + a \alpha \beta$, so if this is the case, it must be the case that $\alpha + \beta = -\frac b a$ (so that $b = -a (\alpha + \beta)$) and $\alpha \beta = \frac c a$ (so that $a \alpha \beta = c$). So we know the sum and the product of the solutions, but not their actual values.

Now, since we know the sum of the solutions, if we can also find the difference of the solutions, we will be able to find each solution, because $(\alpha + \beta) + (\alpha - \beta) = 2 \alpha$ and $(\alpha + \beta) - (\alpha - \beta) = 2 \beta$. And we can find the difference, since we know the product of the solutions, by looking at the squares of the sum and difference. We have $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2 \alpha \beta$ and $(\alpha - \beta)^2 = \alpha^2 + \beta^2 - 2 \alpha \beta$. Therefore

$\begin{array}{rcl} (\alpha - \beta)^2 &=& (\alpha + \beta)^2 + 4 \alpha \beta \\ &=& \left( -\frac b a \right)^2 + 4 \frac c a \\ &=& \frac {b^2} {a^2} + \frac {4 c} a \\ &=& \frac {b^2 - 4 a c} {a^2} \\ \alpha - \beta &=& \pm \sqrt{\frac {b^2 - 4 a c} {a^2}} \\ &=& \pm \frac {\sqrt{b^2 - 4 a c}} a. \end{array}$

Therefore

$\begin{array}{rcl} 2 \alpha &=& (\alpha + \beta) + (\alpha - \beta) \\ &=& -\frac b a \pm \frac {\sqrt{b^2 - 4 a c}} a \\ &=& \frac {-b \pm \sqrt{b^2 - 4 a c}} a \\ \alpha &=& \frac {-b \pm + \sqrt{b^2 - 4 a c}} {2 a} \end{array}$

$\begin{array}{rcl} 2 \beta &=& (\alpha + \beta) - (\alpha - \beta) \\ &=& -\frac b a \mp \frac {\sqrt{b^2 - 4 a c}} a \\ &=& \frac {-b \mp \sqrt{b^2 - 4 a c}} a \\ \beta &=& \frac {-b \mp + \sqrt{b^2 - 4 a c}} {2 a} \end{array}$

and we have the solutions.

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### One response to “Quadratic equations”

1. At one point when I was younger, I boasted of knowing four proofs of the quadratic formula (not counting the method of starting with the formula and plugging it into the original equation). Of course, I probably wouldn’t consider these arguments to be as independent from each other as I thought at the time. Actually, the first proof that I knew of was not the usual one, but the one you gave in this post; when I was a kid and my dad introduced me to the quadratic formula, I asked him to prove it, and he came up with this almost on the spot. Next I learned the standard completing-the-square argument from Algebra The Easy Way. Next, inspired by the method of reduction of the cubic equation, I came up with a third proof which more or less involves making a change of variables x’ = x + b/2a and thus effectively getting rid of the x^1 term. Now I see that this is basically just a rephrasing of the completing-the-square argument, but it generalizes nicely to a translation of x for any nth-degree polynomial which gets rid of the x^{n – 1} term. At some point later, I came up with a fourth proof which was more similar in flavor to the first; I don’t remember it anymore, but probably it wasn’t very independent. I think I had the idea in my mind that there should be dozens of proofs of the quadratic formula as for Pythagorus’ Theorem. But now I figure the art of solving quadratic equations isn’t quite “deep” enough for more than a couple of sort-of-independent approaches.