Applying the quadratic formula on a constant

Question

This is inspired on a few joke videos I've seen. For instance, in one of them you're trying to solve

$$ 8x = 1$$

over the reals. Instead of dividing both sides by $8$, you can transform it into something like

$$ (x) 2^2 + (2x) 2 - 1 = 0$$

and then applying the quadratic formula on the $2$:

$$ 2 = \dfrac{-2x \pm \sqrt{4x^2-4(x)(-1)}}{2x}$$

$$ 2 = \dfrac{ (2x) \left(-1 \pm \sqrt{1+\dfrac{1}{x}} \right)}{2x}$$

Since $2 > 0$

$$ 2 = -1 + \sqrt{1+\dfrac{1}{x}} $$

Adding $1$ on both sides and then squaring

$$ 9 = 1 + \dfrac{1}{x}$$

Subtracting $1$ on both sides and multiplying by $x$

$$ 8x = 1$$

and then dividing both sides by $8$ (lol), we have

$$ x = \dfrac{1}{8}$$

Since its a joke, its clearly useless and improductive applying the quadratic formula on the $2$.

So my question is: is there a problem in which applying the quadratic formula on a constant is actually a "good looking" way of solving it?

Actually, it doesn't matter if its a constant. It would be cool enough if it is being applied in a very unusual and unexpected way to actually solve something.

Answer 1

Not too seriously, either, but suppose the problem is solving the equation in $\,x\,$ over the reals:

$$ x^3 + \pi x^2 + (3 \pi - 2) x + \pi^2 - 4 = 0 $$

This is a cubic equation which can be solved in radicals, but the calculations are tedious.

For a shortcut, instead, consider it as a quadratic equation in the constant $\,\pi\,$:

$$ \pi^2 + x (x + 3) \pi + x^3 - 2 x - 4 = 0 $$

The discriminant turns out to be:

$$ \Delta = x^2(x+3)^2 - 4(x^3 - 2x - 4)=x^4 + 2x^3 + 9x^2+ 8x + 16 = (x^2+x+4)^2 $$

Therefore, the roots in $\,\pi\,$ are:

$$ \pi_{1,2} = \frac{1}{2}\left(-x(x+3) \pm (x^2+x+4)\right) = \begin{cases}-x + 2\\-x^2 -2x -2\end{cases} $$

It follows that the LHS of the equation factors as:

$$ (\pi - \pi_1)(\pi - \pi_2) = (\pi + x - 2)(\pi + x^2 + 2x + 2) $$

The second factor is $\,(x+1)^2 + \pi + 1 \gt 0\,$, so the only real root is $\,x=-\pi+2\,$.

Answer 2

Everyone knows the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

and I think the vast majority of users on this site will immediately tell you it does not apply when $a = 0$, however, I would like to point out a simple extension of this formula that makes sense and seems to have gone unnoticed in the mainstream high school math curriculum. And in particular, this version does allow you to solve linear equations with the quadratic formula.

One can see by some algebra that the latter expression is equal to $$\frac{2c}{-b \pm \sqrt{b^2 - 4ac}}$$ for example by rationalizing the numerator or denominator, respectively. Obviously one does not make sense when $a = 0$, however, the other almost makes sense when $a=0$, in a way that is highly instructive, in my opinion, For if $a=0$, the latter formula reduces to $\frac{2c}{-b \pm |b|}$, and for one of these signs, is guaranteed to involve division by $0$, but for the other choice of sign, results in simply $-c/b$, and this latter is actually exactly what you expect, for in your example you have $0x^2 + 8x - 1 = 0$, and the answer is $1/8$, as expected.

So what gives? What's with this 'almost' functional second formula? Well, it is easiest to understand by using reasoning from calculus, which you may or may not know about, but will probably be exposed to soon. Let's consider, instead of an isolated equation, a family of equations:

$$ax^2 - x + 1 = 0$$

where we think of $a$ as a parameter that we're allowed to play with, but at any given time, it's a fixed number, like a value on a slider. We can check that the equation has two real roots using the discriminant, which will tell us $b^2 - 4ac = 1 - 4a$ must be positive, and therefore $a < 1/4$. But this includes our seemingly degenerate case when $a = 0$, so let's pay attention to what happens to our two real roots as $a$ passes through $0$.

Let's pretend $a$ is a small positive number. When this happens $\sqrt{1-4a}$ is very close to $1$ as well. Thus, the two terms in the latter equation are $2/(1 + \text{very close to } 1)$ a number quite close to $1$, and the other is $2/(1 - \text{very close to } 1)$, a number with denominator very close to $0$, i.e. a very big number. When $a$ is made even smaller, the denominator even bigger, and the root even larger. All this is happening while the first root is staying at $1$. so what's happening is the parabola's getting more and more wide, corresponding to the degeneration of the parabola to a line. It's always passing very close to the point $(1,0)$, and the other root is getting further and further away. So where is it when $a = 0$? It's at infinity.

In the event that there are users reading this thread who know a thing or two about commutative algebra or algebraic geometry, I invite them to think about what this is saying in the projective plane. For a rough sketch, homogenize the quadratic with a new variable, maybe let's call it $z$, and de-homogenize with respect to $x$. This swaps the role of $a,c$, and relates the two formulas again, as the local coordinate on the first chart of $\mathbb{P}^2$ has coordinate $x/z$ when $z \neq 0$, while the latter has coordinate $z/x$. From the viewpoint of the coordinate at infinity, the degeneration doesn't have a root running off to infinity, it has a root of $0$ as the constant term disappears.