Roots and Rationalize
It is a truth universally acknowledged, that a quadratic equation in possession of good coefficients, must be in want of roots.
Such equation, nowadays standardized as
Unfortunately, for many people, the formula often misremembered. Forget to negate the first term, divide by only a constant, swap terms inside the square root, the list goes on and on as wild as mankind’s imagination.
It also takes us out of the realm of beauty, and just doing everything in a robotic manner. Sadden the matter even more.
So what is it, really?
Well, one way to interpret the equation is to lift the restriction that it must equal to zero. Turn the empty side of the equation into another variable, a function if you may, i.e.
It follows that
A parabola from the interpretation of
That is nice and all. But it does not simplify the problem. Since we have to somehow find the roots first, before we can actually draw the curve precisely.
Thus we may look at the function
Let
These functions are more manageable than
is a parabola with the axis of symmetry as -axis, and the focus at ; is a line of slope that intercept -axis at .
A decomposed pair of parabola-line
Family of functions
The choice of choosing a pair of
Realized that
Where
So now it boils down to finding
However, actually dealing with all of
Finding
In other words, the vertex’s position can be written as a function
Which is a congruent parabola to all of
A family of such decomposition, note that the pivot is not a focus of the inverted parabola
Let us take a short break to appreciate the beauty of seamlessly transformation of the problem. And thus ends our sightseeing here.
Geometric construction
Going back to
And we have the trick just for that!
Parabola construction with pin and string method – picture from Wikipedia
However, those tools are not the standard Euclidean construction. Which we are allowed to use only straightedge and compass (SE&C). Actually drawing a parabola with just SE&C would require infinitely many steps.
An envelope of a parabola constructed using origami, which is an adaptation of SE&C
Can we avoid the infinity? Consider the horrible formula
- A base of length
- A hypotenuse of length
- The remaining side, a height, of length
Roots via a pair of right triangles, note that triangle’s apex is not on the parabola
This specific triangle is useless. But again, there are infinitely many triangles in the same manner. That is if we fix the base but vary the height to a variable
Let
A circle touching the focus of the parabola
One point of our interest occur when the circle touch the focus of the parabola
Such an elegant ratio! Thus suggested us a simple construction that bypass parabola drawing entirely. That is we find the point
Using a compass, jot the needle point at
Geometric construction for solving quadratic
This method works best in the purely geometrical setting. That is
Lill’s construction
What if we do not care about the extreme geometric interpretation and just focus (no pun intended) solely on the algebraic aspect, dealing with the coefficients
Nevertheless, let us look closely at the geometric construction again. Consider only a root
What is the last point should we find? If we want to utilize Thales’s theorem, then we have to make a right triangle with one side as a diameter to the circle. The best point is a reflection of parabola’s focus around
Important lengths for the geometric construction
Let us draw three orthogonal line segments that join these important points together. To eliminate ambiguity, we will call them rods, which consists of
- A rod of length
that join the origin to the focus of the parabola. - A rod of length
that join the origin to the -projection of the reflexed focus. - A rod of length
that join the -projection of the reflexed focus to the reflexed focus.
Fix those rods. What if we consider other right triangles such that one vertex of the triangle stayed at the focus. While the opposite vertex falls on the third rod, missing the reflexed focus. It is natural to ask how far it keep missing.
A “missing” distance when we choose a “wrong” answer
Let
In other words,
We can even make the construction simpler. Firstly, notice that the first rod is
It is sensible to choose
Lastly, fixing the last rod and also the function
Lill’s method for solving quadratic
Choose
This interpretation gives us a lot of insight. Such as if the circle is not large enough to cross the (extension of the) middle rod, then the equation must have no solution (in real realm). And when we face a negative value
The ancient Greek, evidently Eratosthenes (prime-sieve guy), known this method for quite a long time, despite limited to polynomial of degree three. It was not until 1867 that Eduard Lill finally generalized the concept to polynomial of any degree.
Arguably, these methods are still in the theme of memorization, like the formula
So why are we still teaching ordinary students with the overly complicated formula
References
- Hull, Thomas C. Origametry: Mathematical Methods in Paper Folding. Cambridge University Press, 2020.
- Mathologer: Why don’t they teach this simple visual solution? (Lill’s method)
P.S. During the SoME2 event, I ran into this excellent video in Chinese. Which also dealing with parabola in the setting of classical physics. Bet you’ll also love it ❤️
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