Here, our goal is to reformulate the Quadratic Formula by appealing to symmetry. We'll start by noting that Quadratic Functions are commonly expressed in three different ways:

1. Standard Form: \(q(x) = ax^{2} + bx + c \)
2. Factored Form: \(q(x) = a(x - r_{1})(x - r_{2}) \)
3. Vertex Form: \( q(x) = a(x - h)^{2} + k \)

We impose conditions on these constants. In particular, \( a, b, c, h, k \in \mathbb{R} \) with \(a \ne 0 \) (to prevent degeneracy) and roots \( r_{1}, r_{2} \in \mathbb{C} \). The existence of these roots is guranteed by The Fundamental Theorem of Algebra.

The high-level plan of attack is to find the vertex in terms of the coefficients. From there, we'll figure out how much to move to the right and left in order to reach the roots.

Since we're interested in the finding the vertex, we'll start by taking a look at the Vertex Form. Upon inspection, it is clear that quadratic functions are symmetric with respect to \( x = h \). We can use this symmetry to our advantage. In particular, we define \( q_{p} = ax^{2} + bx \) (shown in dashes) whose vertex notably shares the same \( x \)-coordinate. We directly compute the roots as \( q_{p} = x (ax + b) = 0 \) which implies \( x = 0, -\frac{b}{a} \) in general. From symmetry, we reason that the vertex lies between these two roots at \( x = -\frac{b}{2a} \) in both cases. Plugging this value, we find \( q \left ( -\frac{b}{2a} \right ) = c - \frac{b^{2}}{4a} \) giving us the \( x \) and \( y \) coordinates of the vertex.

At this point, our task is to find \( \Delta \) such that \( q \left ( -\frac{b}{2a} \pm \Delta \right ) = 0 \). Since this seems a bit cumbersome to compute directly, we'll take another look at the Vertex Form. Upon further inspection, we arrive at the following expansion: \( q(x) = ax^{2} - 2ahx + ah^{2} + k \). Comparing terms from Standard Form, we notice that \( b = -2ah \) and \( c = ah^{2} + k \) allowing us to arbitrarily tune both coefficients by changing the position of the vertex. In other words, \( b \) and \( c \) only influence the position of the vertex and not the overall shape of the graph! We now reason that moving \( \Delta \) units to the left or right of the vertex has the effect of moving up or down \( a \Delta^{2} \) units!

Adding back the \( x \)-coordinate of the vertex and simplifying everything, we arrive at the quadratic formula.

This reformulation not only simplifies the algebra in favour of geometric insights, but it also allows us to attach an elegant interpretation to each component in this formula.

Numberphile is an educational YouTube channel featuring videos that explore topics from a variety of fields of mathematics. Several years ago, one particular video (that has since been remastered) caught my attention. Here, professor Johnny Ball discusses the Mesolabe Compass and its connection to square roots. I'll save you the trouble of watching the entire video and include the main figure for reference.

The question, of course, is why the \( \sqrt{a} \) pops up here. As it so happens, this isn't too difficult to explain. From Thale's Theorem, we know \( \measuredangle ABC \) is a right angle. From here we determine that the two triangles created by dropping the altitude are similar to each other since they share the same angles. Letting \( h \) stand for the altitude, we establish the following relation:

I suppose the second question that comes to mind is why Thale's Theorem holds. One approach that immediately surfaces is to apply the more general Inscribed Angle Theorem to a chord that happens to also be a diameter. Another approach, that I personally find interesting, is to consider the set of points such that the lines connecting \( (-r, 0) \) to \( (x, y) \) and \( (x, y) \) to \( (r, 0) \) are perpendicular to each other. Using point-slope form and enforcing that the slopes are perpendicular (negative reciprocals), we arrive at the following equations.

Solving for \( m \), we rewrite this system.

Amazingly, it turns out that the only set of points that ensure the two lines are perpendicular are those belonging to a circle!

Ron Larson and Bruce Edwards are the authors of an introductory Calculus text. Problem 6 under 1.1 Exercises asks the following:

Secant Lines. Consider the function \( f(x) = \sqrt{x} \) and the point \( (4, 2) \) on the graph of \( f \).

1. Graph \( f \) and the secant lines passing through \( (4, 2) \) and \( (x, f(x)) \) for x-values of 1, 3, and 5.
2. Find the slope of each secant line.
3. Use the results of part 2 to estimate the slope of the tangent line to the graph of \( f \) at \( (4, 2) \). Describe how to improve your approximation of the slope.

The problem itself is not difficult. Here, we will draw our attention to the very last question. When I was assigned this problem for a PSET, I wondered if there was a way to compute the exact tangent line without the machinery of derivatives. As it turns out, you can!

To find the tangent line, we will consider the properties of the tangent line and translate those properties into constraints. The first property is that the line should pass through \( (p, \sqrt{p}) \) for some choice of \( p \). The second property is that the tangent line should only intersect the curve at \( (p, \sqrt{p} ) \) and nowhere else. This relies on the fact that the root function is strictly monotonic.

The first constraint is easy enough to enforce via point-slope form.

To consider the second constraint, we will set the tangent line equal to the curve \( y = \sqrt{x} \) and work out the appropriate choice of \( m \).

We need to choose \( m \) such that the equation above is only satisfied for \( x = p \). Notice that this equation is quadratic in \( \sqrt{x} \). If there is only one solution to this equation, then the discriminant must be set to \( 0 \).

Voila, we have computed the slope of the tangent line for any choice of \( p \). Unfortunately, the method is rather specific to this family of functions with mild transformations.

In 7th grade, my mathematics teacher introduced a rather intriguing result. To my dismay, she declared that \( 0.\bar{9} = 1 \) and presented a variant of William Byer's algebraic argument for justification. The paradox would appear time and time again from the decay of radioactive isotopes to the (ideal) mechanics of a bouncing ball to any number of jokes about mathematicians entering a bar. All of these examples have one thing in common: they follow a geometric series. As it turns out, there is an explicit formula to compute the sum of a geometric series.