# How I Impressed My Wife: Part 6c

This series was inspired by a question that my wife asked me: calculate

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$,

where $C_R$ is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I handled the case of $|b| = 1$ in yesterday’s post. Today, I’ll begin the case of $|b| > 1$.

To find the poles of the integrand, I use the quadratic formula to set the denominator equal to zero:

$z^4 + (4 b^2 - 2) z^2 + 1 = 0$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{(4b^2-2)^2 - 4}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2 + 4 - 4}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}$

$z^2 = 1-2b^2 \pm 2|b| \sqrt{b^2 - 1}$

As shown earlier in this series, the right-hand side is negative if $|b| > 1$. So, for the sake of simplicity, I’ll define

$r_1 = \sqrt{2b^2 - 1 + 2|b| \sqrt{b^2 - 1}}$,

$r_2 = \sqrt{2b^2 - 1 - 2|b| \sqrt{b^2 - 1}}$,

so that the four poles of the integrand are $ir_1$, $ir_2$, $-ir_1$, and $-ir_2$. Of these, only two ($ir_1$ and $ir_2$) lie within the contour for sufficiently large $R$, and so I’ll need to compute the residues for these two poles.

Before starting that task, I notice that

$z^4 + (4 b^2 - 2) z^2 + 1 = (z - ir_1)(z + ir_1)(z - ir_2)(z + ir_2)$,

or

$z^4 + (4b^2 - 2)z^2 + 1 = (z^2 + r_1^2)(z^2 + r_2^2)$,

or

$z^4 + (4b^2 -2)z^2 + 1 = z^4 + (r_1^2 + r_2^2) z^2 + r_1^2 r_2^2$.

Matching coefficients, I see that

$r_1^2 + r_2^2 = 4b^2 - 2$,

$r_1^2 r_2^2 = 1$.

These will become very handy later in the calculation.

The integrand has the form $\displaystyle g(z)/h(z)$, and each pole has order one. As shown earlier in this series, the residue at such pole is equal to

$\displaystyle \frac{g(r)}{h'(r)}$.

In this case, $g(z) = 2(1+z^2)$ and $h(z) = z^4 + (4b^2-2)z^2 + 1$ so that $h'(z) = 4z^3 + 2(4b^2-2)z$, and so the residue at $r_1$ and $r_2$ are given by

$\displaystyle \frac{2(1+[ir_1]^2)}{4 [ir_1]^3 + 2(4b^2-2) [ir_1]} = \displaystyle \frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1}$

and

$\displaystyle \frac{2(1+[ir_2]^2)}{4 [ir_2]^3 + 2(4b^2-2) [ir_2]} = \displaystyle \frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2}$

Finally, to evaluate the contour integral, I simply the sum of the residues within the contour and then multiply the sum by $2\pi i$:

$Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} = 2\pi i \left[\frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1} +\frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2} \right]$

$= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]$

So, to complete the evaluation of $Q$, I’m left with the small matter of simplifying the right-hand side. I’ll tackle this in tomorrow’s post.

# How I Impressed My Wife: Part 5e

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$, I’ll now give a fourth method.
Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

To evaluate this integral, I need to find the four complex roots of the denominator:

$u^4 + (4b^2 - 2) u^2 + 1 = 0$

$u^2 = \displaystyle \frac{ 2 - 4b^2 \pm \sqrt{(4b^2 - 2)^2 - 4}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2 - 4 + 4}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}$

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

To solve for $u$, there are three separate cases that have to be considered: $|b| = 1$, $|b| > 1$, and $|b| < 1$. I’ll begin with the easiest case of $|b| = 1$. In this case, the integral $Q$ is easy to evaluate:

$= Q \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 \cdot [1^2] - 2) u^2 + 1}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + 2 u^2 + 1}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{(1+u^2)^2}$

$=\displaystyle \int_{-\infty}^{\infty} \frac{ 2 du}{1+u^2}$

$=\displaystyle \left[ 2\tan^{-1} x \right]^{\infty}_{-\infty}$

$=\displaystyle \left[ 2 \frac{\pi}{2} - 2 \frac{-\pi}{2} \right]$

$= 2\pi$

This matches the expected answer of $Q = \displaystyle \frac{2\pi}{|b|}$ since I used the assumption that $|b| = 1$.

I’ll continue with this fourth evaluation of the integral, examining the two remaining cases, in future posts.

# How I Impressed My Wife: Part 4c

Previously in this series, I have used two different techniques to show that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
Here’s my progress so far:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}$,

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. Also, $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ (and $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks more complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. For this integral, that means finding the values of $z$ where the denominator is equal to 0, and then determining which of those values lie inside of the closed contour. In this case, that means finding which root(s) of the denominator lie inside the unit circle in the complex plane.

To begin, we use the quadratic formula to find the roots of the denominator:

$z^2 + 2\frac{S}{R}z + 1 = 0$

$Rz^2 + 2Sz + R = 0$

$z = \displaystyle \frac{-2S \pm \sqrt{4S^2 - 4R^2}}{2R}$

$z = \displaystyle \frac{-S \pm \sqrt{S^2 -R^2}}{R}$.

So we have the two roots $r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}$ and $r_2 = \displaystyle \frac{-S - \sqrt{S^2 -R^2}}{R}$. Earlier in this series, I showed that $S > R > 0$ as long as $b \ne 0$, and so the denominator has two distinct real roots. So the integral $Q$ may be rewritten as

$Q = \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}$

Next, we have to determine if either $r_1$ or $r_2$ (or both) lies inside of the contour. I’ll discuss this in tomorrow’s post.

# Engaging students: Completing the square

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Tracy Leeper. Her topic, from Algebra: completing the square.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

Muhammad ibn Musa al-Khwarizmi wrote a book called al-jabr in approximately 825 A.D. He was in Babylon and he worked as a scholar at the House of Wisdom. Al-Khwarizmi had already mastered Euclid’s Elements, which is the foundation for Geometry. So in his book he posed the challenge “What must be the square which, when increased by ten of its own roots; amounts to 39?” or in other words: how to solve he turned to geometry and drew a picture to figure out the answer. By doing so, al-Khwarizmi found out how to solve equations by completing the square. He also included instructions on how he solved the problem in words. His book al-jabr become the foundation for our modern day algebra. The Arabic word al-jabr was translated into Latin to give us algebra, and our word for algorithm came from al-Khwarizmi, if you can believe it. Later on, his work was used by other Arab and Renaissance Italian mathematicians to “complete the cube” for solving cubic equations.

How does this topic extend what your students should have learned in previous courses?

In previous courses my students should have already been introduced to prime factorization, the quadratic formula, parabolas, coordinates graphs and other similar topics. Completing the square is another way for students to find the roots of a quadratic equation. The first way taught is by using nice numbers that will factor easily. Then the math progresses to using the quadratic equation for the numbers that don’t factor easily. Completing the square is just another way to solve a quadratic that does not easily factor. Some students prefer to go straight to the quadratic equation, whereas other students will favor completing the square after they learn how to do it. It gives the students another “tool” for their toolbox on how to solve equations, and will enable them to solve equations that previously were unsolvable, such as the quadratic . By giving students a variety of ways to solve a problem, they can pick whichever way they are most comfortable with, which in turn will boost their confidence in their ability to learn math.

How could you as a teacher create an activity or project that involves your topic?

Usually the simplest way to learn something is to see something concrete of what you are trying to do. For completing the square, I can give the students the procedure to follow, but they probably won’t be able to fully understand why it works. In order to help them visualize it, I would use algebra tiles. One long tile is equal to x, since its length is x and its width is 1. The square is equal to since the length and the width are both equal to x. However, when you try to add to the square by a factor of x, you end up having a corner missing. This is the part that is missing from the initial equation. Then the students see that you don’t have a complete square, but by adding the same amount to both parts, we can get a complete square that can then be factored. Like so…

References:

http://bulldog2.redlands.edu/fac/beery/math115/m115_activ_complsq.htm

# Engaging students: The quadratic formula

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Daniel Littleton. His topic, from Algebra II: the quadratic formula.

A1: What interesting word problems using this topic can your students do now?

The quadratic equation is a formula used to find the solutions of second order polynomials. Meaning, that if f(x) is a polynomial with the highest power of x being two the quadratic equation may be used to find the solution set of the function. This solution set describes the values at which the function crosses the x-axis, resulting in a solution set of (x_1, 0), (x_2, 0). These points on the graph of the function are often referred to as the zeroes of the function.

With this knowledge regarding the information that is derived from the quadratic equation, a student could be asked the following word problem.

The solution set of this equation are the points at which the motorcycle rider would leave the ground for the jump, and fall back to the ground for the landing. It can be easily determined that factoring this quadratic equation is not a feasible option to find the solution set. Therefore, the student would use the quadratic formula with a=(-10/87), b=(11/29) and c=5. After using the quadratic formula the student will arrive at the solution set of (-5.15, 0) and (8.45, 0). The student would interpret this data to mean that the jump begins 5.15 meters back from the center of the arena and ends 8.45 meters ahead of the center of the arena. The rider would also have 6.85 meters of clearance behind him at the start of the jump, and 3.55 meters of clearance in front of him at the end of the jump. These solutions are determined from the knowledge that the total length of the arena is 24 meters and the center of the arena is the origin of the graph.

This is one stimulating example of a word problem that a student could complete in order to engage their interest in the quadratic formula. Word problems following this could vary in complexity and application.

B1: How can this topic be used in your students’ future courses in mathematics or science?

The quadratic equation has multiple applications in solving polynomial equations of the second degree. In future mathematics courses, most likely at the Pre-Calculus level, students will be asked to solve for variables within trigonometric equations. These equations can also be solved using the quadratic equation when the options of using linear interpretations, factoring, or trigonometric identities is not feasible.

For example, a student could be asked to find all solutions of the formula cot x(cot x + 3) = 1. After factoring the equation and setting the answer equal to zero we derive a standard quadratic form of the equation, cot ^2 x + 3 cot x – 1 = 0. The quadratic equation can be utilized in this situation by setting a=1, b=3, c= (-1) and cot x as the variable. After using the quadratic formula we determine the solutions of cot x = (-3.302775638) or (.3027756377). As a calculator cannot be used to find the inverse of cotangent, we use the fact that cot x = 1/tan x and take the reciprocals of the solutions to find that tan x= (-.3027756377) or (3.302775638). By finding the inverse tangent of these values we conclude that x = (-.2940013018) or (1.276795025).

I would like to take this opportunity to note that this is only one set of the solutions to the value of x. In order to express all solutions we need to add integer multiples of the period of the tangent, π, to each of the expressed solutions. Resulting in the final solutions of

x = (-.2940013018) + nπ or (1.276795025) + nπ where n ϵ integers.

This is one example of an occasion when a student would need to apply the quadratic equation in order to derive a solution to an advanced trigonometric formula.

D1: What interesting things can you say about the people who contributed to the discovery and/or development of this topic?

The first efforts to discover a general formula to solve quadratic equations can be traced back to the efforts of Pythagoras and Euclid. Pythagoras and his followers are the ones responsible for the development of the Pythagorean Theorem. Euclid was the individual responsible for the development of the subject of Geometry as it is still used today. The efforts of these two individuals took a strictly geometric approach to the problem; however Pythagoras first noted that the ratios between the area of a square and the length of the side were not always an integer. Euclid built upon the efforts of Pythagoras by concluding that this proportion might not be rational and irrational numbers exist. The works of Euclid and Pythagoras traveled from ancient Greece to India where Hindu mathematicians were using the decimal system that is still in use today. Around 700 A.D. the general solution for the quadratic equation, using the number system, was developed by the mathematician Brahmagupta. Brahmagupta used irrational numbers in his analysis of the quadratic equation and also recognized the existence of two roots in the solution.

By the year 820 A.D. the advancements made by Brahmagupta had traveled to Persia, where a mathematician by the name of Al-Khwarizmi completed further work on the derivation of the quadratic equation. Al-Khwarizmi is the Islamic mathematician given the greatest amount of credit for the development of Algebra as it is known today. However, Al-Khwarizmi rejected the possibility of negative solution. The works of Al-Khwarizmi were brought to Europe by Jewish mathematician Abraham bar Hiyya. It was in the Renaissance Era of Europe, around 1500 A.D., that the quadratic equation in use today was formulated. By 1545 A.D. Girolamo Cardano, a Renaissance scientist, compiled the works of Al-Khwarizmi and Euclid and completed work upon the quadratic equation allowing for the existence of complex numbers. After the development of a universally accepted system of symbols for mathematicians, this form of the quadratic formula was published and distributed throughout the mathematical and scientific community.

This information was collected from the web page http://www.bbc.co.uk/dna/place-london/plain/A2982567

# An unorthodox way of solving quadratic equations

This post concerns an unorthodox but logically correct technique for solving a quadratic equation via factoring. I showed this to some senior math majors as well as graduate students in mathematics; none of them had ever seen this before. Suppose that we want to solve

$6x^2 - 13x - 5 = 0$

without using the quadratic formula. Trying to solve this by factoring looks like a pain in the neck, as there are several possibilities:

$(x + \underline{\quad})(6x - \underline{\quad}) = 0$,

$(x - \underline{\quad})(6x + \underline{\quad}) = 0$,

$(2x + \underline{\quad})(3x - \underline{\quad}) = 0$,

or

$(2x - \underline{\quad})(3x + \underline{\quad}) = 0$.

So instead, let’s replace the original equation with a new equation. I’ll get rid of the leading coefficient and multiply the constant term by the leading coefficient:

$t^2 - 13t - (5)(6) = 0$, or

$t^2 - 13t - 30 =0$.

This is a lot easier to factor:

$(t - 15)(t+ 2) = 0$

$t = 15 \quad \hbox{or} \quad t = -2$

So, to solve for $x$, divide by the original leading coefficient, which was $6$:

$x = 15/6 = 5/2 \quad \hbox{or} \quad x = -2/6 = -1/3$.

As you can check, those are indeed the roots of the original equation.

This technique always works if the quadratic polynomial has rational roots. But why does it work? I’ll give the answer after the thought bubble.

$6x^2 - 13x - 5 = 0$

Let’s make the substitution $x = t/6$:

$6 \displaystyle \left( \frac{t}{6} \right)^2 - 13 \left( \frac{t}{6} \right) - 5 = 0$

$\displaystyle \frac{t^2}{6} - \frac{13t}{6} - 5 = 0$

Multiply both sides by $6$, and we get the transformed equation:

$t^2 - 13t - 30 = 0$

Although I personally love this technique, I have mixed feelings about the pedagogical usefulness of this trick… mostly because, to students, it probably feels like exactly that: a trick to follow without any conceptual understanding. Perhaps this trick is best reserved for talented students who could use an enrichment activity in Algebra II.

# My “history” of solving cubic, quartic and quintic equations

When I teach Algebra II or Precalculus (or train my  future high school teachers to teach these subjects), we eventually land on the Rational Root Test and Descartes’ Rule of Signs as an aid for finding the roots of cubic equations or higher. Before I get too deep into this subject, however, I like to give a 10-15 minute pseudohistory about the discovery of how polynomial equations can be solved. Historians of mathematics will certain take issue with some of this “history.” However, the main purpose of the story is not complete accuracy but engaging students with the history of mathematics. I think the story I tell engages students while remaining reasonably accurate… and I always refer students to various resources if they want to get the real history.

To begin, I write down the easiest two equations to solve (in all cases, $a \ne 0$:

$ax + b = 0 \qquad$ and $\qquad ax^2 + bx + c = 0$

These are pretty easy to solve, with solutions well known to students:

$x = -\displaystyle \frac{b}{a} \qquad$ and $\qquad x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In other words, there are formulas that you can just stick in the coefficients and get the answer out without thinking too hard. Sure, there are alternate ways of solving for $x$ that could be easier, like factoring, but the worst-case scenario is just plugging into the formula.

These formulas were known to Babylonian mathematicians around 2000 B.C. (When I teach this in class, I write the date, and all other dates and discoverers, next to the equations for dramatic pedagogical effect.) Though not written in these modern terms, basically every ancient culture on the globe that did mathematics had some version of these formulas: for example, the ancient Egyptians, Greeks, Chinese, and Mayans.

Naturally, this leads to a simple question: is there a formula for the cubic:

$ax^3 + bx^2 + cx + d = 0$

Is there some formula that we can just plug $a$, $b$, $c$, and $d$ to just get the answer?  The answer is, Yes, there is a formula. But it’s nasty. The formula was not discovered until 1535 A.D., and it was discovered by a man named Tartaglia. During the 1500s, the study of mathematics was less about the dispassionate pursuit of truth and more about exercising machismo. One mathematician would challenge another: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” Then the second mathematician would solve it and challenge the first: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” And so on. Well, Tartaglia came up with a formula that would solve every cubic equation. By plugging in $a$, $b$, $c$, and $d$, you get the answer out.

Tartaglia’s discovery was arguably the first triumph of the European Renaissance. The solution of the cubic was perhaps the first thing known to European mathematicians in the Middle Ages that was unknown to the ancient Greeks.

In 1535, Tartaglia was a relatively unknown mathematician, and so he told a more famous mathematician, Cardano, about his formula. Cardano told Tartaglia, why yes, that is very interesting, and then published the formula under his own name, taking credit without mention of Tartaglia. To this day, the formula is called Cardano’s formula.

So there is a formula. But it would take an entire chalkboard to write down the formula. That’s why we typically don’t make students learn this formula in high school; it’s out there, but it’s simply too complicated to expect students to memorize and use.

$ax^4 + bx^3 + cx^2 + dx + e = 0$

The solution of the quartic was discovered less than five years later by an Italian mathematician named Ferrari. Ferrari found out that there is a formula that you can just plug in $a$, $b$, $c$, $d$, and $e$, turn the crank, and get the answers out. Writing out this formula would take two chalkboards. So there is a formula, but it’s also very, very complicated.

Of course, Ferrari had some famous descendants in the automotive industry.

So now we move onto my favorite equation, the quintic. (If you don’t understand why it’s my favorite, think about my last name.)

$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$

After solving the cubic and quartic in rapid succession, surely there should also be a formula for the quintic. So they tried, and they tried, and they tried, and they got nowhere fast. Finally, the problem was solved nearly 300 years later, in 1832 (for the sake telling a good story, I don’t mention Abel) by a French kid named Evariste Galois. Galois showed that there is no formula. That takes some real moxie. There is no formula. No matter how hard you try, you will not find a formula that can work for every quintic. Sure, there are some quintics that can be solved, like $x^5 = 0$. But there is no formula that will work for every single quintic.

Galois made this discovery when he was 19 years old… in other words, approximately the same age as my students. In fact, we know when wrote down his discovery, because it happened the night before he died. You see, he was living in France in 1832. What was going on in France in 1832? I ask my class, have they seen Les Miserables?

France was torn upside-down in 1832 in the aftermath of the French Revolution, and young Galois got into a heated argument with someone over politics; Galois was a republican, while the other guy was a royalist. More importantly, both men were competing for the hand of the same young woman. So they decided to settle their differences like honorable Frenchmen, with a duel. So Galois wrote up his mathematical notes one night, and the next day, he fought the duel, he lost the duel, and he died.

Thus giving complete and total proof that tremendous mathematical genius does not prevent somebody from being a complete idiot.

For the present, there are formulas for cubic and quartic equations, but they’re long and impractical. And for quintic equations and higher, there is no formula. So that’s why we teach these indirect methods like the Rational Root Test and Descartes’ Rule of Signs, as they give tools to use to guess at the roots of higher-order polynomials without using something like the quadratic formula.

Real references:

http://mathworld.wolfram.com/CubicFormula.html

http://mathworld.wolfram.com/QuarticEquation.html

http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html

http://mathworld.wolfram.com/QuinticEquation.html

http://library.wolfram.com/examples/quintic/

http://library.wolfram.com/examples/quintic/timeline.html

# Engaging students: Solving quadratic equations

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Elizabeth (Markham) Atkins. Her topic, from Algebra II: solving quadratic equations.

D. History: Who were some of the people who contributed to the discovery of this topic?

Factoring quadratic polynomials is a useful trick in mathematics. Mathematics started long ago. http://www.ucs.louisiana.edu/~sxw8045/history.htm stated that the Babylonians “had a general procedure equivalent to solving quadratic equations”. They taught only through examples and did not explain the process or steps to the students. http://www.mytutoronline.com/history-of-quadratic-equation states that the Babylonians solved the quadratic equations on clay tablets. Baudhayana, an Indian mathematician, began by using the equation $ax^2+bx=c$. He provided ways to solve the equations. Both the Babylonians and Chinese were the first to use completing the square method which states you take the equation $ax^2+bx+c$. You take $b$ and divide it by two. After you divide by two you square that number and add it to $ax^2+bx$ and subtract it from $c$.  Even doing it this way the Babylonians and Chinese only found positive roots. Brahmadupta, another Indian mathematician, was the first to find negative solutions. Finally after all these mathematicians found ways of solving quadratic equations Shridhara, an Indian mathematician, wrote a general rule for solving a quadratic equation.

C. Culture: How has this topic appeared in the news?

A. Applications: How could you as a teacher create an activity or project that involves your topic?

Lesson Corner (http://www.lessoncorner.com/Math/Algebra/Quadratic_Equations) is an excellent resource for finding lesson plans and activities for quadratic equations. One lesson (http://distance-ed.math.tamu.edu/peic/lesson_plans/factoring_quadratics.pdf) talking about engaging the students with a game called “Guess the Numbers”. The students are given two columns, a sum column and a product column. They are then to guess the two numbers that will add to get the sum and multiply to get the product. This is an excellent game because it gets the students going and it is like a puzzle to solve. Learn (http://www.learnnc.org/lp/pages/2981) has a lesson plan for a review of quadratic equations.  The students are engaged by playing “Chutes and Ladders”. The teacher transformed it. The procedures are as follows:

1. Draw a card.
2. Roll the dice.
3. If you roll a 1 or a 6, then solve your quadratic equation by completing the square.
4. If you roll a 2 or 5, then solve your quadratic equation by using the quadratic formula.
5. If you roll a 3, then solve your quadratic equation by graphing.
6. If you roll a 4, then solve your quadratic equation by factoring if possible. If not, then solve it another way.
7. If you solve your equation correctly, then you may move on the board the number of spaces that corresponds to your roll of the die.