# 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 comes from my former student Kelsi Kolbe. Her topic, from Algebra: completing the square.

A2) How could you as a teacher create an activity or project that involves your topic?

When students are learning how to complete the square they are usually told the algorithm take b divide it by two and square it, add that number to both sides. To the students this concept seems like a ‘random trick’ that works. This can lead to students forgetting the formula with no way to get it back. However, if we show students how to complete the square using algebra tiles they will be able to understand how the formula came to be (pictured to the left). This will allow the students to be able to have actual concrete knowledge to lean on if they forget the algorithm.

For an engage I would introduce them how to use the algebra tiles by representing different equations on the tiles. I would mix perfect squares and non-perfect squares. I would wait to do the actual completing the square as the explore activity. This way it’s something they can experiment with and really learn the material themselves.

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

Muhammad Al-Khwarizmi was a Persian mathematician in the early 9th century. He oversaw the translation of many mathematical works into Arabic. He even produced his own work which would influence future mathematics. In 830 he published a book called: “Al-Kitab al-mukhtasar fi hisab al-jabr wa’l-muqabala” Which translates to “The Compendious Book on Calculation by Completion and Balancing” This book is still considered a fundamental book of modern algebra. The word algebra actually came from the Latinization of the word “al-jabr” which was in the title of his book. The term ‘algorithm’ also came from the Latinization of Al-Kwarizmi. In his book he solved second degree polynomials. He used new methods of reduction, cancellation, and balancing. He developed a formula to solving quadratic equations. As you can see to the right this is how Al-Khwarizmi used the method of ‘completing the square’ in his book. It is very similar to how we use algebra tiles in modern day. You can really see the effect he had on modern algebra, especially in solving quadratic equations.

E1) How can technology be used to effectively engage students with this topic?

I found a fun YouTube video of the Fort Collins High School Math Department singing a parody of Taylor Swift’s song “blank space”. In the video they are teaching the steps for completing the square. It also addresses imaginary numbers for more complex problems. I think this could be a fun engage to get the students attention. The video incorporates pop culture into something educational. I have always liked watching mathematical parodies videos on YouTube. It not only engages the students, but if they already know the words to the song, they could also get the song stuck in their head, which will help them solve the problems in the future.

References:
Completing the Square. (n.d.). Retrieved September 14, 2017, from http://www.mathisradical.com/completing-the-square.html
Mastin, L. (2010). Islamic Mathmatics – Al-Khwarizmi. Retrived September 14, 2017, from http://www.storyofmathematics.com/islamic_alkhwarizmi.html

# 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 comes from my former student Deborah Duddy. Her topic, from Algebra: completing the square.

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

Applying what is learned in the class is very vital in fact it is a process TEKS that teachers need to use to maximize student’s understanding. “When are we going to use this in real life?” and “Why do we need to know this?” are questions that students ask on a daily basis. Connecting material to the real world helps engage students and develops critical thinking. Describing a path of a ball, how far an item can be tossed in the air and how to maximize profits for a company are just some examples of how quadratics can be used in the real world.

One important event happens during high school; students receive their driver’s license. In their written driver’s test, students must know the distance needed to stop a car at certain speed limits. Using an example like the one below will be interesting for the students and help connect lesson material and real life.

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

To begin class and get students involved with their learning, the class will participate in an activity. Each pair of students will have two different cards such as (x+2)^2 and x^2+4x+4, and any variations of these problems. They can only look at the (x+2)^2 card. Students will work out the problem on paper. Students will be asked to remember how to find the area of a square and then set up a square with the dimensions matching the first card. From there, the pairs would use algebra tiles (after knowing what each tile stands for) and attempt to “complete the square”. This activity will be used as an engage and a beginning explore for the students. This activity will help students see completing a square geometrically.

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

Completing the square is another way of solving/factoring the equation. The process of completing the square is to turn a basic quadratic   equation of ax^2 + bx + c = 0 into a(x-h)^2 + k = 0 where (h,k) is  the vertex of the parabola. Therefore this process is very beneficial because it helps students graph the quadratic equation given. In order to find h and k, students should be able to factor, square a term, find the square root and manipulate the equation.

In solving the equation by completing the square is to subtract the constant off the left side and onto the right side. Then students take the coefficient off the x-term divide it then square it. Students then add this number to both sides of the equations. By simplifying the right side of the equation, students give the perfect square. Then solve the equation left by taking the square root of both sides and determining x.

References:

http://www.classzone.com/eservices/home/pdf/student/LA205EBD.pdf

# 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 Diana A’Lyssa Rodriguez. Her topic, from Algebra: completing the square.

A2. How could you as a teacher create an activity or project that involves your topic?

Using Algebra tiles is a great visual way for students to understand completing the square. The students start with the tiles that correspond to the given problem. The unit tiles are then flipped and moved to the other side of the equal sign. The remaining tiles are positioned into a square shape. The corner piece that appears to be missing will be filled unit tiles. What you do to one side, must be done to the other. Therefore the amount of unit tiles added to the square will also be added to the other side of the equation. Find the zero pairs and take them away. Then, find the corresponding tiles that will outline the square, so when multiplied together equals the equation.

Step 1:

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Step 8:

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

E1. How can technology be used to effectively engage students with this topic?

This video from Khan Academy is a great tool for completing the square. This video explains why we have to take half of the b value and square it (when looking at ax2+bx+c) to obtain the c value. When the students understand why we do something in math, they are more likely to be interested in the topic. The different colors that are used to write out the process allows the students to organize and understand completing the squares better. This particular video is also just long enough to capture the attention of the students but not so long as to lose it. Also, after hearing the same person explain math all the time, students may not understand it as well as they possibly could. So what is said in this video can easily be explained by the teacher but students sometimes need to hear a different voice explain a concept so they can gain a new perspective on the topic.

Resources

# The antiderivative of 1/(x^4+1): Part 6

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

So far, I’ve shown that the denominator can be factored over the real numbers:

$\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$

$= \displaystyle \int \frac{ -\displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx + \int \frac{ \displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx$

$= \displaystyle \frac{\sqrt{2}}{8} \ln \left( \frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx$

To evaluate the remaining two integrals, I’ll use the antiderivative

$\displaystyle \int \frac{dx}{x^2 + k^2} = \displaystyle \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right)$.

To begin, I’ll complete the squares:

$\displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx = \displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} }$

$= \displaystyle \frac{1}{4} \int \frac{ dx }{ \left(x - \displaystyle \frac{ \sqrt{2}}{2} \right)^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 } + \frac{1}{4} \int \frac{ dx }{\left(x + \displaystyle \frac{ \sqrt{2}}{2} \right)^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 }$

Applying the substitutions $u = x - \displaystyle \frac{ \sqrt{2}}{2}$ and $v = x + \displaystyle \frac{ \sqrt{2}}{2}$, I can continue:

$= \displaystyle \frac{1}{4} \int \frac{ du }{ u^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 } + \frac{1}{4} \int \frac{ dv }{v^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 }$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{u}{\sqrt{2}/2} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{v }{\sqrt{2}/2} \right) + C$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{x - \displaystyle \frac{ \sqrt{2}}{2}}{\sqrt{2}/2} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{x + \displaystyle \frac{ \sqrt{2}}{2} }{\sqrt{2}/2} \right) + C$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( x\sqrt{2} - 1 \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( x \sqrt{2} + 1 \right) + C$

Combining, I finally arrive at the answer for $\displaystyle \int \frac{dx}{x^4 + 1}$:

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

Naturally, this can be checked by differentiation, but I’m not going type that out.

# The antiderivative of 1/(x^4+1): Part 3

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

To compute this integral, I will use the technique of partial fractions. In yesterday’s post, I used De Moivre’s Theorem to factor the denominator over the complex plane, which then led to the factorization of the denominator over the real numbers.

In today’s post, I present an alternative way of factoring the denominator by completing the square. However, unlike the ordinary method of completing the square, I’ll do this by adding and subtracting the middle term and not the final term:

$x^4 + 1= x^4 + 2x^2 + 1 - 2x^2$

$= (x^2 + 1)^2 - (x \sqrt{2})^2$

$= (x^2 + 1 + x\sqrt{2})(x^2 + 1 - x \sqrt{2})$.

The quadratic formula can then be used to confirm that both of these quadratics have complex roots and hence are irreducible over the real numbers, and so I have thus factored the denominator over the real numbers:

$\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$.

and the technique of partial fractions can be applied.

There’s a theorem that says that any polynomial over the real numbers can be factored over the real numbers using linear terms and irreducible quadratic terms. However, as seen in this example, there’s no promise that the terms will have rational coefficients.

I’ll continue the calculation of this integral with tomorrow’s post.

# How I Impressed My Wife: Part 2d

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

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

In this series, I’ll explore different ways of evaluating this integral.So far in this series, I’ve shown that

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

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

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

To evaluate this last integral, I complete the square in the denominator. I first factor $(a^2+b^2)$ out of the denominator:

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

$latex = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u^2 + \displaystyle \frac{2 a}{a^2+b^2} u + \displaystyle \frac{a^2}{(a^2+b^2)^2} \right) + \displaystyle \frac{1}{a^2+b^2} – \displaystyle \frac{a^2}{(a^2+b^2)^2} }$

$latex = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{a^2+b^2}{(a^2+b^2)^2} – \displaystyle \frac{a^2}{(a^2+b^2)^2} }$

$latex = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

Next, I employ the substitution $v = u + \displaystyle \frac{a}{a^2+b^2}$, so that $dv = du$. The endpoint of the integral do not change with this substitution, and so

$latex Q = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

I’ll continue with the evaluation of this integral 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

# Two ways of doing an integral (Part 2)

A colleague placed the following problem on an exam, expecting the following solution:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) + C$

However, one student produced the following solution (see yesterday’s post for details):

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) + C$

As he couldn’t find a mistake in the student’s work, he assumed that the two expressions were equivalent. Indeed, he differentiated the student’s work to make sure it was right. But he couldn’t immediately see, using elementary reasoning, why they were equivalent. So he walked across the hall to my office to ask me if I could help.

Here’s how I showed they are equivalent.

Let $\alpha = \displaystyle \sin^{-1} \left( \frac{x-2}{2} \right)$ and $\beta = \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$. Then

$\displaystyle \sin(\alpha - 2\beta) = \sin \alpha \cos 2\beta - \cos \alpha \sin 2\beta$.

Let’s evaluate the four expressions on the right-hand side.

First, $\sin \alpha$ is clearly equal to $\displaystyle \frac{x-2}{2}$.

Second, $\cos 2\beta = 1 - 2 \sin^2 \beta$, so that

$\cos 2\beta = \displaystyle 1 - 2\left( \frac{\sqrt{x}}{2} \right)^2 = \displaystyle 1 - \frac{x}{2} = \displaystyle -\frac{x-2}{2}$.

Third, to evaluate $\cos \alpha$, I’ll use the identity $\displaystyle \cos \left( \sin^{-1} x \right) = \sqrt{1 - x^2}$:

$\cos \alpha = \displaystyle \sqrt{1 - \left( \frac{x-2}{2} \right)^2 } = \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Fourth, $\sin 2\beta = 2 \sin \beta \cos \beta$. Using the above identity again, we find

$\sin 2\beta = \displaystyle 2 \left( \frac{ \sqrt{x} }{2} \right) \sqrt{ 1 - \left( \frac{ \sqrt{x} }{2} \right)^2 }$

$= \sqrt{x} \sqrt{1 - \displaystyle \frac{x}{4}}$

$= \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Combining the above, we find

$\sin(\alpha - 2 \beta) = \displaystyle -\left( \frac{x-2}{2} \right)^2 - \left( \frac{\sqrt{4x-x^2}}{2} \right)^2$

$\sin(\alpha - 2 \beta) = \displaystyle \frac{-(x^2 - 4x + 4) - (4x - x^2)}{4}$

$\sin(\alpha - 2 \beta) = -1$

$\alpha - 2 \beta = \displaystyle -\frac{\pi}{2} + 2\pi n$ for some integer $n$

Also, since $-\pi/2 \le \alpha \le \pi/2$ and $0 \le -2\beta \le \pi$, we see that $-\pi/2 \le \alpha - 2 \beta \le 3\pi/2$. (From its definition, $\beta$ is the arcsine of a positive number and therefore must be nonnegative.) Therefore, $\alpha - 2\beta = -\pi/2$.

In other words,

$\sin^{-1} \left( \displaystyle \frac{x-2}{2} \right)$ and $2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$

differ by a constant, thus showing that the two antiderivatives are equivalent.

# Two ways of doing an integral (Part 1)

A colleague placed the following problem on an exam:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}}$

He expected students to solve this problem by the standard technique, completing the square:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \displaystyle \int \frac{dx}{4-(x-2)^2} = \sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) + C$

However, one student solved this problem by some clever algebra and the substitution $u = \sqrt{x}$, so that $x = u^2$ and $dx = 2u \, du = 2 \sqrt{x} \, du$:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \displaystyle \int \frac{dx}{\sqrt{x} \sqrt{4-x}}$

$= \displaystyle \int \frac{dx}{\sqrt{x} \sqrt{4 - (\sqrt{x})^2}}$

$= \displaystyle \int \frac{2 \, du}{\sqrt{4 - u^2}}$

$= 2 \sin^{-1} \left( \displaystyle \frac{u}{2} \right) + C$

$= 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) + C$

As he couldn’t find a mistake in the student’s work, he assumed that the two expressions were equivalent. Indeed, he differentiated the student’s work to make sure it was right. But he couldn’t immediately see, using elementary reasoning, why they were equivalent. So he walked across the hall to my office to ask me if I could help.

After a few minutes, I was able to show that the two expressions were equivalent.

I’ll leave this one as a cliff-hanger for now. In tomorrow’s post, I’ll show why they’re equivalent.

# 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 comes from my former student Claire McMahon. Her topic, from Algebra: completing the square.

There were a lot of famous mathematicians that contributed to the notion of completing the square.  The first of the mathematicians was that of the Babylonians.  This culture started the notion of not only solving the quadratics but of arithmetic itself.  The Babylonians started with the equations and then proceeded to solve them algebraically.  Back then; they used pre-calculated tables to help them with solving for the roots.  They were basically solving by the quadratic equation at this point.  The man that came along has a very hard name to not only pronounce but to spell, and I will do my best.  I will refer to him as Muhammad from here on out but his full name, or one of the common names to which he is referred is Muhammad ibn Musa al-Khwarizmi.  He developed the term algorithm, which led to an algorithm for solving quadratic equations, namely completing the square.

The notion of completing the square has gone through a series of transformations throughout the history of mathematics.  As mentioned before the Babylonians started with the notion and increased the knowledge by developing the quadratic formula to find the roots of a given quadratic equation.  This spurred the thought that I can solve any equation and find its solution and roots by completing the square.  Muhammad brought this notion to us, of which was mentioned before.  More specifically the text that he developed was “The Compendious Book of Calculations by Completion and Balancing.”  This book of course has been translated several times over but the general idea is laid out in the title.  Modern mathematicians have developed a less compendious form that is now being taught in the math classes today.  They take on many different forms and can be taught with manipulates as well.

The fabulous part to the story is there are a lot of resources that help the kids of today to deal with this “trick” of the math trade.  There are numerous You Tube videos on the different methods of which show every step along the way with encouraging thoughts.  Another great online resource is any of the math websites.  I find it a little unfair that these resources were not readily available when I was struggling with such concepts.  One of my personal favorites is the PurpleMath.com website.  This website breaks everything down to basically that of a fourth grade level.  They have pictures and fun problems to work out on your own.  My favorite part is that you get your answers checked instantaneously to build the self-confidence and self-efficacy it takes to be a successful student.  These particular websites are great tools for teachers as well, as they have a lot of great examples that can be used in the classroom and different ways that a student might present and calculate a problem.