Math Concepts Explained

Convert Polar to Rectangular

2012-01-14T17:33:00.000-08:00

In this post, I am going to describe the theory that allows you to convert polar to rectangular coordinates. This follows up my previous post that describes graphing polar coordinates, where I introduced the concept of this new method of graphing. However, the relationships that I will show you here will hopefully allow you to see the connection between polar and Cartesian coordinate systems, which will make them easier for you to work with!

To quickly refresh what I explained last time, start at the origin (or pole) of your graph and extend the polar axis line out to the right. This is your reference line that will help you describe the location of any other point. Now, pick a point P somewhere, and this is described by its distance (radius), r, from the origin (if you were to connect it to there) and the angle, ɵ, by which that line has rotated away from the polar axis. Whereas points in the rectangular coordinate system are described as P (x, y), points in the polar coordinate system are described as P (r, ɵ).

Now, let's take a closer look at the relationship between P (x, y) and P (r, ɵ). This will use a little bit of trigonometry, so review it in my posts about sine, cosine, and tangent if you need to brush up!

To do this, let's superimpose the two coordinate systems, meaning that we will assume that the origins of each are in the same place, and the polar axis is the same as the positive x-axis. You should be able to see, with the help of the following handy figure, that our point P can be described by P (x, y) and P (r, ɵ). You can also see that, by combining the two coordinate systems, we have formed a triangle which has sides of lengths x and y and hypotenuse r. This triangle will be the basis that allows us to convert polar to rectangular coordinates.

Now, if we apply our rules and identities of trigonometry, the relationships of the triangle sides and angles will connect the polar coordinate system and rectangular coordinate system. And with these relationships, you will be able to convert from polar to rectangular, and also back again to convert rectangular to polar.

The relationships are quite simply the basic trigonometry identities that you already know:

sin ɵ = opposite / hypotenuse

cos ɵ = adjacent / hypotenuse

tan ɵ = opposite / adjacent

Now, if we substitute in the names for our sides, and rearrange to have polar terms on one side and Cartesian terms on the other, we arrive at the following relationships:

sin ɵ = y / r -----> y = r sin ɵ

cos ɵ = x / r -----> x = r cos ɵ

tan ɵ = y / x

Furthermore, we can apply the Theorem of Pythagoras to give us another useful relationship:

r² = x² + y²

And there you have it! Easily derived connections between polar and rectangular coordinate systems. You will find as you work through you studies that sometimes some expressions may be easier to work with in one coordinate system or the other. Keep this in mind, especially when you begin to work with polar equations. Using these quick methods to convert polar to rectangular coordinates may help you to get through your problems a lot faster (and easier)!

In my next post, I'll show you some more things about polar coordinates!

Graphing - Polar Coordinates

2011-12-16T20:55:00.001-08:00

Polar coordinates are a system of describing points on a plane, in a way that is similar yet still quite different from what you have previously used. I am going to explain what polar coordinates are here, and then in a subsequent post, I am going to explain their relationship to the graphing system that you already know about. Hopefully you will find that graphing polar coordinates are not really any harder than what you already know how to do!

Up until now, you have likely only ever worked with rectangular coordinates, otherwise known as the Cartesian coordinate system. This is the familiar horizontal x-axis and vertical y-axis, which intersect at the origin and separate the coordinate plane into four quadrants. Any point on the xy plane can be described by it's location relative to these axes, which requires knowing how far horizontal and how far vertical it is away from the origin. These are the x-coordinate and y-coordinate of the point, and together they form an ordered pair, which completely describes the point's location on the xy plane. (For those looking for extra trivia: the x-coordinate has the technical name "abscissa" while the y-coordinate is called the "ordinate." Try explaining that to your teacher for bonus points! Also of interest, this Cartesian coordinate system is named after René Descartes, who was a French philosopher and mathematician whose work, among other things, first described the connection between algebra and geometry.)

Polar form is quite different from this rectangular coordinate system. It requires a different way of thinking about the points on the graph. Don't think of an x-y axis system for now. To start, just place a point, and then from that point extend a line out to the right (this is conventionally what is done, though technically you could draw it in any direction). The starting point, O, is the origin of this coordinate system, but in this case it may also rightfully be called the pole. The line that you have extended from the pole is the polar axis. We will define points on the plane relative to this axis (as opposed to the Cartesian system, which defines points relative to two axes).

So then, now we need a point P to talk about. Pick any point, and then draw a line (called r, as in radius) that connects it to the pole. The basic premise of a polar coordinate graph is that you describe the point by providing its distance from the pole, and also the angle (ɵ, theta) by which r is rotated away from the polar axis. Knowing these two dimensions is enough information to fully describe where your point is located on the plane. So, while a Cartesian ordered pair looks like (x, y), a polar ordered pair looks like (r, ɵ). So, r and ɵ are referred to as polar coordinates of P, and it is written as P(r, ɵ) which explains that P is the point with polar coordinates (r, ɵ).

To assist you in drawing your polar graphs, polar coordinate graph paper is available. Instead of being labeled as x and y axes, they are labeled with angles (in radians... I'll cover radians in another post as well). However, I don't think you really need to use it until you start graphing more in-depth polar equations. Whenever I draw a polar graph, I just draw my own. :) Briefly, 180°= π radians, 90° = π/2 radians, 270° = 3π/2 radians, and 360° = 2π radians. I'm not going to go into anymore detail than this about radians, because if you're at the point where you're studying polar coordinate graphs, you've already covered radians. :)

Using this new coordinate system and working with polar equations will allow you to draw some really cool polar graphs. Expect to see plenty of "four-leaf clover" graphs and "spiral" graphs, among other really interesting designs. Just try to think of your graph as a series of points along a line that is rotating or sweeping around the pole in the center, moving closer or further as it revolves around it, as described by its equation. I will explain some polar equations in a future post that will highlight some common and cool polar graphs.

In my next post, I want to explain to you how to convert from polar coordinates to rectangular coordinates. With a little bit of logic, you will see that the derivation to connect polar form to rectangular form is quite easy. Similarly, going from rectangular coordinates to polar coordinates follows along the same theory. This connection is a helpful basis to build upon, since by this point you are already very familiar with the Cartesian coordinate system. Layering the polar coordinate system on top is the best way to learn this new approach to graphing.

Now on Google+! Add me to your circles!

2011-12-05T22:19:00.001-08:00

I've finally gotten around to connecting my site to Google+!

I've set up a page specifically for this Math Concepts Explained blog. Visit my Google+ page at https://plus.google.com/114879190180011260820 and add me to your circles! I'm planning on posting brief bits of my blog posts there as I generate new ones, but you're going to have to visit this blog for the full deal! You can also send me emails to ask questions or leave comments, if you don't want to do so on the blog's comment section. Please send your math questions or comments to my new email address at mathconcepts101 [at] gmail [dot] com.

I'm also planning on starting a Math Concepts Explained profile on Facebook soon!

And for those of you who are new to my site, I also have a Twitter account already set up. My profile is MathConcepts on Twitter (click the Follow me button near the top of the page). You can easily connect with me on there as well!

Hopefully these new additions are going to be big for the blog! Thanks to everyone for visiting!

Degree of a Polynomial

2011-11-28T19:40:00.001-08:00

One of the most common questions you will find when first learning about polynomials, "How do you find the degree of a polynomial," sounds incredibly complicated and difficult, but really couldn't be simpler. Sometimes, you don't need to use a bunch of math concepts to find the answer, or to do several mathematical manipulations to arrive at the solution. Often, you don't have to do anything but just LOOK at the equation, and you can figure out the degree of a polynomial. And that is because finding the degree of a polynomial is simple.

The degree of a polynomial is simply the highest exponent of the variable in the equation.

Look at this question, and you will see just how easy it can be:

What is the degree of the polynomial: x³ + x = 5.

Yes, it is as simple as it sounds. The highest power of x in this polynomial is 3, and so that is the degree. See? Told you it can be easy!

Your questions will likely look similar to that one. However, there are a few things to say as well, just in case you get a curve ball question.

Non-zero terms (e.g. integers) have a variable with an exponent of 0 (which means that the variable actually equals 1, and therefore you don't need to show it). If you ONLY have non-zero integer, and therefore assume you have a variable with an exponent of 0, you can say that the degree of this non-zero constant polynomial is 0. That is, constants are zero-degree. However, that is only for non-zero polynomials. Zero itself has an undefined degree.

So, the degree of the non-zero constant polynomial 8 is 0. Easy.

Of course, you will inevitably get harder questions that do actually require a bit of work to arrive at your solution, but that will only be more of the same kind of mathematical concepts you've been studying, such as how to FOIL polynomials, or factoring by grouping. In the end, when you have your polynomial expression, all you need to do is determine what is the highest exponent of the variable, and then state that as the degree. Try some practice questions, and you will find that you won't be asking "how do you find the degree of a polynomial" anymore!

One last thought I'd like you to consider: I have seen people searching my site for help with things like "degree polynomial" or "degree polynomials" and I think that sounds very strange, and suggests to me that this math concept is not being taught very well at all in the classroom. Technically, you don't have "degree polynomials" per se. ALL polynomials, except zero, have a degree which can be anything from third degree polynomials to higher degree polynomials such as ten-thousand degree polynomials (though I certainly don't want to be the one working with a polynomial of that degree!). If you think about it, to say something like a "degree polynomial" is to be redundant!

In any case, if you arrived here after searching for help with finding the degree of a polynomial, or for "degree polynomial," I hope this post has been informative and helpful for you. It's not a math concept that is as hard as it sounds, so hopefully my post makes sense and you can take something away to help teach other students who don't know what "degree polynomials" are. :-) As always, please remember to +1 me below if you liked my post! Also follow me on Twitter with the button above!

Using the Quadratic Formula

2011-11-12T15:05:00.001-08:00

So, now that you know the answer to 'What is the Quadratic Formula,' next I will show you examples of Using the Quadratic Formula. Refer back to my last post to familiarize yourself with what the quadratic formula looks like. I've also explained there the nature of the roots of a quadratic equation. If you haven't read it, I recommend taking a look as it might help you to visualize and to find the solution to a quadratic equation easier.

For my first example of using the quadratic formula to find the roots of a quadratic equation, let's keep it simple.

x² - 2x -3 = 0

Comparing this to the standard form of a quadratic equation, ax² + bx + c = 0, we can equate the letter coefficients to the values provided. That is, we can say that a = 1, b = (-2), c = (-3). Now, we can simply substitute these values into the quadratic formula:

So, we have:

If you follow along with the arithmetic, you can see that we've solved the quadratic formula to show that the roots of the given equation are x = 3 and x = (-1).

Now, remember that I said in a previous lesson that you have to check your answers! Substitute these values back into the original equation, and you will find that they do indeed satisfy the equation. So, these are the correct roots!

Of course, you may have noticed that this question didn't actually require the quadratic formula to solve for the roots. The quadratic formula worked well and got us the answer, but as you saw, it required a bit of work. And more work means more opportunity to make a mistake! You may have noticed that there was actually a faster way of solving the question. If you noticed that you could reduce the question down to (x - 3)(x + 1) = 0, you could simply let each set of brackets equal zero, and then find again that x = 3 and x = (-1) are the correct solutions.

Let's try another one, adding some more of the previous math concepts I've gone over.

Using the quadratic formula, find the roots of: 2x³ + 3x² = 4x

It's looks a little more complicated than the last one, huh? It has higher order exponents, and it doesn't immediately look like a quadratic equation, as the first example did. However, with a little bit of arithmetic, and using your skills from the math concepts I explained in my post about factoring (specifically, GROUPING in my Find Factors: Methods of Factoring post), it will begin to look a bit more familiar and workable.

So then, apply grouping techniques to our question. Let's bring everything to one side first though. Recall that the standard form of a quadratic equation equals zero.

2x³ + 3x² = 4x

2x³ + 3x² - 4x = 0

x(2x² + 3x - 4) = 0

Looks a little better now, right? Maybe, something that might fit into the quadratic formula? Recall that the roots, or solutions, are any values of x that make the expression true. So, what we have derived up to this point is a product of two expressions that equals zero, and therefore the roots will be whatever values of x cause each part of the product to equal zero. The first (potential) root is obvious, from the first of the two expressions in the product: x = 0. (Substitute 0 back into the original equation to verify this is a correct root!) The second part, 2x² + 3x - 4, will require more work, and if we let it equal zero, you can see that it will fit into the quadratic formula perfectly.

To prepare for the quadratic formula, we need to identify our A, B, and C values. They are: A = 2, B = 3, and C = (-4). Now, we just substitute into the formula, do the math, and come up with our root(s) for this part of the question!

So, these are our answers for the two roots to the quadratic expression part of our original question. These are the radical forms of the solutions, so they look way more complicated. But, often the quadratic formula doesn't reduce all the way down to a nice, round number and you will be left with something like this. The last thing you have to do is substitute them back into the original question to verify the roots are true, and that is it! Of course, when you right your answers down, make sure you remember to put the roots from the first part of the question, i.e. the part we created by grouping and solved for x = 0.

That last question goes over a lot of math concepts and is definitely comparable to some of the more complicated math questions you may find in your homework or on exams. Review and study it and make sure you understand it. I'll post another example as well soon, if anyone needs some more examples of using the quadratic formula.

The Quadratic Formula

2011-11-09T10:39:00.000-08:00

Following my posts on How to Solve Quadratic Equations, you will soon find that not all quadratic equations can be solved by quadratic factoring, and you will come to rely on The Quadratic Formula to help you. As a quick refresher, a quadratic equation is one which takes the form of ax² + bx + c = 0, as long as the "a" term is not zero. In other words, a quadratic equation is one in which there is an x². (The "b" or "c" term can be zero.) I have already described the process you should follow if your question can be factored down, and you can express it as a product of two smaller expressions. Then, you can solve for two roots by letting each of the small expressions equal zero. I highly recommend reading my previous post if you need to go over this quadratic factoring technique. Factoring quadratic equation technique works well... when possible.

However, as I said, not all quadratic equations can be solved this way. Sometimes, they are already expressed in a simplest form, or further manipulations just make things messier. In these cases, you can use The Quadratic Formula to solve for the roots of the equations. At first glace, the quadratic formula looks like a beast of a formula to use, and even harder to memorize! But, trust me... commit this formula to memory and learn how to use it, and solving quadratic equations will become so easy for you!

So, what is the Quadratic Formula?

I will go over how to solve it, but first, the Quadratic Formula looks like this:

You can use this for any quadratic expression of the form ax²+ bx + c = 0, where "a" does not equal zero. (If you think about this condition, you can see that if a = 0, then there is no x² term at all, and you are left with a linear equation or something of a higher order. Also, if a = 0, the quadratic formula then has 2(0) in the denominator, which equals 0 and causes the whole expression to be undefined. So, hopefully that short explanation will help you to remember that if a = 0, you cannot use the quadratic formula!)

Working through the math of the quadratic formula isn't as difficult as you may think. To start, all you do is arrange your question into the form of ax²+ bx + c = 0, and then you can easily identify the coefficients for a, b, and c. Then, you simply substitute those values into the quadratic formula, and do the math. One thing to draw your attention to though is the "plus/minus" sign. Basically, the quadratic formula is really TWO formulas, one with a "-b + √....." and one with "-b - √....." These two formulas are what give you your two roots.

You will study this more in the future, but for now you may find it interesting that a quadratic equation, i.e. an equation with an x² term, defines a parabola. The equation of all parabolas have x²as the highest order exponent. As a result, you can imagine that a parabola drawn on an X-Y graph will cross the x-axis twice (at the most). These are the roots or solutions of the equation, and so that is why you cannot have more than 2 roots. Similarly, you can figure out why there may be 1 or even 0 roots, depending on where the parabola is located on the graph.

So, now that you know the answer to 'what is the quadratic formula,' next I will show you how to use it. Examples coming in my next post....

Use Siri for Math on iPhone 4S

2011-10-30T13:31:00.000-07:00

For people who recently bought Apple's new superphone, the iPhone 4S, one of the most compelling new features is the "virtual assistant" Siri. It's a groundbreaking new app that does a pretty good job of understanding natural language spoken to it through the iPhone's mic, and then responding to you with some kind of acknowledgement or action. Some of the cool things that Siri can do are highlighted everywhere in the news: send a text message to someone, call someone, setup an appointment or reminder. In some countries (for now, only a few), Siri can even provide you with directions to a location, or understand things like "i'm hungry" and respond to you with nearby restaurants. However, as useful as these functions are, one of Siri's less advertised features is an interaction with the knowledge engine Wolfram Alpha.

Wolfram Alpha works similarly to Google, but rather than returning a list of search results to a query, it actually attempts to answer your question with data in its gigantic fact database. So, if you ask Wolfram Alpha "what is the fifth planet from the Sun" or "what is the capital city of Belgium," it will respond with "Jupiter" and "Brussels" as well as a boatload of information specifically about those entries. Wolfram Alpha is truly amazing. Now, if you put this together with Siri on the iPhone 4S, what you are able to do is simply ask Siri any factual question (without having to go into a browser or type out your question), it will query Wolfram Alpha for you, and then return the information you're looking for.

So, what does all of this have to do with math? How does this allow you to use Siri for math on iPhone 4S?

Well, if you input math questions into Wolfram Alpha, it will come back with the answer. This works especially well for unit conversions, or looking up constants, or working with math expressions. Having the ability to simply ask Siri your math question, and have the app return the calculation for you is a remarkable shortcut to writing out and working through complex calculatons. You don't even have to punch numbers into a calculator. All you have to do is ask Siri your math questions, and it will return the solutions.

Of course this isn't going to TEACH you how to perform these calculations, but when you know about these math concepts already, this is an absolutely fantastic math shortcut that means you don't have to do the hard work! Check out some of the ways that you can use Siri for math on iPhone 4S. These are just a few examples, and I'm sure you can find even more. (And remember that Wolfram Alpha can handle all kinds of factual questions, so you're not restricted to asking Siri only math questions!) Once you realize how powerful Siri is, you will find that it can be a very powerful, helpful, and convenient calculator!

Show you the value of mathematical constants:

Calculate percentages:

Calculate square roots:

Calculate reciprocals:

Perform unit conversions:

Perform arithmetic on a string of integers:

Perform arithmetic on fractions:

Perform complex calculations:

If you notice in my pictures, Siri doesn't just respond with the answer, but also with a host of other data such as unit conversions or different presentations of data. It will also always show you what it understood for the original question, so you can verify that it understood you correctly and actually performed the computations on the proper input.

Now, admittedly, Siri is not going to help you learn math any easier or faster. You still need to put in time with the usual techniques... i.e. you still have to do your homework and study! And, you won't be allowed to take your iPhone 4S and Siri with you into an exam any time soon. Just imagine everyone in an exam room talking to their iPhones and all the Siri's talking back! However, when you get to a point in your studies that you understand the math concepts being taught, and you've done plenty of practice to understand how to arrive at the correct answer, then having the ability to use Siri for math on iPhone 4S to quickly and accurately find what you need to know is incredibly valuable! (Please hit the Google +1 below if you found this article useful!)

How To Solve Quadratic Equations - Part 2

2011-10-26T22:59:00.000-07:00

This post continues from where my last post left off, on how to solve quadratic equations. I explained the general form that a quadratic equation will take, with the key being that there is an x² term present. To solve them without using the quadratic formula, you need to use a bit of factoring methods to come up with the roots. In particular, one common factoring method to use is the grouping method of factoring. Then, once factored, you consider the property that says "two terms multiplied will equal zero only if one or both of those terms is 0." This may seem like a lot of work, and may sound a bit confusing with all the steps you need to take. But I think with a bit of practice you will come to better appreciate and understand the process you need to follow to arrive at your solution. You will see that you already know the individual steps you need to solve the equation. You just need to become familiar with the order that you use these steps.

Follow along through my example and you will hopefully be able to see what I mean.

Let's consider the equation x² + 7x + 10 = 0

First, we can identify that there is an x² term (with a non-zero coefficient... 1), so we can say that it is a quadratic equation.

To solve a quadratic equation, we want to determine the roots, or what values make the equation true. To help us to achieve this, we want to rearrange the left side so that it is a product of two terms (or expressions). In this way, we can say that "something times something equals zero". And since we need one of those "somethings" to be zero if the product is zero, we essentially break this down to "something #1 = 0" and "something #2 = 0", and by solving these two simpler equations, we will arrive at our roots. So, continuing with our example then, let's factor it. Review my post on methods of factoring if you need a bit of a refresher!

x² + 7x + 10 = 0
(x + 2)(x + 5) = 0

This is what we're looking for... two expressions multiplied together to give zero. Now, we have two equations to work with to find our roots of the quadratic equation. Rewriting, this gives us...

x + 2 = 0 and x + 5 = 0

And quite obviously, these can be solved to show that x = (-2) and (-5). And since we followed that whole process, we can consider these two values to be roots of our original quadratic equation. However, it is VERY IMPORTANT to substitute these values back into the original equation to check! With these values, we can show that:

(-2)² + 7(-2) + 10 = 0
4 - 14 + 10 = 0........ this is true. So -2 is for sure one of the roots. I'll leave -5 for you to verify on your own.

If you find a question and proceed all the way through to find the roots, and you go and plug them back into the original equation, if one of the roots does NOT satisfy the equation, you cannot count it as one of the roots. This sometimes happens when you have an expression in a denominator (eg. (x - 2)), and if you determine through the above steps that your expression gives you a root of 2, by plugging this into your original equation, specifically into the denominator, the denominator will equal 0 and cause the expression to be undefined. Therefore, this root does not satisfy the original equation and you just ignore it.

I hope this has helped to explain the process you need to follow to solve quadratic equations. With practice, they will become second nature. However, despite all of the work required, sometimes it just is not practical or apparent how to factor your quadratic equation. In these cases, you would likely want to rely on the use of the quadratic formula, which I will go over in a future post to explain what it is and how it works. Let me know if this makes sense or if you'd like anything more added. Also, please +1 me below if this was helpful!

How To Solve Quadratic Equations

2011-10-20T21:40:00.000-07:00

Following up my previous post that gave you advice on How To Solve Equations, in this post I would like to go over some strategies on How To Solve Quadratic Equations. Quadratic equations become very common in high school math and college math, and they require a bit more work sometimes to solve. You may already have experience using the quadratic formula, which I will explain shortly and is extraordinarily good to memorize! First though, let's go over solving quadratic equations. To do this, you will commonly rely on factoring quadratics techniques. You can refer to my previous post on Methods of Factoring for some additional tips!

When most students hear "quadratic equation," they usually get anxious because quite often this means having to work with the quadratic formula. This formula is more complicated than most that you have probably encountered up to this point, but factoring quadratics doesn't always rely on the quadratic formula! In fact, they can be quite simple! A quadratic equation isn't just "something that needs the quadratic formula" to solve it. Quite simply, a quadratic equation is just an equation that can written in this form:

ax² + bx + c = 0 where a, b, and c are real numbers and a does not equal 0

See? That doesn't sound so bad. ;) The KEY is that the "a" value is not 0. b and c can be, but not a. You need to have the x² term.

Remember, "solving an equation" means to find the roots or solution... or, what makes the expression true? To do this with quadratic equations, we rely on the property that says "two terms multiplied = 0 only if one or both of those terms is 0." Remember this property! It is key to the quadratic factoring method. If we combine this property with our 'grouping' factoring method, you will see how this all comes together.

More to come in my next post, How To Solve Quadratic Equations - Part 2....

How To Solve Equations

2011-10-13T21:50:00.000-07:00

Knowing how to solve equations is a very important skill to have in mathematics courses. There are all kinds of manipulations and substitutions that could be possible for any given equation, but knowing where and when to apply certain techniques is crucial to solving rational equations correctly. In this post, I am going to go over several concepts that will be useful to you when it comes to solving rational equations.

To start, I will explain first-degree equations in one variable. Quite simply, a first-degree equation is one in which there is only one variable. In general, a first-degree or linear equation has the form:

ax + b = 0, where a and b are real numbers and a does not equal 0

I'm sure you are extremely familiar with this type of equation, though you may not know it by this name. An example of a linear equation would be something like 3x - 12 = 0. Undoubtedly, you can easily see that this equation is true when x = 4. However, it is good to realize that the expression is neither true nor false until you substitute in a value for the variable. Any value that makes the expression correct is called a solution or root of the equation. To further classify this equation, 3x - 12 = 0 is also called a conditional equation, in that it is only true for certain values.

When you have two expressions that have the same solution (or root), these are called equivalent equations. Again, I'm sure you are familiar with the concept, but probably unfamiliar with this name. When you have a first-degree equation in one variable, the general strategy that you typically employ is to express the equation equal to a series of equivalent equations, which you manipulate until you can reduce everything down to the solution to the equation.

The rules for generating equivalent equations are simple and intuitive.

You can add or subtract the same value from both sides of the equation. (A corollary to this is that you can add AND subtract the same value on one side, without changing the other... since adding x and then subtracting x means you really have done nothing!)
You can multiply or divide each side of the equation by the same value.
You can simplify one side of the equation without affecting the other side of the equation.

I think these rules are fairly self-explanatory, so I'm not going to bother going into any examples to demonstrate them.

When you have arrived at your solution / root of your equation, it is ALWAYS smart to take that value and substitute it into the original expression to verify that it is indeed true. It always amazes me how many people arrive at incorrect answers and leave it at that, when a simple review and check can either tell you that you are correct, or your answer needs more work. ALWAYS REMEMBER TO CHECK YOUR ANSWERS!

By checking your answers by substituting the solution into the equation, you sometimes will determine that the solution you have found CANNOT be true, in which case your solution is called an extraneous root. An example of this would be where, when checking your solution, you determine that you have a 0 on the bottom of a fraction (the denominator). A fraction with a zero in the denominator is undefined, and so you can conclude that the root you determined does not satisfy your equation. Extraneous roots may develop especially if you use rule number 2 above, but you multiply both sides by an EXPRESSION rather than a single number. (eg. you multiply both sides by (x + 2))

That is all I am going to say about how to solve equations for now, especially the first-order equations (or linear equations). I will continue in my next post with a discussion of solving quadratic equations.

Principal Axis Factoring

2011-10-09T11:46:00.000-07:00

I'm just going to put a brief mention up here on principal axis factoring, a theoretical statistical method used to analyze common variance. Principal axis factoring, also known as principal factor analysis or common factor analysis, is the most commonly used method in factor analysis. According to Wikipedia, it "seeks the least number of factors which can account for the common variance (correlation) of a set of variables". Factors are determined through the analysis of the common variance.

This concept of factor analysis is beyond the concepts that I would like to explain on my blog. It is much more of a university-level statistics concept than any of the other concepts that I have described so far. If you would like a more thorough explanation of principal axis factoring, I highly suggest that you perform a Google search or find a good reference text book. I mention it here only because of my recent posts on methods of factoring, and I was asked to specifically to reference this factoring method.

AC Method Factoring

2011-10-05T23:08:00.000-07:00

"AC method factoring" is another method of factoring you can use when working with trinomials. That is, adding something of the form ax² bx c factoring. This can be added to my list of methods of factoring that I posted about earlier, but is specifically used for trinomial factoring. The AC method is a simple method that will help you to solve problems faster, and it really isn't all that complicated. However, as with any of the factoring methods, you have to learn and understand the steps to perform it.

AC method factoring requires only a few steps, so I will go over them with you now, and then show you an example or two. To start factoring equations, we need to have a polynomial (trinomial) in the form for adding the terms ax² bx c. We'll start with one demo just using letters, where x is the variable and the other letters are constants:

Ax² + Bx + C

Now, here is what you to for the AC method:

1. The first step is to multiply the A and the C constants. Just take the two numbers (see the example that follows), and multiply them. Keep this value in mind.

2. Next, we want to find two values whose sum equals B, and which multiply to the product we found for AC. This might take some trial and error, but when you find these two numbers, let's call the smaller one M and the bigger one N.

3. Rewrite the original equation (Ax² + Bx + C) by substituting in our values for M and N. That is, we substitute in Mx + Nx = Bx. So then, we now have:

Ax² + Mx + Nx + C

4. Once we have the expression in this form, we can now try to factor by grouping. (Refer to my earlier post about methods of factoring to review grouping, if you need. I will also be posting more about this trick soon.)

5. Rearrange the grouped expression by the distributive property to arrive (hopefully) at your final solution!

This method of factoring will come in quite handy when you come across trinomials, especially ones that have a value for A and don't appear to have an immediately obvious way to factor or simplify. AC method of factoring will come in very handy!

Let's try factoring an example now with real numbers, using AC method factoring. Follow along with the steps I've listed above.

Factor: 2x² + 9x + 9
Step 1: AC = (2)(9) = 18
Step 2: Two numbers to sum to 9 and multiply to 18. Pairs can be 1 and 8, 2 and 7, 3 and 6, etc... this should be evident that the pair we like is 3 and 6. They sum to 9 and multiply to 18. So, M=3 and N=6
Step 3: = 2x² + 3x + 6x + 9
Step 4: = (2x² + 3x) + (6x + 9). This can be factored to x(2x + 3) + 3(2x + 3).
Step 5: Rearrange: (2x + 3)(x + 3).

And this would be your solution. You just used the AC method to factor a trinomial! I hope that demonstrates this method for you and that you can understand the steps you need to perform. Let me know if you'd like more examples, and I can put some up! As always, please +1 me below if this helped you!

Square Roots - Part III (Factoring Square Roots)

2011-10-03T21:52:00.000-07:00

Where I left off in my previous post about irrational numbers, we were trying to solve for the square root of 24 by trial and error. Reducing an irrational number can be a tedious job if you do it this way! Luckily, unless specified otherwise, you are allowed to leave your answer in the "simplified radical form," which you can get by factoring square roots.

"Simplified radical form" is exactly what it sounds like: you simplify your expression and leave it expressed as some radical. But, you have a little bit of work to do to reduce it.

Factoring square roots is quite simple. It has to do with factoring perfect squares. But first, you have to determine the factors of the number under the radical sign, and then if any of those factors are perfect squares, you can pull it (the square root of the perfect square factor) out from underneath the radical sign and put it in front to multiply by it. That sounds awfully wordy and probably isn't the most concise definition, but I think an example will go a long way to helping you understand factoring square roots, and then leaving them in simplified radical form.

Let's continue with the example from my last post, looking for the square root of 24. In this case, let's just reduce it to its simplified radical form, and not bother wasting time trying to find the exact decimal answer.

So, the first step is to ask yourself what the factors of 24 are. (I'm going to ignore the √ for a minute) For this, you can determine this quite easily by trial and error. You can find that the factors of 24 are 1,2,3,4,6,8,12,24. From these now, you want to see if any are perfect squares. Again, you should be able to easily say that 4 is the perfect square. So, 24 can be expressed as 4 x 6. That's what we want. Now, let's look at what we have done so far:

√24 = √(4x6)

Now, as I said before, we can take the perfect square that is under the radical sign, and bring it outside the sign. This is because of the properties of square roots (property #2), which in this case allows us to write:

√(4x6) = √4 x √6

With that property in mind, it should be a bit easier to see why we are interested in the perfect squares, because now, by factoring perfect squares, we can just rewrite the square root of 4 to 2! So, we can finally write our expression in simplified radical form as:

√24 = 2√6

I hope this demonstration has explained to you the basics of factoring square roots, and leaving you answers in the simplified radical form. Let me know in the comments if you'd like another example, and I'll do one for you! Remember to +1 me if this helped you! :)

Find Factors: Methods of Factoring

2011-09-28T23:54:00.000-07:00

For this post, I would like to do one that talks about various methods of factoring which you can use to find factors. There are several different factoring techniques that you can use, depending on the situation. So, I think that I would like to include a variety of methods or factoring tricks... perhaps more of a reference page than one of my usual 'lesson' posts. I will try to update it as we I go along, adding new methods of factoring as I cover them in more detailed posts. I'll also provide links to those lessons for easy access as I post each of the factoring techniques.

First of all, a high level, general definition. What does factoring mean? Factoring means to simplify a mathematic expression by writing the expression as a product of two or more values / expressions. For example, if you factor the expression 12x + 4, you get (4)(3x+1). This is an example of "factoring out" the 4. You will see this phrase "factoring out" very commonly when dealing with these types of problems.

Several factoring techniques are available to you to help you find factors, depending on the question. These are some techniques you can use. Refer to their main post (as they become available) for more descriptive definitions than those given here. You can memorize these factoring tricks and shortcuts and you will save a lot of time with your math work! If you do lots of practice with factoring games or math worksheets, you will become good at these and be able to find factors very quickly.

Common factor
In an expression composed of multiple terms, try to identify if there is one number/variable that is a common factor to each term. Then, after factoring out the common factor, you can rewrite the expression to show multiplying that common factor by the remaining terms.

12x + 4 = (4)(3x + 1)

5x³ + 10x² + 25x = (5x)(x² + 2x + 5)

Difference of squares
Wherever you see a difference of two terms that are perfect squares (either something like x²or 25), you can apply this technique for factoring a difference of two squares. This actually is the same as one of the rules for special polynomial products. In fact, factoring special products follow the same rules you would use to find factors anywhere else. Factoring squares is actually quite simple:

x² - a² = (x - a)(x + a)

You reduce the terms to their square root value, and remember to put one + and one -. Easy as that.

Difference of cubes / Sum of cubes
Factoring perfect cubes (or factoring cubics) is a little trickier, but they follow a strict form that you can memorize and use easily to find factors. Depending on if you are subtracting perfect cubes or adding perfect cubes, you will use the appropriate formula:

x³ - a³ = (x - a)(x² + ax + a²)

x³ + a³ = (x + a)(x² - ax + a²)

They have the same basic form, you just have to pay attention to the signs. Don't mix them up!

Grouping
This is an extension of the common factor method described above. The only difference is that the common factor doesn't have to be common with EVERY term. Group things together, and factor within the groups. Take this example and you should see what I mean:

x³ - x² + x -1

= (x³ - x²) + (x -1)

= x²(x - 1) + (x - 1)

= (x - 1) (x² + 1)

Study that example and it should be fairly self-explanatory how I used grouping to find factors here.

Trial and Error
Sometimes, there aren't any obvious factoring tricks or factoring techniques that you can apply to help you solve your question. Unfortunately, in these situations, you must resort to trial and error. Sometimes you can figure out the numbers that are involved, but you need to test out the signs to get it right. I can't really say anything about this technique except to have patience and keep trying. :(

I will revisit this post shortly to put up a remark about factoring quadratic equations. I will also go over AC method factoring. The AC method of factoring is a factoring method or factoring trick you can use to help you factor expressions in the form of ax² + bx + c (trinomial).

Special Polynomial Products

2011-09-22T23:20:00.000-07:00

Following up on my previous post about Polynomials in Mathematics, I wanted to dedicate a brief post to special polynomial products. These are products that will occur very frequently throughout your mathematics studies, and so it is a very good idea to remember these products.

If you would like to refresh on multiplying together polynomials in general, follow the link to view my previous post about polynomials. Alternately, if you would like to see how to FOIL polynomials, a special method you can use to multiply together binomials, I refer you to my previous post on this as well.

Now, on to the Special Polynomial Products.

These are not overly complicated, and can be derived using basic knowledge of multiplying together polynomials. But, as I said, these will occur so frequently, that you might as well memorize them so you can save time and not have to worry about deriving them every time you see them.

(A - B)(A + B) = A² - B²

This first one is the easiest to arrive at. Notice that the signs are OPPOSITES! Simply apply the FOIL method to the starting expression, and you will arrive at the simplified product. I'll leave that for you to check for yourself.

(A + B)²= A² + 2AB + B²
(A - B)²= A² - 2AB + B²

This set is also quite simple to see, once you FOIL them out. The starting terms are squared, you can easily rewrite them as (A+B)(A+B), etc, apply the FOIL method, and come up with the product above.

(A + B)³= A³ + 3A²B + 3AB² + B³
(A - B)³= A³ - 3A²B + 3AB² - B³

This set, obviously, is a bit tougher to arrive at, but by tougher I only mean 'requires more work'. There aren't any tricks and they aren't any more difficult than the others. I'll derive one for you at the end, after I show you the final set of special products, which is...

(A + B)(A²- AB + B²) = A³ + B³

(A - B)(A²+ AB + B²) = A³ - B³

This set requires the most work to derive, so you can see what I mean when I say it pays to memorize these so that you don't have to mess around every time you see them! I won't derive these ones, but you should try to on your own so that you see how they work and how to arrive at the products IN CASE YOU FORGET THEM! It's ALWAYS a good idea to understand how to do all these things so that you can do them WITHOUT the shortcuts.

Anyways, now I will get you started by showing you how to multiply (A + B)³ and arrive at the special product I showed you above:

(A+B)³.....rewrite this
(A+B)[(A+B)(A+B)]..... note I've isolated 2 binomials so that I can show you to FOIL easier
(A+B)(A² + 2AB + B²)..... now, multiply the two polynomials together. Multiply everything in the second term by A, then multiply everything in the second term by B.
(A)(A²) + (A)(2AB) + (A)(B²) + (B)(A²) + (B)(2AB) + (B)(B²)
A³ + 2A²B + AB² + A²B + 2AB² + B³... Now you can combine like terms.
A³ + 3A²B + 3AB² + B³

And there you have it. It took a little bit of work, but it's not that hard. You just have to keep going at it.

So, those are the special products of polynomials. Try to memorize them to make you homework easier, but MORE IMPORTANTLY, please try to work through the derivations and understand HOW to arrive at the products.

How to FOIL Polynomials

2011-09-22T22:41:00.000-07:00

This post is just a quick refresher of one of my early posts on polynomials that explains how to FOIL polynomials. This is an important lesson, so I just want to reiterate the process once more.

You would use the FOIL method when you are trying to multiply two binomials together...that is, an expression in the form (x + A)(x + B).

The term FOIL stands for First Outer Inner Last (sometimes also using Outside and Inside instead). This refers to the order of multiplying terms together.

I will explain it through an example, which should hopefully be all that is necessary. Let's try to find the product of (x+4)(x+6).

To apply the FOIL method, you have to first identify and understand what terms we will be talking about.

Here is what the letters in FOIL mean:

The term First means "the first terms in each binomial." In this case, x and x are the First terms.
The term Outer means "the outer terms of the product expression." In this case, the first x and 6 are the Outer terms.
The term Inner means "the inner terms of the product expression." In this case, the 4 and the second x are the Inner terms.
The term Last means "the last terms in each binomial." In this case, 4 and 6 are the Last terms.

So then, now that you have identified each of the components of FOIL, now all you do is MULTIPLY the two terms for each letter of FOIL, and then add them up (there will be 4 products).

For this example, we'd have these products:

F: (x)(x) = x²

O: (x)(6) = 6x

I: (4)(x) = 4x

L: (4)(6) = 24

Now, we just add them together, combining like terms wherever possible (here, the inner and outer terms):

x² + 6x + 4x + 24

x² + 10x + 24

And that's all there is to the FOIL method. If you'd like more examples or a better explanation, leave me a comment. But I think that the example should be sufficient for now to demonstrate the concept.

Cool Math Games

2011-09-22T00:02:00.000-07:00

Cool math games are a great way for students to learn valuable math skills without it feeling like studying. These games allow the students to learn and reinforce mathematics concepts that they have been taught, but in the context of a fun game. Obviously, everyone likes having FUN more than having to study. But with educational games, it won't even feel like studying, and they will help develop math skills. Math games can be important learning tools for students of all ages, and should be tried by everyone studying math. \instead of search for math lessons online, trying searching for some of these. Cool math games will make learning math fun!

Math games are especially useful when teaching young kids various math concepts. Younger students may have a tendency to lose focus on new / hard subjects such as math, and so presenting new concepts using traditional methods may not always be the most productive way of doing things. Perhaps a better plan would be to describe the concept as simply as possible, but then instead of assigning repetitive homework problem sets, allow the students to play math games. These fun activities for children will keep their attention much longer than a boring text book. But as opposed to them wasting their time when they need to take a break from regular studying, they will be developing their math skills while playing these educational games. These games shouldn't replace conventional homework, but fun math games are a great way to compliment it.

Here is a sample of the most popular math games on the internet right now (courtesy of Coolmath.com, one of many game websites):

Bloxorz

Boombot

Fraction Splat

Xfactor

Duck

Concentration

(I'd recommend giving their site a visit to see if you can find a game for the subject you are studying. It's a great website for kids.)

Another benefit of playing cool math games is that these games develop focus and concentration skills in order for the student to win or succeed. Furthermore, they can help develop competition and communication between students as they try to score more points or achieve higher goals in the games. This motivates them to want to do better and so will strive to understand the concepts. Compare these benefits to the process of having to repeatedly write out math problems (boring!), or get frustrated with new concepts, possibly while studying alone and not in a social environment. From this point of view, cool math games provide many benefits over traditional written homework and studying.

Math games are increasingly becoming more popular amongst both students and teachers. While there are still many teachers who focus on the old-school teaching methods and styles that they know and themselves were taught with, many are now seeing the benefits of fun math games on the students' ability to learn new mathematics concepts. In the end, it shouldn't be about HOW the student is taught these skills, but rather it should only be a matter that they ACHIEVE these skills. Cool math games are a great way to assist in this goal.

Zero Exponent and Negative Exponents

2011-09-18T12:09:00.000-07:00

Just a short post to follow up on my last one about the properties of exponents. In that post, I explained to you various properties of exponents that allow you to do arithmetic with them to simplify them into easier to work with expressions. However, I only used positive exponents in my explanations. Here, I need to briefly explain two more concepts: zero exponents and negative exponents.

Zero exponents are simple. They follow one single rule that applies wherever you will see them. No matter what the base is, if it is as simple as x, or as complicated as (45xyz / (412ab+26c)), if you raise the base to a power of zero, the expression equals 1.

a⁰ = 1

It is a very handy shortcut, so whenever you see it, you can make life a lot easier for yourself if you just simplify the expression to 1. For example, you don't want to get stuck rearranging to solve for a in that 45xyz... example above, when you can simplify the whole thing to 1.

So that is zero exponents.

Next is negative exponents. They aren't QUITE as simple, but the trick to working with them is. And here it is:

Whenever you see a negative exponent, all you do is make it positive, but put the whole thing on the bottom of a fraction under 1, like so:

a^-n = 1 / aⁿ

Once it's like that, you can apply any or all of the properties of exponents to it without having to deal with the negative anymore.

Just to put some numbers to it, here is an example for you to see what I mean:

(a²b³)^-4
= 1 / (a²b³)⁴
= 1 / a⁸b¹²

Hopefully you can see what I mean with these examples. They are just a few shortcuts that will make working with exponents a lot easier for you. Let me know if you need any more explanation about these topics, but for now I will leave it at that.

Properties of Exponents

2011-09-18T11:53:00.000-07:00

When working with exponential notation, it is useful to understand some of the basic properties of exponents. This is probably review for many of you, but I will post this here now for those who are just learning it for the first time or need a bit more help to understand it.

First, let's examine the form of exponents.

If we have a real number, a, and a natural number, n, we can define the exponential expression aⁿ as:

aⁿ = a x a x a ... a (n factors of a),

where we say that a is the base and n is the exponent. The exponent is also frequently referred to as the power.

Now, knowing the basic form of an exponential expression, let's look at some of the properties of exponents that will prove to be very helpful to you in the future! These will be used over and over again, and will be the basis of future lessons, so it will be very good for you to understand these well. There are really only four of them, so it shouldn't be too hard to memorize them.

1. The first property defines how to multiply exponents. If we have a real number, a, and natural numbers n and m, we can say:

a^maⁿ = a^m+n

So, what this basically says is that when you have exponential terms to multiply together, if they have the same base number, you can simply ADD the exponents. To demonstrate this, we can write out an example. If we want to simplify a²a³, we can expand it to show [a x a] x [a x a x a], which is a x a x a x a x a, which is a⁵. The base a was common, so we could just add the 2 and 3 in the exponent. Same thing applies when the terms are bigger. Try for yourself to prove that (x+1)²(x+1)⁵ = (x+1)⁷.

2. On the other hand, a similar property exists to divide exponents. If you want to divide exponential terms, you just subtract the exponents, like this:

a^m/ aⁿ = a^m-n

Prove to yourself that a³/ a²= a, in the same way I demonstrated above.

3. If we want to raise an exponential term to another power, such as with (a^m)ⁿ, we can simplify it to a^mⁿ. Again, try it yourself by writing it out like I did above to demonstrate this with the expression (a⁵)² = a¹⁰.

4. Finally, similarly to multiplying a coefficient times everything inside a set of brackets, if we want to raise a product or quotient expression by a power, the exponent can be written for each term, like so:

(ab)^m = a^mb^m, or

(a/b)^m = a^m / b^m

In this case, it should be simple to see that this means something like (ab)² = a²b². If you want to start combining these properties, you can see something like (a⁵b²)² = a¹⁰b⁴.

Hopefully with these examples you can see that the properties of exponents are not that complicated to work with, and once you get the hang of the basics and can start combining them, they are actually very easy to use!

Polynomials in Mathematics

2011-09-10T15:53:00.000-07:00

In this post, I would like to briefly refresh the concept of Polynomials in Mathematics, as at the start of the school year there are always students looking for extra polynomials help.

I would like to explain some basic terms and definitions that will hopefully make working with polynomials less confusing.

The first two terms which are fairly obvious are "variable" and "constant". You probably don't need me to tell you that constants are terms that do not change (they are fixed... they don't always have to be known, but they do not change), whereas you can change the value of the variable to arrive at a different solution to your problem. As in, if you vary the length of the square, you calculate a different area of the square. The general practice is to assign letters near the start of the alphabet as constants (a, b, c), and at the end for variables (x, y, z). These terms are self-explanatory, so I will leave them at that.

Substituting values into your variables only makes sense in some cases. E.g. sub in -4 for x, when the expression is square root of x. It doesn't make sense. Therefore, the set of numbers that DO make the expression make sense is called the "domain of the variable".

An example of this concept can be seen here. Find the domain of the expression 1 / (x-5).

You can substitute any number in for x and calculate a solution, except for one. You cannot have a zero as the denominator, as that makes the expression undefined. So, in this case, you have to ask yourself "what makes the denominator equal zero?" You can calculate in this case that x - 5 = 0, and so x=5. Therefore, the domain of the original expression is all real numbers, except x cannot equal 5.

The term "polynomial" is given when you have a series of "terms" in an expression with different powers of x as long as the powers are not negative. In general, a polynomial looks like this:

a_nxⁿ + a_n-1x^n-1 + ... a₁x + a₀

Each x value is contained in a "term". The a values are "coefficients", which are constants. They do not vary. Overall, the "degree" of the polynomial is the highest power of x. General convention is to right the terms from the highest degree of x to the lowest.

Since a polynomial is a generic collection of many (poly = many) terms, specific forms of polynomials have special names. A "monomial" is a polynomial that only has one term, such as 2x. A "binomial" has two terms, such as 4x² - 1x. And a "trinomial" has three terms (guess what it looks like!).

Polynomial terms of the same degree can be combined.

In the expression 3x + 2x + 1, you can combine the terms so that you simplify to 5x + 1. You can do this with x²terms and higher as well.

There are a couple of special polynomial products that will appear very frequently in your studies. I will also go over these in my next post so that you can memorize them to apply them easily. Hopefully, when you see these, you will understand how they work and will not need to look for polynomials help anymore!

Radicals and Rational Exponents

2011-09-05T23:03:00.000-07:00

Having discussed the topic of radicals and nth roots in my previous set of posts, we can apply that information to the concept of fractional exponents. These rational exponents are no more difficult to work with, and once you identify just what they are telling you, you can easily apply your rules for exponents to simplify them and solve your equations. Factoring fractional exponents may look tricky, but you will see that they follow the same rules that you already know.

Consider the expression 2^1/3. Knowing your rules for exponents, you can see that if we cube this so that we get (2^1/3)³, this is equal to 2¹, which is just 2. However, what you will realize is that what we have just done is apply property #1 of the properties of nth roots. That is, (ⁿ√x)ⁿ = x. So, what we have just shown is that 2^1/3 is the same as ³√2.

So then, with that example we can now extend what we've done to come to a definition of rational exponents and roots.

Part 1: If we let b be any real number, and n is a natural number, b^1/n can be defined by ⁿ√b. (if n is even, b must be greater than or equal to 0).

Furthermore, we can extend our definition of exponents fractions to include any fraction, not just 1/x type.

Part 2: If we let m/n be a rational number, n is positive, and ⁿ√b exists, we can say that b^m/n = ⁿ√(b)^m or (ⁿ√b)^m. (Both forms say the same thing and work the same way.)

It is important to realize and understand that these rational exponents follow the same properties as integer exponents, which you are likely already familiar with (I'll do a post soon to refresh you, or for those who have not learned that topic yet).

So now that you know what rational exponents are, let's take a look at a few examples of how you work with radical and rational exponents. Simplifying rational exponents takes some practice, but it is all the same math that you have already learned.

1. Simplify the following expressions:
a) 64^1/2
b) -64^1/2
c) (-64)^1/2
d) 64^-1/2

a) This is simply the square root of 64, √64, which is 8.
b) Recall the notation with negative signs and nth roots. This means -(64^1/2) = -√64 = -8.
c) As there is no such real number that gives -64 when squared, this expression cannot be simplified to a real number.
d) 64^-1/2 indicates √64^-1, and a negative exponent means that you take the inverse of it. So,64^-1 means 1/64, and then the square root of this is 1/8.

2. Simplify (4x^2/3)(3x^1/4)
(4x^2/3)(3x^1/4) = 12x^{(2/3) + (1/4)} (by the exponent rules)
= 12x^11/12

3. Simplify (√x)(³√y⁴)
= (x^1/2)(y^3/4)
Alternatively, if we give the fractions a common denominator, we can rewrite as follows:
(x^1/2)(y^3/4) = (x^2/4)(y^3/4) = (⁴√x²)(⁴√y³) = ⁴√(x²y³)

So, there you have it. Working with radicals and rational exponents is not that different from working with any other exponents, once you know what you're doing. Go over my examples again to be sure you understand the concepts, and practice some more on your own, and you will be an expert in no time. Drop me a line in the comments if you have any problems.

One final note: irrational or radical exponents follow the same rules as well, and I will cover some of those examples in a future post.

Rationalizing Denominators

2011-09-01T22:59:00.000-07:00

In my previous posts, about nth roots and the properties of nth roots, I mentioned that I would post next about rationalizing denominators, and how to rationalize the denominator. This may be a new concept to some students, so I will explain it first and describe where it will come in handy, before I actually show you how to do it.

Rationalizing denominators is a way of simplifying radicals, specifically fractions that have a radical on the bottom (in the denominator). You may come across expressions like these, for example, when dealing with property #3 of nth roots. In fact, you will soon find that equations are often not considered 'completely simplified' if there is still a radical in the denominator. To get rid of this, you must rationalize the denominator.

Here's how you do it. I think you will find it much easier than its name makes it sound!

First, let's consider a couple of points.

The first one is an obvious one, and you will see why in a second... that is, x / x = 1. Right? Anything divided by itself is 1 (except zero). 3 / 3 is 1. 100 / 100 is 1. 57483 / 57483 is 1. That's how fractions work: if you have all of the pieces of the pie, out of all the pieces of the whole pie, you have the whole pie. Easy, probably doesn't even need to be said. But keep it in mind.

The second point is a refresher of the properties of nth roots... specifically, property 1 which says (ⁿ√x)ⁿ = x. So, for n = 2, we have (²√x)² = (√x)²= (√x)(√x) = x

So then, with those two points in mind, here is the trick to rationalizing the denominator when there is only a radical down there. Quite simply, you take the expression that you have, and whatever radical is in the denominator, you multiply both the top and the bottom by that. Let me show you what I mean:

Take 2 / √5. To rationalize the denominator, we multiply the top and the bottom by √5. So, this is what we get:

Hopefully, you can easily see how we really haven't really changed anything at all. We multiplied the whole thing by 1, and then simplified the top and bottom using simple multiplication and division rules that you already know. And that would then be considered to be fully simplified. However, I should point out that this is all that's necessary when there is ONLY a radical in the denominator and nothing else.

If we had √5 + 2 in the denominator, that needs a bit more work. It's still rationalizing the denominator, because we 'make the denominator rational' by getting rid of the radical. But, when there's a plus or minus down there as well, there is an extra twist.

We still multiply the whole thing by 1 (that is, by something over something), but now it's a bit more complicated. If you try to multiply top and bottom by √5 + 2, YOU ARE INCORRECT. You can try (just follow the typical FOIL rules for multiplying binomials), but you will find that you end up with more of a mess on the bottom... more terms than you start with! It doesn't simplify anything at all!

To properly get rid of this radical, you multiply the top and the bottom by √5 - 2. That's right, notice that the sign in the middle is flipped! This term is called a conjugate, and the two terms with the opposite sign in the middle are conjugates of each other. If you FOIL out (√5 + 2)(√5 - 2), you end up with 5 - 4 = 1 on the bottom! (It won't always be 1, but you will always get rid of the radical and can simplify further!).

So, let's try another example to put it all together.

Rationalize the denominator and simplify: 5 / (2 + √6)

And that would be your simplified answer! Note that I multiplied the top and bottom by -1 (still not changing the equation!) to switch the negative signs around so that I didn't have a negative on the bottom. Not always necessary, but it looks much cleaner. (Technically, some teachers would say the proper answer is (5√6 / 2) - 5. That works too.)

So that is all there is to rationalizing the denominator, and using conjugates to do it. Since this is now a fairly long post, I'll leave it there. But if you'd like another example or two with variables instead of numbers, or nth roots higher than squares, leave me a note in the comments and I'll either update the post or do another one with more!

Properties of nth Roots

2011-08-31T22:06:00.000-07:00

Having discussed the concept of nth roots in my last post, here I want to show you some very important Properties of nth Roots that will help you to understand and solve you homework questions with nth roots. There are only a few, and I think you will find that they seems pretty easy. It's good for you to see them though, and realize that there are specific properties and rules that apply to nth roots.

Before we get into it, you may or may not have been taught this already, but I will briefly mention it here because I think it may help clear up some confusion with the rules of nth roots. When you have a root, you can express that as an exponent and it means the same things. And you do that by considering the index to be a fraction exponent. Probably not a very good choice of words, but this should be clear with this short demonstration:

²√x = x^1/2

If you get confused or muddled up with the following properties and rules, try thinking of them in terms of this example, and then you should be able to apply the rules of exponents that you already know that help you sort things out. Square root notation is just a fancy, specialized way of working with fractional exponents.

So, here we go.

To start with, we have to state some assumptions and generalizations. We let x and y be real numbers, and m and n are natural numbers. In this case, the following properties are true as long as the expressions are not undefined:

1. (ⁿ√x)ⁿ = x
Like I said, most of these make sense if you think about them for a second. In this case, consider that most (if not all) arithmetic functions have opposites. Addition is opposite to subtraction, multiplication to division. In this case (let's assume n = 2 for a second) square root is opposite to squaring. Or, cube root is opposite to cubing. And so on. Square root of 16 is 4. 4 squared is 16. They cancel each other out, and you're left with x. Makes sense? E.g. √x²= x

2. ⁿ√xy = ⁿ√xⁿ√y
Again, this one kinda makes immediate sense. Kind of like multiplication of a polynomial, you multiply everything inside the brackets by what is on the outside (ie. 2 x (x + 3) = 2x + 6 ). In this case, the root is applied to everything beneath the root sign. It is important to make sure that you realize that this law holds only for multiplying (and dividing... see #3) terms beneath the root sign, but does not work for adding or subtracting terms. So, ³√xy = ³√x³√y

3. ⁿ√(x/y) = (ⁿ√x) / (ⁿ√y)
This is a follow-up to #2. As long as the terms beneath the root sign are multiplying or dividing, you can re-write the expression like this, with a radical sign on the top dividing by a radical sign on the bottom. Again, this doesn't apply if you are adding or subtracting x and y beneath the radical sign.

4. a) if n is even:   ⁿ√xⁿ = |x|
This is connected to how you can square a positive or negative number and get the same POSITIVE result. So, if you try ²√(5)² you get positive 25. Same thing if you try -5... positive 25. If the n term is even, you will always get the positive, or ABSOLUTE VALUE, of x. However...

b) if n is odd:   ⁿ√xⁿ = x
In this case, you CAN finish with a negative number. If you try  ³√(5)³ you get 5, but if you try ³√(-5)³you get -5. Write it out long-hand to see why. -5 x -5 x -5 = -125. Cube root of -125 is -5. There is only one answer.

5.   ^m√(ⁿ√x) = ^mn√x (Sorry, this one is hard to type out)
Once again, this one is kind of like multiplication, and the rules of exponents. If you have (x,²)², the rules of exponents say that you can multiply the exponents to have x⁴. In this case, the numbers are in a different place, but the same kind of rule holds. ²√(³√x) = ^2x3√x = ⁶√x.

You will likely find most use of these properties in simplifying larger expressions, making them easier to work with. Such as, pulling out perfect square or cube from a larger number:
E.g. √72 = √9√ 8 = √9 x √4 x √2 = 3 x 2 x √2 = 6√2

In my next post, I will explain a concept that always seems to give students difficulties... conjugates, and rationalizing the denominator.

nth Roots

2011-08-28T22:33:00.000-07:00

This brief post is going to explain to you what nth roots are. Following posts will show you how to work with them in your equations.

Going back to elementary algebra lessons, you were taught the concepts of square roots. For example, the square root of 25 is 5. This notion was explained, but likely never generalized much past saying "what times what gives you your number?" However, like many things in the mathematics world, there is far more to roots than this.

Enter the concepts of "nth roots".

Everyone knows the symbol for square root, and that it means "what times what gives the number underneath this symbol". Some of you may have seen a tiny 3 written on the top left just above the root sign. What this means is a touch more complicated: "what times what times what gives the number underneath this symbol?" It is called the cube root.

It's always simpler to go forwards before going backwards, so let me show you one:

2 x 2 x 2 = 8

You can say 2³ (2 to the power of 3) equals 8. Straight-forward, right? Now, go backwards. Find the cube root of 8. Well, in this case, "what times what times what" is "2 x 2 x 2", so the cube root of 8 is 2.

So then, we now understand square roots, and cube roots. Now, naturally, we can expand on those. In general terms, nth roots can be defined as this:

a^n = b (a to the power of n equals b)

and therefore a is the nth root of b,

where n is a natural number, and a and b are real numbers.

Some examples should hopefully clear up the complexity of what I just wrote.

3 and -3 are square roots of 9, because 3² = 9

4 and -4 are fourth roots of 64, because 4⁴=64

2 and -2 are sixth roots of 64, because 2⁶=64

Further, what these examples should demonstrate to you is that all EVEN roots of postive numbers occur in pairs, with there being one positive and one negative number for each. Therefore, to distinguish between the two, there has been accepted a common notation:

The positive, or principal, nth root is designated as the root sign with the nth root number over it: ⁿ√
E.g. ⁴√16 = 2, and NOT -2
To indicate -2, you write the initial expression as -ⁿ√
E.g. -⁴√16 = -2

On the other hand, all ODD roots only occur singly. Such as 2 is the fifth root of 32, whereas -2 is the fifth root of -32. (Write it out to see the differences. -2 x -2 x -2......)

A little bit more nomenclature, just so you always know what is going on now:
What you have always knows as the "square root sign" is technically called the radical sign.
The number underneath the radical sign is called the radicand, and the number n used to indicate the root is called the index.

To summarize all of this, we can revise our original definition to this:

If n is a natural number, and a and b are non-negative real numbers, then

ⁿ√b = a, if and only if b=aⁿ

^{The number a is the prinicpal nth root of b}

Also:

If a and b are negative and n is an odd natural number, then

ⁿ√b = a, if and only if b=aⁿ

^{So there you have it. You now know what is meant by nth roots. In my next post, I will go over some of the properties of nth roots that will make your homework questions a lot easier.}

How to Show Inequalities and Absolute Values on a Number Line

2011-08-17T23:01:00.000-07:00

This will be a short post, but I wanted to quickly go over something I touched on in my previous post about properties of absolute value, which is how to graphically show absolute value and inequalities on a number line. I mentioned that there are a couple of acceptable ways to do this, so I will try to explain them here.

I want to first go over the concept of inequalities briefly. It is all very intuitive and common sense, but you will see why I want to point this out now before we put inequalities on to a number line.

If I give you the example x < 2, we know exactly what that means. In fact, the words you say when you read that mathematical expression pretty much says it all: x is less than 2. Great. Similarly, the example x < 2 is also as obvious: x is less than or equal to 2. That was the easy part, but the key is to always pay attention to what the sign is saying. Does it say "less than" or "less than or equal to" something. Let me show you why this distinction is important when it comes to graphing inequalities on a number line.

Let's take that first examples again, x < 2. To put this on a number line, there are 3 important things here: the point in question (2), the direction on the number line that makes the expression true (less than), and does the inequality contain the point, or exclude it (less than, or less than or equal to).

If the sign being used is either < or >, without the equals part, you can represent this on a number line as an empty circle at that point. Alternatively, you could show it with a round bracket. This notation implies that you are talking about points that get infinitely close to your number (2, in our example... such as 1.9999999999), but not exactly 2. In addition to that, for all the values that make the expression true you put a line over the number line, or draw a thicker line on top of the number line. If the true values go all the way to infinity, you can put an arrowhead on the line. So for our example of x < 2, we would represent it like this:

or like this:

Now, when we are talking about an expression that has an equals part to it, that is simply represented by a filled circle on the number line, or a square bracket. Note that both kinds of brackets open towards the direction that the expression is true. So, for our second example of x < 2, we could show this on a number line like this:

or like this:

I should probably apologize for the shoddy artistic skills. ;)

By looking at these number lines, and the ways that we have represented the inequalities, it is very easy to see that we are talking about x < 2 in the first pair, and x < 2 in the second pair.

So, that is most of what I wanted to explain about inequalities and number lines. Now, to apply these ideas to solving an absolute value question, it should be straight forward. And when in doubt, just substitute in numbers for x to see if the inequality is true or false!

If we are talking about |x| < 2, we saw in my original post on absolute values that this simply means that x is 2 units away from zero. It doesn't specify only positive numbers, or negative numbers, or any type of restrictions on the question. Therefore, 2 units away from zero in both directions encompasses the interval of -2 to 2. So, to show this on a number line is simply two open circles (on -2 and 2), and a thick line connection to two points to represent that all number within that interval make that inequality true. Try it and see for yourself! 1.999934958... true. -1.21111... true. 0.00000000003... true. The absolute value of these values for x are all less than 2.

I don't think I need to give anymore examples to explain this any further. It's pretty simple stuff, right? If you do need a bit more explanation or you would like additional examples, don't hesitate to drop me a line and I'll see what I can do!