Simplifying Radical Expressions

My Algebra I students are having lots of problems with simplifying radical expressions. It seems that they have run across a concept that eludes them. I am posting about this topic to provide several more examples.

√18

First, we find the prime factorization of 18. If you need help with this concept, go here.

√18 = √(2•3•3)

Since the 3 is repeated twice, we can pull it out from under the square root symbol. Therefore,

√18 = 3√2

Now my students would be asking what happens if you have variables?

√24x³y²

√24x³y² = √(2•2•2•3•x•x•x•y•y)

Pull out everything that is repeated twice - 2, x, y. Leave everything else under the radical symbol.

2xy√(6x)

Some other places to visit for more details on this concept are here and here and here.

If you are one of my students and would like some extra credit, please visit those sites and comment here on whether or not they were helpful.

Polynomial Operations and Factoring

Here are some websites that might help clarify polynomial operations and factoring.

Polynomial addition and subtraction - http://math.about.com/library/blpoly.htm

Adding polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_add.htm

Subtracting polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_subt.htm

Multiplying polynomials - http://www.purplemath.com/modules/polymult.htm and http://www.jamesbrennan.org/algebra/polynomials/multiplication_of_polynomials.htm

Factoring polynomials - http://www.jamesbrennan.org/algebra/polynomials/factoring_polynomials.htm and http://www.purplemath.com/modules/simpfact.htm

In order for my students to earn extra credit, they must visit at least 2 of the sites and comment on both.

Pythagorean Theorem

The Pythagorean Theorem is used to find a missing side length of a right triangle. It only works for right triangles (one with a right angle).

The formula to use is a² + b² = c². A really good explanation of the formula and its derivation can be found here.

In order to use the theorem, you must know the parts of the triangle. Parts a and b of the theorem are the legs of the triangle. These are the side lengths that make up the right angle. The third side (longest side) is noted as c in the theorem.

Now substitute the values given into the theorem (formula) and solve the equation.

Example: A triangle has a side length of 6 cm and a side length of 8 cm. What is the length of the hypotenuse?

6² + 8² = c²
36 + 64 = c²
100 = c²
√100 = √c²
10 = c

Example: A triangle has a side length of 6 cm and a hypotenuse length of 12 cm. What is the length of the missing side?

a² + 6² = 12²
a² + 36 = 144
a² + 36 – 36 = 144 – 36
a² = 108
√a² = √108
a = √108
a = 6√3

Picture of triangle above from http://en.wikibooks.org/wiki/File:Right_triangle_shows_hyp_legs.PNG

Rate-Time-Distance Word Problems Part 2

Now for the second part of the rate-time-distance problems. This example refers to opposite direction travel.

Let’s look at the following example:

Train 1 leaves the train station travelling south at a rate of speed that is 10 mph more than train 2 which is travelling north. After 3 hours, they are 450 miles apart. Find the rate of train 2.

We are going to use another table to organize the rate, time and distance information.

Train	Rate	Time	Distance
1	r +10	3	3 (r + 10)
2	r	3	3r

Since the trains are moving in opposite directions, we are going to add their distances together to get 450 miles.

3 (r + 10) + 3r = 450
3r + 30 + 3r = 450
6r + 30 = 450
6r = 420
r = 70

Therefore the rate of train 2 is 70 mph.

Problems with Starting Homework?

Here's a tip that I use for any job that I continue to put off (procrastinate about). Set a timer for 5, 10 or 15 minutes. You can do anything (even Algebra work) for those periods of time. Sit down and get started. When the timer goes off, stop all work and rest your brain for 10-15 minutes. If there's still more to do, set the timer again and go again. Continue this until you have completed that homework.

Don't forget that there are answers in the back of most books. Checking those answers to see if you are on the right track should not be considered cheating. I encourage my students to check those answers. This will help boost your confidence if you are getting the problems right. It will also encourage you to look back over notes, examples in the textbook, etc. if you are getting the answers incorrect.

Ever Wonder About Dividing by Zero?

If you've ever wondered why you can't divide by zero, here is an explanation that uses what you learned in elementary school to help you understand this concept.

In elementary school, you learned that division is just multiplication fact families in an alternate order. In other words, 6/3 = 2 because 2 x 3 = 6.

We can use fact families to find out about dividing by zero.

0/7 = ?? The answer is 0, because 0 x 7 = 0.

What about 7/0? If you put this in your calculator, you'll get an error. Why?

Well, if we use fact families, 7/0 = ?? What number can you multiply by 0 to get 7? There isn't any number. So the answer is undefined.

Now you may be wondering about 0/0? The calculator still gives you an error. Shouldn't it have given you a 1 or maybe a 0? Isn't 7/7 =1? And 2/2=1?

Let's try fact families again to see what the answer should be. 0/0 = ??

What number multiplied by 0 = 0? Well, every number would work wouldn't it, so the answer is indeterminate. You can't determine what the answer should be.

Hopefully, this explanation has helped to clear up division by zero and why is doesn't work.

New Blog Created

I've created this new blog to help answer some of those tricky algebra questions that high school students have. I will be including articles that seem to give lots of students trouble. I currently teach Algebra I and have been doing so for several years. I'd like to provide assistance to those who need it so here goes! Welcome and please visit often.