Tuesday, October 20, 2009

Factoring (Math Part 3)

Today I'm going to look in the chapter about factoring. This chapter doesn't scare me as much as quadratic equations, but I'm still going to have my pen and paper handy.

This chapter begins with the description of terms for algebraic expressions. After naming the expressions "a one-term expression  a monomial, a two-term expression a binomial, and a three-term expression a trinomial," the professor describes the roots of the terms, mono=one, bi=two and tri= three. He also gives examples of the roots, like monorail (one rail), bicylce (two wheels) and tricycle (three wheels). These examples remind me of a Greek and Latin root class I took in college, the roots of the words can help you figure out the meaning, this applies to math as well as English.

As I keep reading, I'm having trouble keeping up with all the terminology, "multiply a monomial by a binomial." I stopped, read it over and went back to my notes. Then I thought of writing out an example of a monomial by a binomial. I wrote: 3 x (4x + 7) which I solved = 12x + 21. Now that I had numbers to visualize the words, I understood what they were talking about.

Next, we came across multiplying two binomials. This lead to the FOIL method. I was reminded of my last blog, and how I had to look up the FOIL method to help me with quadratic equations. I think this would have been a helpful chapter to review before doing quadratic equations.

Now to do some practice with factoring.

Monday, October 19, 2009

Quadratic Equations (Part 2)

Today I decided to try something different while reading. I took the book and a piece of paper and pencil and wrote down the numbers from the problems on the blank piece of paper.

The narrative that explains the math makes me think of educational computer games I used to play when I was younger. There would be characters and they would seem to just do math equations for fun.

When Recordis suggested using the FOIL method I was reminded of my 8th grade class. I remember solving problems using this method, however I forgot what the letters stood for. After writing the problem down, I looked up "FOIL."

First - multiply the first term in each set of parenthesis: 4x * x = 4x^2

Outside - multiply the two terms on the outside: 4x * 2 = 8x


Inside - multiply both of the inside terms: 6 * x = 6x


Last - multiply the last term in each set of parenthesis: 6 * 2 = 12


I used this to help me look at the problem (x + 5)(x - 6) = 0

I tried it myself and got x^2- x - 30= 0, just like the next step in the book.

However, the next paragraph said to go back to the original equation, even though I was frustrated and confused, I decided it would be best to just keep reading.

I understood the next few sections on how to solve for 0 and get the two answers -5 and 6.


Then we were brought back to the quadratic equation. This really confused me, I thought we had the answer and that was it, what's the point in going back to the beginning? I kept reading hoping to find out.


The narrative about the King and his court was really starting to confuse me. I just didn't understand the point of all this assuming and guessing they did to solve problems. I was reminded of my 8th grade teacher telling me that I would have to keep plugging in numbers to find the two answers. I remember hating this explanation because to me guessing does not seem like a mathematical way of solving problems.

I kept reading until I got to a point that was clear to me. "You will never be able to solve every quadratic equation this way! Your trial-and-error factoring method will soon become helpless when you meet the equations I shall give you." I made a personal connection here, because this is how I felt when I was reading the earlier passages.


After a lot of trial-and-error, the King and his court determine the answer and create an equation to solve future problems quicker: The Quadratic Formula.

My only problem with this is, why all the explanation in the beginning of the chapter? All that back-story to the formula just confused me. Is that ever going to come in handy? Why couldn't they say: this is the formula, plug in the numbers to solve.

The rest of the chapter is practice problems, I'm going to go try a few of them now that I'm equipped with the formula.

Tuesday, October 6, 2009

Quadratic Equations (Part 1)

On page 162, in the first word problem, the professor says that Pal can only catch the ball when it is exactly 10 units about the ground. I wonder why does it have to be exactly 10 units? What if he bent down to get the ball, or if he jumped up? When I've seen baseball games, the outfielder uses different strategies to try to catch the ball at different heights, so I wonder why Pal can only catch the ball at exactly 10 units.

When the king described the function for the ball, that x is the time since the ball left the bat, and y is the height of the ball above the ground, I could already see a graph forming in my mind. I could see an imaginary line measuring time going from the bat to the outfield, I could see another imaginary line from the group up towards the ball, measuring its height.

On the next page , they explain that after hitting a lot of fly balls, they came up with the equation f(x) = -x^2 + 7x. I don't understand how the data was transferred into this function. I re-read the passage a few times and I still feel like there was a step skipped.

The professor then said that f (x) = 10. I'm confused why this is, but I guess you just have to trust him.

After plugging that in we got that x^2 - 7x +10 = 0.

After finding this equation, we could use it to make a graph. I could see the lines of the graph, with the x and y axis. After having an image in my head, I thought that if we plug in numbers for x, we will get the y coordinates, and then we can graph the equation. Looking at the graph I was confused to what the curve meant, at what point could Pal catch it? What did all these numbers mean?

I re-read the entire problem. I realized that when f (x) = 10, that was where Pal could catch it, so we were trying to figure out the time. After plugging in numbers, we found that there were 2 different points on the graph that solved the equation. I couldn't figure out which was the right one, even after re-reading the problem and looking at the graph. I double checked my numbers and they still worked.

I think that was enough math for one night, so I'm going to put it aside and hopefully return refreshed another time.