Quads, right?

Please actually do the problems. Please don't let the hours I spent on this blog go to waste. Please have a Happy Hew Year. Please study hard.

Yeh. I realize I'm not the best at explaining things. Far from it.
So, like, I F Y O U N E E D H E L P Y O U C A N A S K M E O R E M I L E O R S E A N O R D O C O R P H I L L I P. Y E S S O A S K.

Mhmm.

Factoring/Solving Quadratics practice:

Find ALL SOLUTIONS to the following.
(x-3)(x-5)=0
(x+3)(x-6)=0
(x-2009)(x+2009)=0
(x-3)^2=0

THINK THESE OUT*
(3x-6)(4x-8)=0
(x+ 4/5)(5x+10)=0
(x-3)(2x-5)=0
(3x)(x+4)=0

Factor and solve:
x^2-2x-3=0
x^2+5x+6=0
4x^2-9=0
***x^2+2008x-2009=0
***3x^2+66x=-363

Give an example of a quadratic equation with no real solutions. (HINT* Think about what x^2 CANNOT be equal to)

Hey guys

Doc, your volume 2 warmups and workouts are SOOOO HARD.

Anyways
If you ever need to find a hotel anywhere:
http://www.hotelscombined.com/
:)

New problem:
Mike walks 3/10 as fast as Joe walks. They start at the same time and walk the same path from Chinatown to Kaimuki, and then back to Chinatown on that path. When Joe reaches Kaimuki, he stubs his toe and walks back 1/5 as fast as normal. When Joe meets Mike on the return trip, what fraction of the way to Kailua is Mike?

(The sum of the numerator and denominator of this simplified fraction is 43).

FOILing (and cleaner expansion methods)

Expand (a+b)(c+d). You get ac+ad+bc+bd. FOIL (Front Outer Inner Last) is a mnemonic for that. You multiply the FRONT ones first, a and c, then the OUTER ones, a and d, then the INNER ones, b and c, and last the LAST ones (oh coincidence), b and d.

The order you multiply in doesn't matter of course, it's just for easy memorization.

Here's why it comes out. (It took me years to get this :P)
(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd.
Just some simple ol' distribution.

An expression with two terms is called a binomial. FOILing is one way to think of how to expand a product of binomials. It's not the only one.

You can do "multiply out all combinations", "distribute step by step", etc, etc.

Foiling only forks for two binomials. But the "multiply out all combinations" and "distribute step by step" expansion methods work for all products of expressions. Bottom line. They're better.

Tchea.

Quadratics Intro and Factorization.

A quadratic is an expression in the form . When solving a quadratic equation, we're basically trying to find values of x such that
.

Factoring
Factoring is taking an expression and writing it in terms of the product of two or more expressions, or factors.

Factoring can be really simple, but also really complicated. For example, it's obvious that by factoring out the 9 in each term, and , because we "factored out" the x. THen again, it's not so obvious that (this is a famous factorization called the Sophie Germain Identity).

All the factoring this post will focus on is one called quadratic factorization (phew). It involves, well, quadratics.

Factoring is, by far, the simplest way to solve quadratics, and a very useful tool. It all starts in expansion... (Please do not use FOIL in the following problems, please distribute step by step. If you don't know FOIL, swell. :) )

Pre-problems
Expand out the following:
.
What's the pattern here? (Wait until you get it before reading on).

So, every expression in the form of (x and y can be anything) is a quadratic in terms of x and y and takes the form of (note that this might not nessescarily be a quadratic in terms of other variables, say, if x=p^2, this would not be a quadratic in terms of p, but still a quadratic in terms of x and y).

To avoid confusion, only x and y will be used as variables from here on.

Expand the following:
.

Notice that each expansion is a quadratic. This will always be the case for such expressions (try to see why).

Carefully look at each step you take expanding, and come up with a general expansion for , where a and b are constants (numbers).

So, you should have gotten this.

This, then, is the basis of factoring. Basically, when you factor, you look for numbers a and b that satisfy .

Here are some examples.
.
We're looking for numbers a and b such that a+b=5 and ab=4. Simple guess and checkwork shows that a=1 and b=4 (or a=4 and b=1, it doesnt make a difference in this case).
Therefore, we know that . Expand it out and see if it works ;).

Now, let's try .
We're looking for numbers a and b such that a+b=-2 and ab=1. If ab is positive, either a and b are both positive, or they're both negative. But, a and b cannot have a negative sum if they're both positive, so a and b both must be negative. Guessing a=-1 and b=-1 works. Therefore, .

But wait, there might be a better way to do this! The answer gives it away, but noticing that this equation was in the form of by setting y to -1, would tell you that this expression is , right off the bat. Saved a whole bunch of effort and thinking we don't all like to do :P.
So...
REMEMBER YOUR FORMULAS AND FACTORIZATIONS.

There are also some clever "special factorizations", that are so useful that it's good to remember them all so you can recognize when to use them.
You already know the square of a binomial (two-piece expression) factorization, so the difference of squares factorization is the only new thing. Try factor . The factorization is this.

That's just it for factorization. It's not only limited to quadratics, or one variable, which can make things très compliqué, really. But as far as quadratic factorization goes, that's just it. You now have your single most important tool in solving quadratics. Congratulations.

Factorization is useful because not only is it useful, it also helps you notice patterns in math.

Now I still haven't answered the question, how do you solve quadratics with factorization? Let's say I factored a quadratic equation into . Well, it's actually really simple. All you do is take the a and b (-3 and 55), and swaps the signs around on each. So, your two (there's a max of exactly two solutions per quadratic, some have 1 and 0) solutions are 3 and -55. This works, because setting x=3 or x=-55 in the above quadratic makes one of the "factors" 0. And anything times 0 is 0. Any time you have a factored quadratic (in the form of ), just take the a and b and swap their signs (because you are looking for the value of x that, when added to a or b, makes 0).

Warning: NOT ALL QUADRATICS MAY BE FACTORED IN THIS WAY!!! Some quadratics just can't be factored- the ones that don't have any roots (solutions) to "factor into". Don't worry about those though, you will probably NEVER have to actually solve one (maybe manipulate one). Also, if the quadratic has big numbers in it, try not to try too hard :). There are other ways.

So, in summary and conclusion, factoring a quadratic is one way to solve a quadratic. It is usually done by examining what the coefficients of the quadratic have to be in terms of constants a and b, and solving for the constants, and using them to make a factorization. It this sense, quadratic factorization is "reverse distributing" or "reverse expansion".

Problems:.
Factor the following:
.


.
For what value of x makes this NOT TRUE?

.

Hey, it's break. Do some wkshts.

Break time and Problem 2.

Problem 2:
It's BREAK!!! BREAK!!! BREAK!!! I'm gonna rush through all the Warmups and stuff in one day!

MERRY CHRISTMAS DUDES!

Yep, and no one's gonna read this until January anyways... Shucks.

So I guess you guys'll do this when you get the chance ;).

Problem 2: This is a classic simple geometry problem.
A circle is inscribed in a square. In the area in one of the four corners of the square, in the area outside the circle, is inscribed another circle. If the square had side length 2, what is the radius of the smaller circle?

Lesson 1 Part 1: Counting Independent Events

Starting off...

Lesson 1: Counting

Part 1: Counting Independent Events

Let's say we were trying to solve the following problem:

There are 5 pickles, 9 flags, 2 laptop cases, and 11 Christmas trees. How many ways can you choose one pickle, one flag, one laptop case, and one Christmas tree.

How would you do this problem?

Well, there are 5 ways to choose a pickle, 9 ways to choose a flag, 2 ways to choose a laptop case, and 11 ways to choose a Christmas tree. Let's pick a pickle first. There 5 possibilities. Now, let's choose a flag. There are 9 possibilities for flag choices for each of the 5 pickles we could have chosen, so there is 5x9=45 ways to choose a pickle and a flag. The same logic applies to each of these pickle-and-flag combinations, with there being 2 ways to choose a laptop case for each of these (45x2=90 combinations in all), and then the Christmas tree, with 11 ways to choose a Christmas tree for each of those 90 possible combinations of the other three. So, there are 90x11=990 ways to choose a pickle, a flag, a laptop and a Christmas tree.

Well, that was painfully obvious, that would would multiply the number of possibilities for each to come up with the answer. However, this is important in the sense that it is the BASE for ALL counting problems. Since the factors or "events" are completely independent of each other (one does not change the other in some way), we call this counting independent events. When counting independent events, we are not concerned with the factors in itself, but with the number of possibilities each factor gives (note that the number might remain the same while the factors chance!). If you want a prime example, try do #*3 in the problems below.

You can also show counting independent events with a probability tree, with the first event branching out in each of its possibilities, and then the second event coming out from each branch.

Problems:
Counting independent Events

1. Emile wants to eat ice cream. The ice cream shop offers 5 flavors of ice cream, and 3 scoop sizes. He can choose to eat it on a sugar cone, a plain cone, or without a cone. Finally, the shop has 3 flavors of sprinkles in 4 different shapes each. How many possible different ways can Emile order his ice cream?

Counting independent probabilities
2. Emile wants to play the flute. There is an equal chance that he will rent his flute from the school or he will buy his flute. There is a 9/17 chance that he will get a Pearl flute. There is a 1/99 chance that he will get a silver, open-holed flute. Out of all silver open-holed Pearl flutes, 1/3 is the exact variety Sean plays. There is a 1/8 chance that there will be a random sentence in the problem to throw you off. What are the odds he the exact same kind of flute as Sean?

Trickier counting, by a bit.

*3. A painter named Emile wants to paint his long 1x9 grid wall with a different color on each grid, under the condition that no two adjacent grids be painted the same color. The painter has 4 colors of paint to choose from. How many ways can he paint his wall?