Monday, January 30, 2017

Week of January 30,2017


Solve Inequalities
How do you solve a linear inequality and sketch the solution on a number line?
How do you solve an absolute value inequality and sketch the solution on a number line?
Graphic Organizer
Online Extra Practice
Textbook Video Tutorials
Regents Prep Text Tutorials

Operations on Complex Numbers

Complex numbers are "binomials" of a sort, and are added, subtracted, and multiplied in a similar way. (Division, which is further down the page, is a bit different.) First, though, you'll probably be asked to demonstrate that you understand the definition of complex numbers.
  • Solve 3 – 4i = x + yi
  • Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. In other words, 3 = x and –4 = y.
To simplify complex-valued expressions, you combine "like" terms and apply the various other methods you learned for working with polynomials.
  • Simplify (2 + 3i) + (1 – 6i).
    (2 + 3i) + (1 – 6i) = (2 + 1) + (3i – 6i) = 3 + (–3i) = 3 – 3i
  • Simplify (5 – 2i) – (–4 – i).
    (5 – 2i) – (–4 – i)
      = (5 – 2i) – 1(–4 – i) = 5 – 2i – 1(–4) – 1(–i)
      = 5 – 2i + 4 + i= (5 + 4) + (–2i + i)
      = (9) + (–1i) = 9 – i
You may find it helpful to insert the "1" in front of the second set of parentheses (highlighted in red above) so you can better keep track of the "minus" being multiplied through the parentheses.
  • Simplify (2 – i)(3 + 4i).
For the last example above, FOILing works for this kind of multiplication, if you learned that method. But whatever method you use, remember that multiplying and adding with complexes works just like multiplying and adding polynomials, except that, while x2 is just x2i2 is –1. You can use the exact same techniques for simplifying complex-number expressions as you do for polynomial expressions, but you can simplify even further with complexes because i2 reduces to the number –1.

Adding and multiplying complexes isn't too bad. It's when you work with fractions (that is, with division) that things turn ugly. Most of the reason for this ugliness is actually arbitrary. Remember back in elementary school, when you first learned fractions? Your teacher would get her panties in a wad if you used "improper" fractions. For instance, you couldn't say " 3/2 "; you had to convert it to "1 1/2". But now that you're in algebra, nobody cares, and you've probably noticed that "improper" fractions are often more useful than "mixed" numbers. The issue with complex numbers is that your professor will get his boxers in a bunch if you leave imaginaries in the denominator. So how do you handle this?
Suppose you have the following exercise:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
  • Simplify  3/(2i)
    This is pretty "simple", but they want me to get rid of that i underneath, in the denominator. The 2in the denominator is fine, but the i has got to go. To do this, I will use the fact that i2 = –1. If I multiply the fraction, top and bottom, by i, then the i underneath will vanish in a puff of negativity:
      3/2i = –(3/2)i
    So the answer is -(3/2)i
This was simple enough, but what if they give you something more complicated?
  • Simplify 3/(2 + i)
    If I multiply this fraction, top and bottom, by i, I'll get:
      3/(2 + i) = 3i/(-1 + 2i)
    Since I still have an i underneath, this didn't help much. So how do I handle this simplification? I use something called "conjugates". The conjugate of a complex number a + bi is the same number, but with the opposite sign in the middle: a – bi. When you multiply conjugates, you are, in effect, multiplying to create something in the pattern of a difference of squares:
      (a + bi)(a - bi) = a^2 + b^2
    Note that the i's disappeared, and the final result was a sum of squares. This is what the conjugate is for, and here's how it is used:
      3/(2 + i) = (6/5) - (3/5)i
    So the answer is (6/5) - (3/5)i
In the last step, note how the fraction was split into two pieces. This is because, technically speaking, a complex number is in two parts, the real part and the i part. They aren't supposed to "share" the denominator. To be sure your answer is completely correct, split the complex-valued fraction into its two separate terms.

Monday, January 23, 2017

Week of January 23, 2017


TASK: What is a polynomial? Explain what a polynomial is and provide an example. (Look at your flashcards if you need a reminder)

We can combine polynomials by using the four operations (addition, subtraction, multiplication, and division). This lesson will discuss how we add and subtract polynomials. The next lesson will show you how to multiply polynomials. You will learn how to divide polynomials in Unit 4.

The most important thing to remember when working with polynomials is that they are a group of numbers. This means that each polynomial needs to be in parentheses at the beginning of the problem. However, most of the time there are not a lot of like terms to combine within the parentheses. Therefore, after we combine all of the like terms within the parentheses, we need to get rid of the parentheses.

TASK: How do we get rid of parentheses?

Example: Find the sum of adding_polynomials_1.jpgand adding_polynomials_2.jpg.

The key word "sum" indicates that we have to add these two polynomials, so we can set up our problem like this:

Remember to put the parentheses because polynomials are a group!

Since there are no like terms inside each set of parentheses that we can combine, we have to use the distributive property to get rid of the parentheses. However, there's no number outside the parentheses for us to distribute.

TASK: When there's no number outside the parentheses for us to distribute, what number can we put there? Why does this work?

So we get:


When we distributed the positive 1s to both sets of parentheses, we got rid of the parentheses and everything else stayed the same. Then, we combined the like terms to get out answer.

Be careful when combining like terms because the signs of the numbers can get tricky. The easiest way to remember it is that whatever sign is in front of the term is the sign that goes with it. For example, in the problem above, we combine positive 9 and negative 12 to give us the negative 3 in the answer.

Subtracting polynomials is the same as adding polynomials, except there's one step that's a little trickier. Let's look at the same example from above, but instead of adding them, we're going to subtract them. We set up the problem like this:

We still have no numbers outside the parentheses, so we put the ones:


This is where the problem gets a little trickier because this time instead of distributing a positive 1 to each set of parentheses, we're distributing a positive 1 to the first set of parentheses, but a negative 1 to the second set of parentheses. This is going to change the signs of each term in the second polynomial. So we get:


Take a look at one more example. It follows the same steps, but it looks a little different:


TASK: Explain the steps done in the example problem above.

 ALGEBRA 2 Difference of Two Squares

Before I show you any special guys, you need to be very familiar with some basic perfect squares:
1^2 = 12^2 = 43^2 = 94^2 = 16
5^2 = 256^2 = 367^2= 498^2 = 64
9^2 = 8110^2 = 10011^2 = 12112^2 = 144
And some perfect cubes:
1^3 = 12^3 = 83^3= 27
4^3= 645^3 = 125
You should know these cold. If my cat comes into you room some night, wakes you out of a dead sleep and yells, "meow meow meow meow!" OK, pretend that he speaks English and yells, "64! What is it? WHAT IS IT?!" Without even thinking, you should yell, "A perfect square! Don't hurt me!"
(Don't worry. My cat isn't allowed out at night.)
OK -- let's go!
Special Guy # 1:
The difference of two squares
x^2 - a^2 = ( x - a ) ( x + a ) ... difference of squares equation
Check it out:
Factor   x^2 - 9
Write it as
x^2 - 3^2 ... the 3 is the a in the difference of squares equation
x^2 - 9 = x^2 - 3^2 = ( x - 3 ) ( x + 3 )
Check using FOIL -- Believe me?

y^2 - 16

Here's another one:
Factor   25b^2 - 1
Rewrite it as squares...
25b^2 - 1 ... the 25b^2 becomes 5 * 5 * b * b ... which gives ( 5b ) ^2 - 1^2 = ( 5b - 1 ) ( 5b + 1 )

Factor  4x^2 - 81
What about this guy?
x^2 + 4
What did we call special guy # 1 ?
The DIFFERENCE of two squares!
x^2+ 4  is a sum, not a difference... and we do NOT know how to factor him yet! ( Later!)

Monday, January 9, 2017

Week of January 9, 2017

Algebra 1

1.1 Intercept Terms, Factors, and Coefficients
Algebra 2
Assessments: Pretest on Thursday, you do not study for this item.  A quiz is likely late in the week.  Unit test is Tuesday or Wednesday of the following week.

Today, we look at quantitative data and Categorical Data. Also, we look at sample types.
HW: None

Today, we look at sample types again and Explanatory and Response distinction
HW: None

Moving along in the week, a discussion of biased and unbiased sampling and correlation and causation
HW: None

Pretest today, do not study.   we look at mean, median, mode, range and skewness of graphs
HW: None

 Yes, it is Friday! We will review today and test is on Tuesday.
HW: None