Sample Review Problems for Quad Formula and Imaginary Numbers

If you missed class today (Mon Dec. 16th) please check the video links below on how to solve problems with the quadratic formula and how to simplify expressions with imaginary numbers.

How to Solve with Quadratic Formula:


How to Simplify with Imaginary Numbers:

Review Week for the Semester Exam (12/12 Through 12/18)

Hello algebra two students! We are done learning new material for the nine weeks and have entered the review period. Semester exams are the 19th and 20th. Here is our review schedule:

Thurs 12/12: Graphing review (parabolas, piecewise, inequalities, 3-d coordinate plane)
Fri 12/13: Matrices review (determinants, inverses, identities, solving for variables)
Mon: 12/16: Quadratic formula and imaginary numbers
Tues 12/17: Factoring (GCF, trinomials, dividing trinomials)
Wed 12/18: Exponent Rules, semester wrap-up

If you need help AT ALL please come and see me during a time that we are both free. I am more than happy to help you out, just communicate to me that you need it. Review sheets are going to be due on Wednesday. I will release some practice problems on Wednesday to help you study more. Good luck and happy studying!

Negative Rational Exponents and Solving Rational Equations

Negative rational (fraction) exponents are just like regular negative exponents. To get rid of a negative exponent, we put it on the other side of the fraction bar. Likewise, whenever we see a negative fraction exponent, move it (and the base number) to the other side of the fraction bar to make it positive, then solve as usual. Easy!

To see an example of this, click below:

Rational Equations are equations with powers in them. For example, we could see something like x^5 = 243. To solve these types of equations, we need to think OPPOSITES:

x^2 is the opposite of radical 2.
x^3 is the opposite of radical 3.
x^4 is the opposite of radical 4.
x^5 is the opposite of radical 5.....and so on. Do the opposite to each side until you get x by itself.

To see an example of how to solve a rational equation, click below:

Rational exponents

Rational exponents are exponents that are fractions. We re-write them so that the top number is an exponent and the bottom number is a radical. Then, we simplify both the exponent and the radical to get our answer. Note that order doesn't matter - you can do either the exponent or the radical first and you will get the same answer.

For a video example on how to do this, click here:
http://www.educreations.com/lesson/view/rational-exponents/14387309/?s=CO1gBl&ref=app

Multiplying Polynomials

A polynomial is a term that has at least one variable, a plus or minus sign, and at least two terms.  When multiplying Polynomials, we need to remember the exponent rules for multiplication: multiply the base and add the exponent. Distribute your terms to everything on the inside of the parentheses.  For two full length examples on how to do this, please click the links below. One is an example of multiplying polynomials and the other is what to do if you a polynomial expression to the third power. Enjoy!

http://www.educreations.com/lesson/view/multiplying-polynomials/13770520/?s=rq90dK&ref=app

http://www.educreations.com/lesson/view/binomial-cube/13770838/?s=zVmwt3&ref=app

Exponent Rules

Today we discussed the three exponent rules that everyone should know: the multiplication rule, the power rule, and the division rule.  When multiplying with exponents, we multiply the bases (big numbers) and add the exponents of like terms. When multiplying with exponents on the outside of parentheses, put your bases to the power of the outside exponent and multiply your exponents inside with the one outside. When dividing with exponents, divide the bases and subtract the exponents.

Note: if you get a negative exponent when dividing, change it to a positive exponent by putting it in the denominator.

To see video examples on how to use these three rules, click here:

http://www.educreations.com/lesson/view/multiplication-rule/13607398/?s=v2LhdR&ref=app
Multiplication Rule

http://www.educreations.com/lesson/view/power-rule/13607503/?s=KnSbkN&ref=app
Power rule

http://www.educreations.com/lesson/view/division-rule/13607615/?s=RBSuTS&ref=app
Division Rule

Simplifying with Imaginary Numbers

Imaginary numbers are slightly different than other variables. We know that i is the same thing as the square root of -1, which doesn't make any sense, so we leave it in the problem. But what if we see i to the second power? We plug in a -1 instead! Why is this important? Because we cannot have anything higher than i in the solution to a problem. Therefore, if we see i squared in a problem, we must change it to -1 instead and then simplify.

To see a few examples on how to simplify expressions with i in them, please click the link below:

http://www.educreations.com/lesson/view/simplifying-radicals/13379407/?s=5goBUq&ref=app

The quadratic formula and imaginary answers

W know that the quadratic formula is used to find the zeroes, or x-intercepts, of a parabol on a coordinate plane. However, not all parabolas have functions that cross the x-axis!  Therefore, when we solve these types of questions with the quadratic formula, we will get imaginary answers that contain the variable "i" meaning that the solutions are not truly existing on the coordinate plane.

For s full-length example on how to solve a quadratic formula problem using imaginary numbers, please click the link here:

http://www.educreations.com/lesson/view/quad-formula-imaginary/13379176/?s=EINdGl&ref=app

Angry Birds Project, Day 1 and 2

For those of you who have been in and out of class during this 3-day week, we did a 2-day project on how parabolas and quadratic functions related to the game Angry Birds. Each bird launched makes a parabola - upside-down, of course, because the path of the bird travels in an n-shape rather than a u-shape.

For the first day, we analyzed a graph of a traveling angry bird. We compared the path that the bird travels with the vertex, horizontal (x-axis) distance traveled, and the vertical (y-axis) height of the parabola.

In the second day, we designed our own Levels for the angry birds. Each level had to contain at least three targets, and you had to give me a quadratic equation (in vertex form) that would get the bird to hit each target in your Level. Shout out to Mia Goodwin, Hunter Ezell, and Tayda Galloway for some amazing-looking Levels! Shout out to Bruce Johnson for making his targets angry-squids! And shout out to Kindall because all of my 7th period students really liked your Level - great job!

Graphing a Parabola in Vertex Form

Vertex form is different from standard form, it looks like: y = 2(x - 3)^2 + 4. The 3 inside the parentheses tells us where to move on the x-axis, but the OPPOSITE because it's inside parentheses. The 4 at the end tells us where to move on the y-axis (not opposite because it's not inside parentheses). Together, (3, 4) makes the vertex of the parabola.

We can treat the 2 on the outside of the parabola like a slope. From your vertex, go up 2, over one on both sides of the function. Connect them with a curve at the bottom to make your parabola.

There are a few special cases that we should note:

y = 1/2(x + 3)^2 + 3 - the 1/2 in front of the parentheses tells us the parabola will get wider, not thinner.
y = (x + 3)^2 - this means the vertex will remain on the x - axis
y = 3x^2 + 1 - this means the vertex will remain on the y-axis
y = x^2 - when there are no numbers, the parabola remains on the origin of the coordinate plane.

Graphing a Parabola in Standard Form

Standard form of a parabola is when we see f(x) = x^2 + 5x + 4: a quadratic term with x-squared, a linear term with x, and a constant term which can be any number. When we do this, our goal is to first factor our function by grouping with parentheses. Ours would factor into (x + 1)(x + 4). Then, set them equal to zero and solve: this gives you the x-intercepts. x + 1 =0 and x + 4 = 0. When you solve to get x by itself, you get x = -1 and x = -4. Graph these and find their midpoint on the x-axis (x = -2.5). This is the x-coordinate of the vertex of the parabola. Then, use the formula y = -b/2a to find the y-coordinate of the vertex. Ours would be y = -5(2)(1) = -2.5. Thus, our vertex is (-2.5, -2.5) and we can plot that point on the graph. These three points form the foundation of the parabola: connect them and graph.

Note that this vertex would be at the bottom, so it would be a minimum of the function: it is the lowest point in the function. If the parabola was upside-down, the vertex would be a maximum because it would be at the top of the function.

Parabola Vocabulary

Parabolas are functions that contain x-squared within them in some form. They can have up to three terms: a quadratic (x-squared) term, a linear term (x), and a constant term (a number), so that the parabola takes the form f(x) = ax^2 + bx + c. For example, f(x) = 2x^2 + 5x + 9. However, this does not have to be the case. As long as the function contains x-squared within it, it can be a parabolic function.

A parabola is shaped like a "U" if the number in front of x-squared is positive, and an upside-down U (n) if the number in front of the x-squared is negative. The middle of a parabola is the axis of symmetry, and the top/bottom of the axis of symmetry is called the vertex. The larger the number in front of the x-squared term, the thinner the function will be. The smaller the number in front of the x-squared term, the wider the function will be.

Systems of Inequalities

This is the last topic in systems of equations! Systems of inequalities are very similar, except instead of finding x and y, we are graphing both inequalities on a coordinate plane and shading the solution area for both.

For example:
y > 2x + 1
y < -3x - 3

To solve this question, graph both systems on the same coordinate plane and shade the side of the graph that tests true when you plug in (0, 0) to both inequalities. The area of shading that overlaps for both inequalities will be your solution region. This means that ANY point that you plug in within that region will work for both x and y.

Review for the Systems of Equations Quiz

Tomorrow is the last quiz of the nine weeks! Be sure to bring your 1/2 page cheat sheet and study your examples from today so that you can finish the nine weeks strong! This quiz is not meant to be tricky or confusing. I hope that you all do well and get A's - it is exactly like the review we did in class today.

In case you missed class, we went over the following topics for the quiz tomorrow:
-Solving with substitution
-Solving with elimination
-Solving systems with 3-variables using Cramer's rule (determinants)
-Plotting a point in 3-D
-Plotting a line in 3-D (looks like a triangle shaded in on a coordinate plane).

Know all of this? Then you should be all set to ace the quiz tomorrow! If you put a BOX around your name on the top of your page I will add TWENTY BONUS POINTS to your quiz score!

Plotting Points and Graphing Lines in 3-Dimensions

A 3-D coordinate plane looks like this:

where the x-axis comes out towards you, the y-axis is now the horizontal axis, and the z-axis is now the vertical axis. To plot a point in 3-D, you must first go over by your x, then over by your y, then up/down by your z. Your answer might not look like it makes sense - AND THAT'S OKAY! We're trying to graph something in 3-D on a 2-D piece of paper, so naturally, it's not going to look 1005 perfect. Just make sure you get the point in the general correct area of the graph.

To graph a line in 3-D, such as 3x + 4y + 3z = 12, we substitute zeroes in for every variable that we're not trying to find. For example, when trying to find x, I would keep x and plug in 0's for y and z, like this: 3x + 4(0) + 3(0) = 12. The zeroes would cancel out, and when solving for x, divide by 3 on both sides to get x = 4. Plot this on the 3-D plane. Repeat this process for the other two variables and plot them as well. Your final answer should be three points: connect them together and make a triangle, then shade in the triangle to represent the plane of answers.

Solving 3-Variable Systems of Equations

There are many ways to solve 3-variable systems of equations. The three most common are substitution, elimination, and Cramer's Rule. I chose to teach Cramer's rule because it builds off of previous concepts, and because it is the most straightforward (in my opinion) of the three methods.

To begin, you'll have a problem that looks something like this:
2x + y - z = 5
3x + 5y - 4z = 10
x - 2y + 2z = 12

where you will need to solve for x, y, and z, and your answer will be a coordinate triplet, meaning that it can be graphed on a 3-D coordinate plane. First, make a 3 x 3 matrix of all the coefficients to the left side of the equals sign. Find that determinant, and label it "D."

Next, make another 3 x 3 matrix but leave out the column of x's (so your first column should be blank). Where the x coefficients used to be, insert the numbers to the right of the equals sign. Find that determinant, label it "Dx." Repeat this process for both y and z, leaving out those variables' numbers and replacing them with the numbers to the right of the equals sign for each matrix. Find those determinants and label them "Dy" and "Dz."

Almost done! To get your final values of x, y, and z, simply divide the following to get your answer:
x = Dx/D,      y = Dy/D,      z = Dz/D

Solving 2-Variable Equations using Elimination

We solve 2-variable systems of equations using elimination whenever both x and y are on the same side of the equals sign for both equations. The idea of "elimination" is that we eliminate, or cross out one of the variables so that we can solve for one of them. Then, once you have one variable, plug it back in and solve for the other one. Express your answer as a coordinate pair.

Example: Solve the system of equations.

3x + 4y = 2
-3x - y = 7        -Cross out your x's first because 3x cancels out with -3x. Add the remaining terms.
       3y = 9       -Solve for y
         y = 3

Now, plug y = 3 into either of the equations and solve for x.
3x + 4(3) = 2
3x + 12 = 2    -Multiply
3x = -10         -Subtract 12 on both sides
x = -3.33        - Divide by 3 on both sides

Answer: (-3.33, 3) Express it as a coordinate pair.

Guest Speaker and Random Meeting

We had a guest speaker yesterday from Arkansas State University! Shy the master's student told us about the Oral Communications class that she teaches to college freshman, and what it's like to be a freshman on the A-State campus. Expectations are a lot different in college - especially when your class only meets one day a week but you have it for 3-hours at a time!

I was pulled from my second period class today (just as we were about to graph y = sinx - boo!) to attend a training session at the Junior High. My apologies for this one - I didn't know about it in advance. A BIG THANK YOU to everyone who practiced their math ACT while I was gone and a HUGE Shout Out to my aid Bre Womack for setting everything up while the sub was there!

Solving 2-Variable Systems of Equations by Substitution

We're done with matrices and are moving onto systems of equations. These are linear equations (so we expect at least two variables) that intersect each other at some point. The point is the answer to our system of equations. With two variables, that means it has to have at least an x answer and a y answer. We display our answer as a coordinate pair (x,y).

To solve using substitution, we have two equations, and expect one variable to be by itself on one side of the equals sign. We substitute that variable, solve for the remaining variable, and find the solution to one variable. Then, once we have that solution, we plug it back into the first equation to get the answer to our other variable.

FOR EXAMPLE:

Solve the system of equations:
y = 2x
6x - 2y = 12

First, we know that y = 2x. That means whenever we see a "y" in the second equation, we can put "2x"

Second equation: 4x - 2(2x) = 12    - substitute
                           6x - 4x = 12         - multiply
                                   2x = 12        - combine like terms
                                     x = 6          - divide by 2 on both sides to get x = 6

Then go back to the 1st equation y = 2x. Since we know what x is (we just found it, x = 6), we can plug the 6 in for the x and solve for the y:

y = 2x
y = 2(6)  - plug in x = 6
y = 12    - multiply to solve

Note that it doesn't matter which variable we find first - the x or the y - because we're going to need both of them to represent the answer. Write your answer as a coordinate pair: (6, 12) since it represents a point on the coordinate plane where your lines intersect.

Matrix Test and Finding the Intersection of Two Lines

Good job on those matrix tests! People had the most trouble with the "solving matrix equations" questions, but other than that, I'd say that the average person did fairly well on everything else and at least half of you managed to answer one of the bonus questions correctly - great job!

Our next unit of study will be systems of equations in two and three dimensions. To prepare for this, I left some subwork today (I was on the field trip to Little Rock) where you had to plot two lines on the same coordinate plane and then tell where those two lines intersected, or crossed, one another. More information on that tomorrow...

If I didn't get your review sheet then tomorrow is the last late day that you can turn it in for partial credit, otherwise, just keep it for no credit.

Sub Day (extra day to study matrices!!)

I was at a 1/2 day meeting today with the math department, so, you lucky ducks, your matrix test is pushed back until tomorrow. The study guide given to you by the sub was due back by the end of the class period and should have helped refresh your memory for some of the key concepts you'll need for tomorrow.

Review sheets are also due tomorrow - bring them back!

Review for the Matrices Test

Today we reviewed for the matrix test - there are 14 questions and 2 bonus questions. If you want to bring a half-sheet of notes to your test, feel free to do so. You will get the whole period to complete your test.

The district told me that I have a math department meeting tomorrow, so you lucky ducks, you get one more day to study. The things that you need to know to succeed on the test are:

1. How to add/subtract matrices
2. How to multiply matrices
3. How to find the determinant of matrices and show work for a 3 x 3 matrix
4. How to find the inverse of a matrix
5. How to prove that two matrices are inverses of one another
6. How to solve matrix equations by both 1) adding/subtracting and 2) multiplying

Good luck! If you write your favorite color next to your name on the test I will add ten bonus points to your score!

Protein and Matrices

Today we served as health consultants and made matrices containing the protein and fat content for professional athletes. We walked around the room and looked at actual food labels to find this information - real world math! Then, we organized the data into matrices and found:

1) The determinant of the matrix
2) Solved an equation with the matrix
3) Expanded the matrix to include 9 clients rather than 1 client.

Shout out to Marcola Bobo, Brittany Ivory, and Nikki McKinney who impressed our class visitor very much! Another HUGE shout out to Aaron Houston for helping half of the class with their work - AMAZING JOB!

Solving Matrix Equations

Today was one of the longer days we've had in a while. First, we solved equations that involved adding and subtracting to solve for the missing variable. This was easy - just add or subtract the matrix on both sides to get the answer. Note that your answer represents an ENTIRE matrix, not just a number.

Equations involving multiplication were trickier. We know the opposite of multiplication is divison. However, we cannot divide matrices. So, as an alternative, we need to:

1. Find the inverse of the matrix next to the variable. To do this, recall that you must find the determinant, then multiply every element in the matrix by 1/determinant.

2. Once you find the inverse, multiply both sides of the equation by the inverse. The two matrices on the side with the variable cross each other out because they are the inverse of each other. It doesn't matter that they don't contain the same numbers - they're still inverses, so they still cross out. Bring the variable down.

Then, multiply the two matrices on the other side. Remember, you must multiply the first row with the first column, then first row with the second column, then the second row with the second column, then the second row with the second column. The resulting matrix will be your answer.

Review Sheet

I was in the library today doing a training for the ipads - you used today to work on your review sheets. Shout out to Marcola Bobo who was the only one excused from this assignment since she already finished it - nice work Marcola!

Inverse Matrices

We know from yesterday that, to prove that a matrix is an inverse, the matrix and it's inverse must be multiplied together, and the answer must be an identity matrix (1's on the main diagonal, 0's everywhere else).

Today, we calculated the inverse of a given matrix. As we know, the word "inverse" in math means the opposite, so the answer that we calculate will be a matrix that is, mathematically, opposite from the one that we started with.

To do this, we
1. find the determinant of the matrix (remember, the matrix must be a perfect square!)
2. multiply every element in the matrix by (1/determinant). This means that, in many cases, the inverse matrix will contain fractions/decimals in it.

The answer is your inverse matrix!
The only time that this DOESN'T work is when your determinant is zero, because (1/0) is undefined.

Proving Inverse Matrices

To prove an inverse matrix, we must multiply two matrices together. If their products form an identity matrix, then the two matrices are inverses to one another. If not, then Matrix B is not an inverse to Matrix A.

An identity matrix is one containing all 1's on the main diagonal and 0's everywhere else in the matrix. We can only solve for inverse and identity matrices if the matrices we are working with are perfect squares (example: 2 x 2 matrix or 3 x 3 matrix).

More practice on this tomorrow!

Multiplying Matrices

Multiplying matrices is a little different than adding/subtracting matrices. We must make sure that:

# columns in 1st matrix = # rows in 2nd matrix. 

Otherwise, we can't multiply them (and therefore, just write CAN'T SIMPLIFY). If we can multiply, then multiply each column element with each corresponding row element, then add them together. Repeat this process until you don't have any more columns or rows left.

This kind of problem requires a lot of practice to make perfect. If you need some extra examples I will be happy to give you some!

Matrices: Adding, Subtracting, and Solving for Missing Variables

Matrices order information into rows and columns. We always label matrices with a capital letter variable (such as, Matrix A). To add or subtract matrices, they must be the same size and shape. If there is a number in front of the matrix, such as 2A + B, we must first multiply everything in Matrix A by 2 before we add it to everything in Matrix B. Our answers should also be in the form of a matrix.

When solving for missing variables, we must set the element of the matrix with the variable equal to the number in the other matrix in the same corresponding position. This creates a new equation, which we can solve for the value of the variable.

Chapter 2 Test

The chapter 2 test was taken in the lecture hall today. Generally, the grades on this test were pretty high with a couple of exceptions. If you didn't take the test or you are dissatisfied with your grade, please tell me so that you can retake the test after school.

Our next topic is matrices!

Review for Chapter 2 Test

Your chapter 2 test on linear functions is tomorrow! The test contains 14 questions and 2 bonus questions. All answers are free response and you must show your work for everything.

The following will be on the test: be sure to check your notes or this blog to review anything that you're not 100% on. Bring your 1/2 page cheat sheet tomorrow too for extra help!

1. Graphing equations of lines - begin with the starting point, up/down and over by the slope
2. Graphing absolute value equations - plot the vertex, then graph both sides based on the slope
3. Absolute value translations - write the equation and graph the translation
4. Linear Inequalities - graph the line, dotted or solid depending on the sign, and test point (0,0) to shade
5. Piecewise Functions - graph the asymptote, determine which graph goes on which side, then graph
6. Step functions - graph each function on the y axis depending on what it can be between on the x axis.

Step Functions

Step functions get their name because they look like stairs. We graph one by first locating the y values. Then, we look to see which x values they are between. We typically see an open circle on one side of the graph and a closed circle on the other side of the graph.

Step functions have a domain of all real numbers - for every step function that you're going to see. The range of a step function can be found by locating the smallest and the largest number in the y-values that the problem gives to you.

Step functions only LOOK hard because there are many signs and numbers. Once you've gotten the hang of it - it's very easy!

Piecewise Functions, Day 2

We practiced more with piecewise functions today. They are, by far, the most difficult thing that we've done so far in algebra 2. Remember the steps for graphing:

1. Determine WHERE the asymptote of the function should be on the x-axis
2. Determine which function goes on what side of the graph: left or right.
3. Graph both of your functions ON THE CORRECT SIDES OF THE ASYMPTOTES. Remember, your line can only exist on one side of the graph - not both!

Shout out to everyone from 5th period who graphed some REALLY difficult functions today. I'm bringing you a surprise tomorrow in class. You all rock!

Piecewise Functions

Piecewise functions have more than one function within them. We first must determine the asymptotes of the graph - this helps us determine the sides that each function will go on. Then, we graph the function in the appropriate region of the graph. Sometimes, we cannot graph the function right away because the y-intercept is not in the correct region of the graph. When this happens, go up and over or down and over by your slope until you get into the correct region of the graph, then begin to draw your line. Remember - two points are needed to draw your line!

Also, don't forget about open circle for greater than, and closed circle for greater than or equal to.

Shout out to Jamal Turner for attempting to teach a student how to solve piecewise functions after finishing early in class - terrific work!

Thursday: Translating Absolute Value Functions

Knowing the base equation of absolute value is y = |x|, we can alter this function to change the slope, direction, or vertex of the graph. If the translation is horizontal, we put the number of units inside the absolute value and flip the sign. For a vertical translation: put the number of units outside the absolute value and keep the sign. Reflections are marked by a - in front of the slope, and shifts in the slope are placed in front of the x in the absolute value function.

Absolute Value Functions

We've been looking at absolute value functions for the past two days. They look like a "V" and contain a vertex, which is the starting point. Don't forget that when you look at the number inside the absolute value, you must flip the sign and go the opposite way on the x axis. We don't flip any numbers outside of the absolute value.

We also translated functions on the coordinate plane. We usually start with the base function y = |x| unless it specifies otherwise in the problem. Go over and backwards by your x, and up or down on the y.

The second review sheet was given out today! The due date is Sept 12 (Thursday) with the early bird deadline the day before. Shout out to Shulondria Smith for tutoring students in class today when they didn't understand, and shout out to Gabby Foster for staying after class to talk about math problems!

Algebra 1 Review

Your quiz is tomorrow! Remember that you are allowed to bring 1/2 sheet of notebook paper with whatever notes you want for the quiz: front side only. The quiz has 14 questions and two bonus questions.

We reviewed today with the algebra top 10 list and a make-your-own-study-guide activity. Study your notes and your examples so that you don't get stuck. Practice makes perfect!

Review sheets were due today. Any review sheets that I don't yet have can be turned in tomorrow for half credit. I'm not taking review sheets that are more than two days late.

Domain, Range, and Graphing Functions Review

Today we looked at a few examples involving functions, equations of lines, and graphing on a coordinate plane. Remember that functions have input and output, so therefore, they must contain at LEAST two variables. Two variables mean that we can graph them on two axes (x axis and the y axis).

We can easily graph equations in slope-intercept form, which uses the equation y = mx + b. m is your slope, and b is your y-intercept. To graph, we start at the y-intercept and we go up/down and over depending on what our slope is. Recall that slope is "rise over run," so make sure to look at your y-value (in the numerator) first and your x-value (in the denominator) second.

Shout out to TJ Davis for getting really excited about graphing things on sticky graph paper, and shout out to everyone sitting near him for putting up with the excitement. Shout out to Kelsey Hodges for extreme focus and perseverance throughout today - nice work!

Absolute Value Equations

Absolute value equations have two answers: one on the positive side and one on the negative side. We need to get the absolute value by itself on one side of the equation. Then, we need to solve the equation twice: once regularly, and once by flipping the signs of everything on the opposite side of the equation. We will have two answers when we are finished. NOTE: just because you are making two equations, one with a positive side and one with a negative side, does not mean that you have to have both a positive and a negative answer. You can have two negative answers, two positive answers, or one of each.

Five extra bonus points if you solve and graph this equation correctly and bring it into me by Friday, August 31st:      4|2c + 2| > 4c + 10. Good luck!

Shout out to Zadie Williams for her incredibly neat work and perseverance in a room full of upperclassmen! Great job! Shout out to Destiney Barnes for being the smartest person at Marion High School last year!

Inequalities Review

Inequalities are statements using greater than or less than - they contain multiple solutions. We use an open circle on the number line to represent < or > and a closed circle to represent greater than or equal to, less than or equal to.

Don't forget to flip the sign of the inequality if you are multiplying or dividing by a negative number!

In compound inequalities, "and" is when the arrows come together on the number line, and "or" is when they go in the opposite direction.

In equations or inequalities that contain fractions, you can always multiply out the denominator rather than changing it into a decimal. Not necessary but it's a nice trick to use if you like working with whole numbers rather than decimals.

Factoring/Solving for a Variable

When factoring, find the common factor (can be a variable OR a number!) and take it out of ALL your terms. It MUST be in all terms in order for you to take it out. For example, we cannot take the x out of the expression 2x + 5y + 7x because the 5y does not have an x.

To solve for a variable, use PEMDAS backwards and take variables/numbers away from each side, step by step, until the variable you are solving for is by itself. You should have the SAME VARIABLES in your answer as when you began.

Shout out to Sandra Hatcher and LeDarius Henderson for getting the ultra-challenging candy question, and shout out to Shanice Prater, Candace Latham, Khadijah Hamilton, Kassie Tittle, Kyndra Hunter, Brittany Ivory, and Regina Jordan for staying after class to learn how to solve it!

Equations Review

We reviewed some tougher equations today and may/may not have became frustrated by solving equations with fractions (Hint: It might be helpful to turn them into decimals first). We then looked at a really tricky problem involving a sidewalk, a pool, and perimeter.

In problems where our answer is a number = the same number, like 2 = 2, remember that the equation has infinite solutions! Any number that we plug in will work for the answer!

Shout Out to Shannon Reynolds for completing her ENTIRE review sheet already - wow, absolutely blown away by your hard work and dedication! You're awesome!

Pre-Test Day

We took the pre-test today - GREAT EFFORT on everyone's part to remember concepts and skills from Algebra 1! Seriously impressed here. If you didn't take the pre-test then you have until next Tuesday, August 27th, to make it up, otherwise it's a zero. A great time to do that would be during my service period (3rd), during my lunch (3rd lunch) or after school.

The first review sheet was given out today. You have until Thursday, August 29th to get it done. If you give it to me a day early, I will give you 5 bonus points.

Shout out to Michael Clayton for giving certain people in 7th period a pep talk about keeping their heads off the desk during class! Shout out to Regina Jordan for moving seats without complaining - thanks! I appreciate it :)

Welcome to Algebra 2! I'm so excited to be teaching this class this year, and I know that there will be tons of learning going on. My goal is for everyone to master enough Algebra 2 to achieve a 25 or higher on the math section of their ACT college entrance exam. To do this, you'll need to put in a lot of hard work, but I promise it will be worth it in the end.

We went over the syllabus in class today and filled out a student survey. I explained my expectations for the class as well as what it takes to be successful in room 215. If you missed class, please grab a syllabus tomorrow.

Homework: Bring back signed syllabus to class. +5 bonus points if I get it Tues. Aug. 20th.  Regular credit if I get it Wed. Aug 21st. No credit if it's late.