Algebra (Period 1)

Unions, Sets & Intersections 9-1

Conjunctions & Disjunctions 9-2

Conjunction = and --> the graph is an intersection ("yo") and inequality looks like our domains and ranges on our projects
EXAMPLE: 5 < x < 10 
open dots; between 5 and 10 is colored in 

Disjunction = or --> graph will go opposite ways ("dorky" dancer) and inequality looks like this:

x < -2 OR x > 4

open dots; one arrow goes right at 4 and the other arrow goes left at -2

Equations and Absolute Value 
1 VARIABLE
 9-3
Solve the equation twice - Once with the solution positive and once with it negative

I2x - 4I = 10

Solve it twice:

2x - 4 = 10 or 2x - 4 = -10

x = 7 or x = -3


If there is a term on the same side of the equation as the absolute value, move that to the other side of the equation first 
(just like we did with radical equations!) 
Then solve twice.


REMEMBER THAT THE SOLUTION GIVEN CANNOT BE NEGATIVE (the null set)




Inequalities and Absolute Value 
1 VARIABLE
 9-4
There are 2 possible types of inequalities - less than and greater than

For less thAND:

These are conjunctions and so you solve it twice and the solution ends up between them

I3xI < 15 is equal to -15<3x<15 
so x is greater than -5 and less than 5 
For greatOR than:

These are disjunctions and are solved twice with the solution infinitely in different directions

I3xI >15 is equal to 3x < -15 and 3x > 15


Algebra (Period 1)

SOLVING SYSTEMS OF EQUATIONS 8-1 TO 8-3

(2 equations with 2 variables)
You cannot solve an equation with 2 variables - you can find multiple coordinates that work
TO SOLVE MEANS THE
ONE COORDINATE THAT WORKS FOR BOTH EQUATIONS

There are 3 ways to find that point:
1. Graph both equations: Where the 2 lines intersect is the solution
2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other

Example solved all 3 ways:
Find the solution to the following system:
2x + 3y = 8 and 5x + 2y = -2

1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point [you should get (-2, 4)]

2. SUBSTITUTION:
Isolate whatever variable seems easiest
I will isolate y in the second equation: 2y = -5x - 2
y = -5/2 x - 1
Plug this -5/2 x - 1 where y is in the other equation
2x + 3(-5/2 x - 1) = 8
2x - 15/2 x - 3 = 8
4/2 x - 15/2 x - 3 = 8
-11/2 x - 3 = 8
-11/2 x = 11
-2/11(-11/2 x) = 11(-2/11)
x = -2
Plug into whichever equation is easiest to find y

3. ADDITION:
Multiply each equation so that one variable will "drop out" (additive inverse)
I will eliminate the x, but could eliminate the y if I wanted to
5(2x + 3y) = (8)5
-2(5x + 2y) = (-2)-2

10x + 15y = 40
-10x - 4y = 4

ADD TO ELIMINATE THE x term:
11y = 44
y = 4
Plug into whichever equation is easiest to find x

NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!

Pre Algebra (Period 2 & 4)

PYTHAGOREAN THEOREM
 FOR RIGHT TRIANGLES ONLY! 11-2

2 legs - make the right angle - called a and b
(doesn't matter which is which because you will add them and adding is COMMUTATIVE!)

hypotenuse - longest side across from the right angle - called c

You can find the third side of a right triangle as long as you know the other two sides:

a² + b² = c²

After squaring the two sides that you know, you'll need to find the square root of that number to find the length of the missing side (that's why it's in this chapter!)


EASIEST - FIND THE HYPOTENUSE (c)

Example #1 from p. 510

8² + 15² = c²

64 + 225 = c²
289 = c² Take the SQ RT of each side

c = 17



A LITTLE HARDER - FIND A MISSING LEG (Either a or b)

Example #5 from p. 510

5² + b² = 13²

25 + b² = 169

b² = 169 - 25

b² = 144 Take the SQ RT of each side

b = 12



CONVERSE OF PYTHAGOREAN THEOREM

If you add the squares of the legs and that sum EQUALS the square of the longest side, it's a RIGHT TRIANGLE.


If you add the squares of the 2 smallest sides and that sum is GREATER THAN the square of the longest side, you have an ACUTE TRIANGLE.


If you add the squares of the 2 smallest sides and that sum is LESS THAN the square of the longest side, you have an OBTUSE TRIANGLE.
Check out the following website and the applet that shows why this theorem works!!
Website for the Pythagorean Theorem and its many proofs
and

Applet

Algebra (Period 1)

The Quadratic Formula: 13-4

Make sure to read the first post on the Quadratic Formula BEFORE or AFTER you watch these You Tube Videos....

How about creating your own video!! (For extra Credit!!)






Now if Emily can recite the Quadratic Formula after singing Twinkle, Twinkle Little Star... shouldn't you?

Algebra (Period 1)

The Quadratic Formula: 13-4

Method 5:
THE QUADRATIC FORMULA

Let's all sing it to "Pop Goes the Weasel!

"
x = -b plus or minus the square root of b squared minus 4ac all over 2a

x =
-b ± √ (b² - 4ac)
2a

Notice how the first part is the x value of the vertex -b/2a


The plus or minus square root of b squared minus 4ac represents 
how far away the two x intercepts (or roots) are from the vertex!!!!



Very few real world quadratics can be solved by factoring or square rooting each side.


And completing the square always works, but it long and cumbersome!

All quadratics can be solved by using the QUADRATIC FORMULA.

You will find out that some quadratics have NO REAL solutions, which means that there are no x intercepts - the parabola does not cross the x axis!
Think about what kinds of parabolas would do this....ones that are smiles that have a vertex above the x or ones that are frowns that have a vertex below the x axis.
You will find out in Algebra II that these parabolas have IMAGINARY roots




So now you know 5 ways that you know to find the roots:


1. graph


2. factor if possible


3. square root each side


4. complete the square - that's what the quadratic formula is based on!


5. plug and chug in the Quadratic Formula -
This method always works if there's a REAL solution!




DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA!


You should know that for the quadratic formula, you don't need the "a" coefficient to be positive!

That's important if you use factoring, SQRTing each side, and completing the square.


But for the quadratic formula, either way, you'll get the same roots!


EXAMPLE: x² + 8x = 48 


You can move the 48 over or move the x² + 8x over and you'll get the same answers!
Let's look at:


"a" coefficient positive *****vs***** "a" coefficient negative



x² + 8x - 48 = 0 ****VS**** -x² - 8x + 48 = 0


First x² + 8x - 48 = 0
-8 ± √[64 - 4(1)(-48)]

2(1)



-8 ± √(64 + 192)

2



-8 ± √(256)
2



-8 ± 16
2



(-8 + 16)/2 and (-8 - 16)/2



8/2 and -24/2
4 and -12

Now let's compare -x² - 8x + 48 = 0


8 ± √[64 - 4(-1)(48)]

-2(1)

8 ± √(64 + 192)

-2

8 ± √(256)
-2


8 ± 16
-2

(8 + 16)/-2 and (8 - 16)/-2
24/-2 and -8/-2
-12 and 4

So all the signs are simply the opposite of each other and therefore the answers are the same