Algebra Period 3 (Friday)

How about one more video..
let me know if you find one you think everyone in our class would like


Solving Rational Expressions: 13-5

YOU NEED TO GET RID OF ALL DENOMINATORS!
After you do that, you may end up with a Quadratic that you can solve using any of your methods.

Easy one first

(a+1)/2 = 1/a
You could find the LCM of the denominators which would be 2a but what about using cross products—as you did with proportions—last year.
then the above becomes
a(a + 1) = 2(1)
0r
a² + a = 2
a² + a - 2= 0 using ZERO products property
Factor
(a + 2)(a-1) = 0
so a = -2 and a= 1


When you MUST Find the LCM of the denominators:
MULTIPLY EACH & EVERY TERM ON BOTH SIDES BY THE LCM!
You should end up with NO DENOMINATORS!
(Otherwise, you don't have the right LCM!)

EXAMPLE:
4/x – 4/(x+2) = 1
In this case the LCM is discovered just by multiplying both of the existing denominators. It is as if they are relatively prime!!
The LCM is x(x + 2)
Multiply each term on BOTH sides by x(x + 2)
[x(x +2)4]/x – [4 x(x +2)]/(x+2) = 1 [x(x +2)]

Simplify the denominators with the numerators:
(x + 2) 4 - x(4) = x(x + 2)
SIMPLIFY more:
4x + 8 - 4x = x² + 2x

USE THE ZERO PRODUCTS PROPERTY:
x² + 2x - 8 = 0
FACTOR OR QUADRATIC FORMULA:
(x + 4)(x - 2) = 0

x = -4 and x = 2


Solving Radical Equations 13-6
THIS IS A REVIEW OF CHAPTER 11

You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!

x – 5 = SQRT (x+ 7)
squaring both sides gives us
(x -5)² = [SQRT(x + 7)]²

x² - 10x + 25 = x + 7
x²- 11x + 18 = 0
(x – 9)(x -2) = 0
or x = 9 and x = 2

BUT when you check you discover
x -5 = SQRT ( x + 7)
9 – 5 =?= SQRT (9 + 7)
4 = 4
BUT

x -5 = SQRT ( x + 7)
2 – 5 =?= SQRT (2 + 7)
-3 DOES NOT EQUAL 3

so only solution is x = 9

Solve
[SQRT ( 27 - 3x)] + 3 = x
move the 3 to the other side using transformations

[SQRT ( 27 - 3x)] = x -3
NOW squaring both sides

[SQRT ( 27 - 3x)]² = (x -3)²
27 – 3x = x² - 6x + 9
move everything to the right side – to use the ZERO PRODUCTS PROPERTY

0 = x² - 3x – 18
0 = ( x – 6) (x +3)
so x = 6 or x = -3

WE MUST CHECK again:
Start with the ORIGINAL equation
[SQRT ( 27 - 3x)] + 3 = x

Plug in for x = 6
[SQRT ( 27 – 3(6))] + 3 =? = 6
[SQRT ( 27 - 18)] + 3 =? = 6
[SQRT ( 9)] + 3 =?= 6
3 + 3 = 6 YES

But we discover

[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = -3
[SQRT ( 27 – 3(-3))] + 3 =?= -3
[SQRT ( 27 +9)] + 3 =?= -3
[SQRT ( 36)] + 3 =?= -3
6 + 3 DOES NOT EQUAL -3
so only x = 6 is the solution

Algebra Period 3 (Thursday)

The Quadratic Formula

Wait until you see and hear these videos--
This is the one from class





this is to the Flintstones



pretty funny stuff




So what do you think? Create one of your own...

THE QUADRATIC FORMULA:
-b + or - SQRT b² - 4ac
2a

-b plus or minus the square root of b squared minus 4ac all over 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! ax² + bx + c = 0



DISCRIMINANTS - a part of the Quadratic Formula that helps you to understand the graph of the parabola even before you graph it!
the discriminant is b² - 4ac

(the radicand in the Quadratic Formula, but without the SQRT)

Depending on the value of the radicand, you will know
HOW MANY REAL ROOTS IT HAS
1) Some quadratics have 2 real roots (x intercepts or solutions) - Graph crosses x axis twice
2) Some have 1 real root (x intercept or solution) - Vertex is sitting on the x axis
3) Some have NO real roots (no x intercepts or solutions) - vertex either is above the x axis and is a smiley face (a coefficient is positive) or
the vertex is below the x axis and is a frown face (a coefficient is negative)

In both of these cases, the parabola will NEVER CROSS (intercept) the x axis!
b² -4ac > 0 if it's positive, 2 roots
b² -4ac = 0 if it's zero - 1 root
b² -4ac < 0 if it's negative - no real roots

Math 6 H Periods 1, 6 & 7

Review of Solving Combined Operations 8-5
Follow these general procedures to solving equations:

Step 1: Simplify each side of the equation

Step 2: If there are still indicated additions or subtrations, use the inverse operations to undo them

Step 3: If there are indicated multiplication or divisions involving the variable, use the the inverse operations to undo them

YOU MUST ALWAYS PERFORM THE SAME OPERATION ON BOTH SIDES OF THE EQUATION.

DO ONTO ONE SIDE WHAT YOU WOULD DO TO THE OTHER... hmmm... where have you heard that before??

Balance, balance , balance... it's all a balancing act!!


Solve the equation 3/2(n) + 7 = 22
subtract 7 FROM BOTH SIDES
(3/2)n = 15
Multiply BOTH SIDES by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)
simplify before you multiply out
n = 10

40 - (5/3)n = 15
There are two ways to solve this one
add (5/3)n to both sides
40 = (5/3)n + 15
then subtract 15 from BOTH SIDES
25 = 5/3(n)
Multiply both sides by 3/5, the reciprocal of 5/3
(3/5)25 = 5/3(n)(3/5)
Simplify before multiplying
15 = n

OR
40 - (5/3)n = 15
subtract 30 from both sides
- (5/3)n = 15 -40
need to ask yourself who wins? the negative... and by how much 25
- (5/3)n = -25

multiply by -3/5, which is the reciprocal of -5/3
(-3/5)-(5/3)n = -25(-3/5)
n = 15
we arrived at the same solution

Algebra Period 3 (Tues/Wed)

Solving Quadratics by Completing the Square 13-3
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!

When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!

This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!

EXAMPLE: x² - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square

Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)

Step 2: Square b/2
(-5 x -5 = 25)

Step 3: Add (b/2)² to both sides of the equation
(x² - 10x + 25 = +25)

Step 4: Factor to a binomial square
(x - 5)² = 25

Step 5: Square root each side and solve
SQRT (x - 5)² = SQRT 25
x - 5 = + and - 5

ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x² - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10

x²-4x -7 = 0
add 7 to both sides
x²-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x²-4x + 4 = 7 +4
x²- 4x + 4 = 11
(x -2)² = 11
take the square root of both sides
SQRT(x -2)² = + or – SQRT 11
x - 2 = + or - SQRT 11

IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:

Example: 2x² - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x² - 3x = 1
Divide each term by the "a" coefficient:
x² - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]² = (-3/4)²
= 9/16
x² - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)² = 8/16 + 9/16
(x - 3/4)² = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4

When our textbook ask you to complete the square
as in x² - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)² = 9
so the answer is
x² - 6x + 9