Algebra Honors ( Periods 6 & 7)

Completing the Square: 12-2
METHOD 4:


Now this is completely new to you!!!

When does the square root = ± square root method work well?

When the side with the variable is a PERFECT SQUARE! We saw that in the previous section.
perfect square = k ( when k ≥ 0)

So what if that side is not a perfect BINOMIAL SQUARED?

It may be possible to trasform it into one... by COMPLETING THE SQUARE
You can follow steps to make it into one!

Why is this good? 

Because then you can just square root each side to find the roots!


THIS METHOD ALWAYS WORKS!


EXAMPLE:
x² - 3x -18 = 0

(Head's up-- I wouldn't use this method here because I can see that it factors easily... into (x-6)(x+3) = 0 so the solution set is {-3,6}
However, knowing what I need to get for my solutions might be a good way to practice Completing the square..
so

x² - 3x = 18

Step 1) b/2

That is, take half of the b coefficient ,
or in this case (- 3/2)



Step 2) Square b/2

(b/2)²
(-3/2 ⋅ -3/2 = 9/4)


Step 3) Add (b/2)² to both sides of the equation

x² - 3x + 9/4 = 18 +9/4

Step 4) Factor to a binomial square

(x - 3/2)² = 18 + 9/4
 = 81/4


Step 5) Square root each side and solve

√(x - 3/2)² = √ 81/4

x - 3/2 = ± 9/2
x = 3/2 ± 9/2
x = 3/2 + 9/2 AND x = 3/2 - 9/2
x = 6 and x = -3
{-3, 6}
We got the same solutions !! Yay!!


x² - 10x = 0


Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED.

But here's how you can make it one!


Step 1) b/2

That is, take half of the b coefficient ,
or in this case (- 10/2 = -5)


Step 2) Square b/2

(b/2)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

√(x - 5)² = √ 25

x - 5 = ± 5

Step 6) ADD 5 TO BOTH SIDES

x - 5 = ± 5
+ 5 = +5
x = 5 ± 5
Step 7) Simplify if possible

x = 5 + 5 and x = 5 - 5

x = 10 and x = 0



So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.
YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10.
I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a)
Notice that I could also factor x² - 10x = 0 to get the solution more easily.
So don't complete the square if the quadratic factors easily!

IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:
x² - 10x - 11 = 0
x² - 10x - 11 + 11 = 0 + 11
x² - 10x = 11
NOW COMPLETE THE SQUARE AS ABOVE:
x² - 10x + 25 = +25 + 11
(x - 5)² = 36
√ (x - 5)² = √ 36
x - 5 = ± 6
x = 5 ± 6
x = 11 and x = -1

Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!
Now an example that DOES NOT FACTOR: x² - 10x - 18 = 0
x² - 10x - 18 = 0
x² - 10x - 11 + 18 = 0 + 18
x² - 10x = 18
NOW COMPLETE THE SQUARE AS ABOVE:
x² - 10x + 25 = +25 + 18
(x - 5)² = 43
√ (x - 5)² = √ 43
x - 5 = ± √ 43
x = 5 ± √ 43
x = 5 + √ 43 and x = 5 - √ 43
When there is an IRRATIONAL square root, always SIMPLIFY if possible!

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 find the completing the square term and add it to both sides:
[(-3/2)(-3/2)]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
√[(x - 3/4)² ] = ±√ [17/16]

x - 3/4 = ±[√17] /4

x = 3/4 ±[√ 17] /4

x = (3 ± [√17]) /4
or written better x =
3 ± √ 17
4

Math 6A ( Periods 1 & 2)

Adding Integers 11-2

Rules: The sum of two positive integers is a positive integer.
The sum of two negative integers is a negative integer.

So- if the two numbers have the same sign, use their sign and just add the numbers.

-15 + -13 = - 28

-10 + -4 = -14

Rules: The sum of a positive integer and a negative integer is :

POSITIVE… IF the positive number has a greater absolute value

NEGATIVE… IF the negative number has a greater absolute value

ZERO… IF both numbers have the same absolute value

Think of a game between two teams- The POSITIVE TEAM and The NEGATIVE TEAM.

30 + -16 … ask yourself the all important question…
“WHO WINS?
in this case the positive and then ask
“BY HOW MUCH?”
take the difference 14

14 + - 52…
“WHO WINS?”
the negative… “BY HOW MUCH?”
38
so the answer is -38


(-2 + 3) + - 6 you can work this 2 ways

(-2 + 3) + - 6 = 1 + -6 = -5 or
using all the properties that work for whole numbers
Commutative and Associative properties of addition
can change expression to (-2 + -6) + 3 or -8 + 3 = -5 you still arrive at the same solution.

You want to use these properties when you are adding more than 2 integers.
First look for zero pairs—you can cross them out right away!!
3 + (-3) = 0
-9 + 9 = 0

Then you can use C(+) to move the integers around to make it easier to add them together rather than adding them in the original order. In addition, you can use A(+) to group your positive and negative numbers in ways that make it easier to add as well.

One surefire way is to add all the positives up… and then add all the negatives up.
At this point ask yourself that all important question… WHO WINS? …
use the winner’s sign..
and then ask yourself..
BY HOW MUCH?

example:

-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13

Taking a good scan of the numbers, do you see any zero pairs?
YES—so cross them out and you are left with
-4 + 5 + (-4) + 13
add your positives 5 + 13 = 18
add your negatives and use their sign – 4 + -4 = -8

Okay, Who wins? the positive
By how much? 10
so
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13 = 10

Math 6A ( Periods 1 & 2(

Negative Numbers 11-1

On a horizontal number line we use negative numbers for the coordinates of points to the left of zero. We denote the number called ‘negative four’ by the symbol -4. The symbol -4 is normally read ‘ negative 4’ but we can also say ‘ the opposite of 4.’

The graphs of 4 and -4 are the same distance from 0—>but in opposite directions. Thus they are opposites. -4 is the opposite of 4.

The opposite of 0 is 0

Absolute Value is a distance concept. Absolute value is the graph of the distance of a number from 0 on a number line. The absolute value of a number can NEVER be negative!!

Counting (also known as Natural) numbers: 1, 2, 3, 4, ….
Whole numbers 0, 1, 2, 3, 4….
Integers are natural numbers and their opposites AND zero
…-4, -3, -2, -1, 0, 1, 2, 3, 4….

The opposite of 0 is 0.

The integer 0 is neither positive nor negative.

The farther we go to the right on a number line--- the bigger the number. We can compare two integers by looking at their position on a number line.

if x < 0 what do we know? x is negative number if x > 0, what do we know? x is a positive number

We have been practicing representing integers by their graphs, that is, by points on a number line.

Make sure that your number line includes arrows at both ends and a line indicating where zero falls on your number line.
The graph of a number MUST have a closed dot right on the number line at that specific number.
Please see our textbook page 366 for an accurate example.

Algebra Honors ( Period 6 & 7)

Quadratic Equations with Perfect Squares 12-1

In Chapter 5 you learned how to solve certain quadratic equations by factoring and in Chapter 11 you learned how to solve quadratics in the for
x² = k
as in x² = 49
x = ±7

This lesson extend to any quadratic equation involving a perfect square

if we have x² = k
if:
k > 0 then x² = k has two real-numbered roots with x = ±√k
if k = 0 then x² = k has one real numbered root x = 0
if k<0 nbsp="" sup="" then="" x="">2
  = k has no real numbered roots
m² = 49
√(m²) = ±√49
m = ±7
{-7, 7}


5r²= 45

5r² /5= 45/5
r²= 9
√r²= ±√9
r = ±3
{-3,3}

(x + 6)² = 64
√(x+6)² = ±√64
x+6 = ±8

be careful here
you now have
x = -6±8 which means

x = -6-8 = -14 AND x = -6+8 = 2
{-14, 2}

9r² = 121

9r²/9 = 121/9
r² = 121/9
√r²= ±√121/9
r = ±11/3

Make sure to check your solutions to insure that they both work

(x-3)² = 100
√ (x-3)² = ±√100
(x-3) = ±10
x = 3 ±10
x = 3 +10 = 13 and x = 3-10 = -7
{-7, 13}

5(x-4)² = 40
5(x-4)²/5 = 40/5
(x-4)² = 8
√ (x-4)² = ±√8
Now you need to simplify the Radical

√ (x-4)² = ±2√2
(x-4) = ±2√2
x = 4 ±2√2

{4 - 2√2, 4 + 2√2}



7(x - 8)² = -28
7(x-8)²/7 = -28/7
(x-8)² = -4
WAIT!!! that can't happen no real numbered solutions...

An equation that has a negative on one side and a perfect square as the other has NO REAL NUMBER Solutions

y² + 6y + 9 = 49
(y+3)² = 49
√(y+3)² = ±√49
(y+3) = ±7
y = -3 ±7
y = -3-7 = -10 and y = -3 +7 = 4
{-10. 4}

Perfect Squares like (x -4) ² and ( y + 3) ²

The square of any real number is always a non negative real number.
2(3x-5)² + 15 = 7 becomes
2(3x -5)² = -8
or (3x -5)² = -4... NO REAL number solution!!

Algebra Honors ( Periods 6 & 7)

Introduction to Quadratic Equations

We learned from Chapter 8 that a quadratic function is a function that can be defined by an equation of the form
ax² + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax² + bx + c = y we have an equation of the form ax² + bx + c = 0. An equation that can be written in this form is called a quadratic equation.

STANDARD FORM is ax² + bx + c = 0
4x² + 7x = 5 write in standard form and determine a, b, and c
4x² + 7x – 5 = 0
a= 4
b = 7
c = -5


CHAPTER 12 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic

We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s where y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers

2) Set y or f(x) = 0 and then factor (we did this in Chapter 5)
but...some quadratics are not factorable

3) Square root each side (+ or - square root on the answer side) We did this in Chapter 11
but...sometimes the variable side is not a perfect square (it's irrational)

4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated

5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)

REVIEW:

Reading the x intercepts from a graph or factoring and solving using the zero products property.

METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.

METHOD 2:
When we factored in Chapter 5 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!

y² – 5y = 6 = 6y – 18
first put this in standard form
y² – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3

Substitute to verify that 8 and 3 are solutions!!

METHOD 3:

You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!

DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!

This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!

3x² = 18
divide both sides by 3 and get: x² = 6
square root each side and get x = + SQRT 6 or - SQRT 6


(x - 5)² = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2

(x + 2)² = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7

METHOD 4:
The Quadratic Formula!!
We will get to this one...

FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t² + v₀t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t² + v₀t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t²+v₀t + c We can find when it hits the ground by solving 0 = -5t² + v₀t + c



For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t² + 9t + 2 or
-5t² + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics