Algebra Honors (Periods 6 & 7)

Imaginary Numbers

x²= -16

No real number solutions

Imaginary number i us defined to be √-1

√-1   2³   ∑   ∏

I ate some pie!

All imaginary numbers involve i

3i        -5i/2   and 2i√3

i   =√-1

i²  = -1

so

x²= -16  solved over the set of imaginary numbers

x= √-16 = √16∙√-1 = 4√-1 =4i

If r > 0 then √-r = i√r

I = √-1

i² = -1

i³ = i∙i² =i(-1) = -i

i⁴ = i²∙i² = (-1)(-1) = 1

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

Algebra Honors ( Periods 5 & 6)

 Solving Problems Involving Quadratic Equations 12-6

You can use quadratic equations to solve problems...
We used the examples in the book to start with:

The park commission wants a new rectangular sign with an area of 25 m² for the visitor center. The length of the sign is to be 4 m longer than the width . To the nearest tenth of a meter, what will be the length and the width of the sign?
Always make sure you check before AND after-- to see what the problem is really asking for...
Let x = the width in meters
then x + 4 = the length in meters

Use the formula for the area of a rectangle to write an equation

x(x+4) = 25
solve it
x(x +4) = 25 becomes
x² + 4x = 25
You can use two methods: the quadratic formula would be my second choice since completing the square works easily here

x² + 4x + 4 = 25 + 4
(x + 2)² = 29
x + 2 = ±√29
You can use your calculator, or the table of square roots... or approximately easily using the method taught in class earlier this year
but to the nearest tenth you get
-2 + √29 ≈3.4
-2 - √29 ≈-7.4

Since you can't have a negative root since a negative length has no meaning... you know the width must be about 3.4 meters and therefore the length is 3.4 +4 or approximately 7.4 meters

Problem 2:
The sum of a number and its square is 156. Find the number

Let x = the number
then
x² + x = 156

x² + x - 156 =0
Using the skills you have for factoring
(x +13)(x-12) = 0
so
x = 13 and x = 12
You have two solutions to this question!!

Problem 3:
The altitude of a triangle is 9 cm less than the base. The area is 143 cm²
What are the altitude and base?

Remember the formula for the area of a triangle is A = ½bh

Let b = the length of the base
then the altitude ( the height) is b-9
so
½(b)(b-9) = 143
b² -9b = 286
b² -9b - 286 = 0
(b -22)(b +13) = 0
b = 22 and b= -13
You can't have a negative length so
the base is 22 cm and the altitude is 13 cm


An object that moves through the air and is solely under the influence of gravity is called a projectile. The approximate height (h) in meters of a projectile at t seconds after it begins its flight from the ground with initial upward velocity v₀ is given by the formula
h = -5t² + v₀t
We can find when such a projectile is at ground level (h=0) by solving
0 = -5t² + v₀t.
If a 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 h=0 or
0 = -5t² + v₀t +c.

When a projectile is thrown into the air with an initial vertical velocity of r feet per second, its distance (d) in feet above the starting point t seconds after it is thrown is approximately
d = rt – 16t²

Algebra Honors (Periods 5 & 6)

The Quadratic Formula: 12-3

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