Algebra Honors ( Periods 5 & 6)

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 5 & 6)

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)

Quadratic Functions 8-8

A QUADRATIC FUNCTION IS NOT y = mx + b
(which is a LINEAR function),
but instead is
y = ax² + bx + c
OR
f(x) = ax² + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared (degree of 2)

Quadratics have a squared term, so they have 2 possible solutions (roots)
You already saw this when you factored the trinomial and used zero products property.

If the domain is all real numbers, then you will have a PARABOLA which looks like a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.


Let's look at f(x) = x² - 2x - 2

When we plotted a few points, we discovered that definitely was NOT a straight line.
We used f(0), f(1), f(2), f(3), f(4), and then f(-1), f(-2)
we connect with a smooth curved line  a U shape with arrows at either end--> because the parabola continues without end. Do not put the equation of the parabola on it, however.
Notice this parabola opens upward and has a minimum point or lowest point  at (1, -3)
the y coordinate at this point is the least value of the function.
Domain: {x Ι x = R} This is read as "the Domain is x such that x equals all real numbers."  and the
 Range: { y Ι  y  ≥ -3} This is read as "the Range is y such that y is greater than or equal to -3."

the vertical line x = 1 contains this minimum point and is called the axis of symmetry.   If you fold the graph along the axis of symmetry the two halves coincide. You could always just plot points-- but how would you know which points to pick?

Let's look at f(x) = -x²  + 2x + 2
Just by looking at this function I can tell that it opens downward. That (0,2) is where one side of the parabola crosses the y-intercept.

f(x) = ax²  + bx + c ( a ≠ 0 )   is a quadratic function.  Why can't a =0?
If the domain of f is the set of all real numbers, then the graph is a parabola
if a in the function f(x) =ax²  + bx + c is positive, the graph opens upward. It's a happy face!
If a in the function f (x) = ax²  + bx + c is negative, the graph opens downward. It's a sad face!

The minimum or maximum point is called the vertex of the parabola.
Notice that all the other points ( except the vertex) occur in pairs with the same y-coordinate-- and the same distance away from the axis of symmetry. That makes sense, doesn't it?

The x coordinate of the vertex of a parabola f(x) = ax²  + bx + c
is  -b/2a
YES... you do need to memorize this!
The axis of symmetry is the line  x = -b/2a

H:x  --> 2x²  + 4x -3
Find the vertex
Remember -b/2a is the x coordinate of the vertex
so
-b/2a = -4/2(2) = -1
The x coordinate of the vertex is -1. Plug in this x value to find the corresponding y value for the vertez
f(-1) = 2(-1)²  + 4(-1) -3 = 2 -4 -3 = -5
so the vertex is
(-1, -5)
and
x = -1     is the line of symmetry or  the AXIS OF SYMMETRY

What happens as the "a" coefficient gets really big or really small (fraction/decimal)? We'll look at that together on my graphing calculator.

But think, what happened when the "m" (slope) coefficient got big?

The slope got steeper.

So now think that both sides of the U get steeper at the same time.

What's happening to the shape of the U???

Now think, what happened when the "m" (slope) coefficient got tiny?
The slope was a bunny slope.
So now think that both sides of the U are bunny slopes at the same time.
What's happening to the shape of the U???

Putting in standard form:
Standard form is:
y = ax² + bx + c

OR

f(x) = ax² + bx + c

You can't read the sign of a, b, or c until it's in standard form (just like y = mx + b!)

Graphing quadratics:

You can graph quadratics exactly the way you graphed lines
...by plugging in your choice of an x value and using the equation to find your y value.



Because it's a U shape, you should graph 5 points as follows:

First MAKE SURE THE EQUATION IS IN STANDARD FORM!

y must be isolated on one side and then you can read the a and b coefficients.

y = ax² + bx + c


Point 1) the vertex - the minimum value of the smile or the maximum value of the frown


The x value of the VERTEX = -b/2a


Plug that into the equation and then find the y value of the vertex



Next, draw the AXIS OF SYMMETRY :

x = -b/2a


a line through the vertex parallel to the y axis


Point 2) Pick an x value IMMEDIATELY to the right or left of the AXIS OF SYMMETRY and find its
y by plugging into the equation.



Point 3) Graph its mirror image on the other side of the AXIS OF SYMMETRY by counting from the axis of symmetry


Points 4 and 5) Repeat point 2 and 3 directions with another point ONE STEP FARTHER from the AXIS OF SYMMETRY.



JOIN YOUR 5 POINTS IN A SMOOTH "U" SHAPE ( not a V shape!)

AND 
EXTEND LINES WITH ARROWS ON END



Parabolas are functions whose domains are ALL REAL NUMBERS.


Their ranges depend on where the vertex is and also if the ‘a’ coefficient is positive or negative



EXAMPLE: f(x) = -3x²

(or y = -3x² )

the ‘a’ coefficient is negative so it is a frowny face


The vertex is called the maximum.


The x value of the vertex is -b/2a 


a = -3 and b = 0 (it's missing!)


The x value of the vertex = -b/2a = -0/2(-3) = 0


Plug the x value of 0 back into the function to find the y value of the
vertex:


y = -3(02) = 0
So the vertex is (0, 0)

The domain is all real numbers.


The range is y is less than or equal to zero (It's a frowny face)



To graph this function:


1) Graph vertex (0, 0)


2) Draw the AXIS OF SYMMETRY –
a dotted line at x = 0 (actually this is the y axis!)


3) Pick x value immediately to the right of axis of symmetry, x = 1


Plug it in the equation to find the y value: y = -3(1) = -3


Plot (1, -3)


4) Count the same 1 step from axis of symmetry on the other side of the axis
and place another point to the LEFT of axis at the same y value
(-1, -3)


5) Pick another x value to the right 2 steps away from the axis of symmetry, x = 2

Plug it in the equation to find y:
y = -3(22) = -12
Plot (2, -12)


6) Count 2 steps from axis of symmetry on the other side of it and
place another point to the LEFT of axis at the same y value
(-2, -12)



JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END