Transcript Class 2: Objectives

Class 2: College Algebra Objectives



Solve Quadratic equations
Add, subtract, multiply, and divide
complex numbers.
Solve Quadratic equations in the
complex number system.
Quadratic Equations


Quadratic Equations can be
identified by having a degree of 2
(squared variable).
Several methods to solve




Factoring
Square Root Method
Quadratic Formula
POLY
Multiply the following: (2x + 3)(3x – 7)







Distribute the first quantity through the
second quantity
2x(3x – 7) + 3(3x – 7)
6x2 – 14x + 9x – 21
6x2 – 5x – 21
Notice the first term comes from the
factors at the front of the parenthesis.
Notice the last term comes from the
factors at the back of the parenthesis
Notice the middle term comes from
adding the products of the inside terms
and the outside terms
Process

Regardless of what method of solving the
quadratic equation must first be put into
standard form.
ax2 + bx + c = 0

Factoring






Check for like terms
Factors of first term go at front of parenthesis
Factors of last term go at back of parenthesis
Check for Middle Term
Set each factor equal to zero and solve
Check answer/s
Examples: Solve x2 + 6 = 5x
Standard Form: x2 – 5x + 6 = 0
Check for like terms None
Factors of first term go at front of
parenthesis: (x
)(x
)=0
Factors of last term go at back of
parenthesis: (x – 2)(x – 3) = 0
Check for Middle Term: -2x + -3x = -5x
Set each factor equal to zero and solve
x – 2 = 0 x – 3 = 0 -> x=2 or x=3
Check answer/s 22 + 6 = 5(2)
10 =
10
32 + 6 = 5(3)
15 = 15
Example: Solve 2x2 = x + 3









Standard Form: 2x2 – x – 3 = 0
Check for like terms
Factors of first term go at front of
parenthesis: (2x
)(x
)=0
Factors of last term go at back of
parenthesis: (2x – 3)(x + 1) = 0
Check for Middle Term: -3x + 2x = -1x
Set each factor equal to zero and solve
2x – 3 = 0
x+1=0
2x = 3
x = -1
Check answer/s
Example: Solve 9x2 – 6x + 1 = 0








Standard Form:
Check for like terms
Factors of first term go at front of
parenthesis: (3x )(3x
)=0
Factors of last term go at back of
parenthesis (3x – 1)(3x – 1) = 0
Check for Middle Term: -3x + -3x = -6x
Set each factor equal to zero and solve:
3x – 1 = 0
3x – 1 = 0
3x = 1
3x = 1
x = 1/3
x = 1/3
Check answer/s
Double Root: have two of the same
answer
Square Root Method





Can only be used when one variable
exists and that variable is squared.
Move numbers away from squared
term.
Take the square root of each side
(remember the +- sign)
Solve the remaining equation
Check answer/s
Example: Solve x2 = 49
X2 = 49
Only one variable that is squared.
Can take a square root to cancel square with
variable.
Remember that a variable could be a
positive or negative value whose square is
positive.
Sqrt (x2) = + Sqrt 49
Square and square root cancel with variable
leaving:
X=+7
Check Answers
72 = 49
(-7)2 = 49
Example: Solve (x – 2)2 = 25








Only one variable so can solve by “doing
opposite”.
Take square root of both sides
(remember + )
Sqrt(x-2)2 = + sqrt (25)
Solve for x by moving numbers away
from variable.
X–2=5
x – 2 = -5
X=7
x = -3
Check
(7 – 2)2 = 25
(-3 – 2)2 = 25
Example: Solve (x + 1)2 - 3 = 1









Isolate squared term
(x + 1)2 = 4
Take square root of both sides
Sqrt (x + 1)2 = + sqrt 4
Solve for x
X+1=2
x + 1 = -2
X=1
x = -3
Check
(1 + 1)2 – 3 = 1
(-3 + 1)2 – 3 = 1
Quadratic Formula

Quadratic Formula was developed
from a process called “completing
the square”.
Quadratic Formula:
X = (-b + sqrt(b2 – 4ac)) / 2a


Discriminant: portion under the
square root b2 – 4ac
Discriminant Rules: We will use these rules when
solving quadratic equations in the complex number
system (next class).



b2 – 4ac > 0 : two unequal real
solutions
b2 – 4ac < 0: two complex
conjugate solutions (no real
solutions)
b2 – 4ac = 0 : one root which
occurs twice - a root of multiplicity
2 (double root)
Examples: Solve 3x2 – 5x = -1
1. Standard Form
3x2 – 5x + 1 = 0
 2. Identify a, b, and c
a=3
b = -5
c=1
 3. Plug values into the quadratic formula
and simplify
(5 + sqrt(25 – 4(3)(1)))/ 2(3)
(5 + sqrt 13) / 6
 4. Check solutions: use calculator and
decimal approximations to check
solutions.

3 ((5 + sqrt 13) / 6)2 – 5((5 + sqrt 13) / 6) = -1
3 ((5 - sqrt 13) / 6)2 – 5((5 - sqrt 13) / 6) = -1
Example: Solve
(25/2) x2 – 30x + 18 = 0









Multiply both sides by common
denominator to eliminate denominator
2(25/2)x2 – 2(30)x + 2(18) = 0
25x2 – 60x + 36 = 0
Standard Form: 25x2 – 60x + 36 = 0
Identify a, b, and c:
a = 25, b = - 60, c = 36
Solve with quadratic formula – once you
have a solution click enter to check your
answer.
X = 6/5
Check answer
12.5(6/5)2 – 30(6/5) + 18 = 0
Solve: 9 + 3/x – 2/x2 = 0








Multiply by common denominator to
eliminate fractions
x2( 9 + 3/x – 2/x2) = x2(0)
9x2 + 3x - 2 = 0
Put in standard form: 9x2 + 3x - 2 = 0
Identify a, b, c
Use quadratic formula to solve
x = -2/3
x = 1/3
Check answer
Solve: 2x2 – 5x + 3 = 0


Solve this problem on your own.
Once you have an answer click
“enter” to check your answer. If
you do not have the correct answer,
please see Mrs. Dorshorst for help.
x = 3/2
x=1
Application

From each corner of a square piece
of sheet metal, remove a square of
side 9 centimeters. Turn up the
edges to form an open box. If the
box is to hold 144 cubic
centimeters, what should be the
dimensions of the piece of sheet
metal?
From each corner of a square piece of sheet metal, remove a square
of side 9 centimeters. Turn up the edges to form an open box. If the
box is to hold 144 cubic centimeters, what should be the dimensions
of the piece of sheet metal?



Volume: Amount that container
holds.
Volume = length x width x height
x – 18
X - 18
From each corner of a square piece of sheet metal, remove a square
of side 9 centimeters. Turn up the edges to form an open box. If the
box is to hold 144 cubic centimeters, what should be the dimensions
of the piece of sheet metal?

Volume:











Length: x – 18
Width: x – 18
Height: 9
V = 9 (x – 18)(x – 18)
144 = 9(x - 18)2
144/9 = (x – 18)2
+ 12/3 = x - 18
+ 4 + 18 = x ->
22 = x
- 4 + 18 = x
->
14 = x
Discard the 14 as the dimension of the side.
Why?
Sides would be 22 cm by 22 cm.
Example:

A ball is thrown vertically upward
from the top of a building 96 feet
tall with an initial velocity of 80
ft/sec. The distance s (in feet) of
the ball from the ground after t
seconds is s = 96 + 80t – 16t2


After how many seconds does the ball
strike the ground?
After how many seconds will the ball
pass the top of the building on its way
down?
A ball is thrown vertically upward from the top of a building 96
feet tall with an initial velocity of 80 ft/sec. The distance s (in
feet) of the ball from the ground after t seconds is
s = 96 + 80t – 16t2





How long before ball is on the ground?
When the ball is on the ground the
distance is 0.
0 = 96 + 80t – 16t2
Put in standard form then solve by
quadratic formula
a = -16, b = 80, c = 96
t = 6 sec
A ball is thrown vertically upward from the top of a
building 96 feet tall with an initial velocity of 80
ft/sec. The distance s (in feet) of the ball from the
ground after t seconds is s = 96 + 80t – 16t2






How long before ball reaches top of
building?
Reaches top of building when s=96
96 = 96 + 80t – 16t2
Can solve by factoring or by using
quadratic formula.
Get in standard form first then
solve.
t = 5 sec
Assn: Page 106

#9, 11, 15, 21, 39, 49, 61, 65, 79,
85, 95, 99, 105