Download Lesson # 18 Aim: How do we complete the square? - mvb-math

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Subject Class Calendar Fall 2009
Date
Day
Lesson
Lesson #1 Aim: How do we perform operations
involving monomials and polynomials?
Lesson #2 Aim: How do we divide polynomials?
Lesson #3 Aim: How do we solve first degree
inequalities
Lesson #4 Aim: How do we solve compound linear
inequalities ?
Lesson #5 Aim: How do we graph absolute value
relations and functions?
Lesson #6 Aim: How do we solve equations
involving absolute values?
Lesson #7 Aim: How do we solve absolute value
inequalities?
Lesson #8 Aim: How can we factor polynomials?
Lesson #9A Aim: How do we factor the difference
of two perfect squares or polynomials completely?
Lesson #9B Aim: How do we solve quadratic
equations by factoring?
Lesson # 10 Aim: How do we graph the parabola y =
ax2 + bx + c?
Lesson #11 Aim: How do we solve and graph a
quadratic inequality?
Lesson #12 Aim: How can we use the graph of a
parabola to solve quadratic inequalities in two
variables?
Lesson #13 Aim: How do we solve more complex
quadratic inequalities?
Lesson #14 Aim: How do we simplify radicals?
Lesson #15 Aim: How do we add and subtract
radicals?
Lesson #16 Aim: How do we multiply and divide
radicals?
Lesson #17 Aim: How do we rationalize a fraction
with a radical denominator (monomial or binomial
surd)?
Lesson # 18 Aim: How do we complete the square?
Lesson #19 Aim: How do we apply the quadratic
formula to solve quadratic equations with rational
roots?
Lesson #20 Aim: How do we apply the quadratic
formula to solve quadratic equations with irrational
roots?
HW
Comple
te
Yes
Yes
Yes
Yes
Yes
Yes
Partial
Yes
Yes
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
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Lesson #21 Aim: How do we apply the quadratic
formula to solve verbal problems?
Lesson #22 Aim: What are complex numbers and
their properties?
Lesson #23 Aim: How do we add and subtract
complex numbers?
Lesson #24 Aim: How do we multiply complex
numbers?
Lesson #25 Aim: How do we divide complex
numbers and simplify fractions with complex
denominators?
Lesson #26 Aim: How do we find complex roots a
quadratic equation using the quadratic formula?
Lesson 27: Aim: How do we use the discriminant to
determine the nature of the roots of a quadratic
equation?
Lesson #28 Aim: How do we find the sum and
product of the roots of a quadratic equation?
Lesson #29 Aim: How do we solve systems of
equations using the graphing calculator?
Lesson #30
Aim: How do we solve systems of
equations
Lesson #31 Aim: How do we reduce rational
expressions?
Lesson #32 Aim: How do we multiply and divide
rational expressions?
Lesson #33 Aim: How do we add and subtract
rational expressions with like or unlike
denominators?
Lesson #34 Aim: How do we add and subtract
rational expressions with unlike polynomial
denominators?
Lesson #35 Aim: How do we reduce complex
fractions?
Lesson #36 Aim: How do we solve rational
equations?
Lesson #37 Aim: How do we solve rational
inequalities?
Lesson #38 Aim: How do we evaluate expressions
involving negative and rational exponents?
Lesson #39 Aim: How do we find the solution set for
radical equations?
Lesson #40 Aim: How do we find the solution set of
an equation with fractional exponents?
Lesson #41 Aim: What are relations?
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
Partial
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Lesson #42 Aim: What are functions?
Partial
Lesson #43 Aim: How do we use function notation?
Partial
Lesson #44 Aim: What are compositions of
functions?
Lesson #45 Aim: How do we find the inverse of a
given relation?
Lesson #46 Aim: What is an exponential function?
Partial
Lesson #47 Aim: What is the inverse of the
exponential function?
Lesson #48 Aim: How do we find the log b a?
Partial
Partial
Partial
Lesson #49 Aim: How do we use logarithms to find
values of products and quotients?
Lesson #50 Aim: How do we use logarithms for
raising a number to a power or finding roots of
numbers?
Lesson #51 Aim: How do we solve exponential
equations?
Lesson #52 Aim: How do we solve exponential
and logarithmic equations?
Lesson #53 Aim: How do we solve verbal
problems leading to exponential or logarithmic
equations?
Lesson #54 Aim: What is a transformation? How
do we do Transformations involving reflections?
Lesson 55
Aim: What are geometric translations,
dilations and rotations?
Lesson #56 Aim: How do we perform transformations
of the plane on relations and functions?
Lesson #57 Aim: How do we graph and write the
equation of a circle?
Lesson #58 Aim: What is direct and inverse variation?
Lesson #59 Aim: How do we find the roots of
polynomial equations of higher degree by factoring
and by applying the quadratic formula?
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Lesson #1 Aim: How do we perform operations with polynomial expressions containing
rational coefficients?
Students will be able to
1. Add and subtract polynomials with rational coefficients
2. Multiply and divide monomials with exponents and rational coefficients
3. Simplify parenthetical expressions (nested groupings)
4. Multiply polynomials with integral and rational coefficients
5. Explain the procedures used to add, subtract, multiply and divide monomials and
polynomials
Homework:
Do Now: Simplify
3xy+2xy+10xy=15xy
9ab(2ab3)=18a2b4
(x+5)(x+4)=x2+9x+20
Definitions:
Monomial -An expression which contains only one term, for instance (2a,3xy,5, x2)
Binomial - A polynomial of two terms such as 10a+12b is a binomial What is a bigomist?
Man with 2 wives. What is a bisexual? A person who goes with 2 different sexes.
Trinomial - A polynomial of three terms such as x2+3x+2
Polynomial- is a sum of monomials. What is a polygon? A polygamist?
Coefficient- If a terms has a variable, the numerical factor is called the coefficient
 Example –1x+5Y+17; -1X and 5Y are terms; -1 and 5 are coefficients, 17 is a
constant term.
Term- An algebraic expression that is a number, a variable or the product or quotient of
numbers or variables. Not separated by a + or a -.
Example 5, 2x, -4/7y2, .7a2b3 , 5x+5, 3xy, 5ab,7x2
Example of separate terms: 2x+y, 3x2+y, 2x+5
Like Terms- Two or more terms that contain the same variable, with corresponding
variables having the same exponents.
 Example: 6k and k; 5x2 and –7x2; 9ab and 2/5ab
Unlike Terms- Two or more terms that contain different variables, or the same variables
with different exponents.
 Example 3x and 4y; 5x2 and 5x3; 2xy and 2x
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Standard Form- The terms decrease in terms from left to right.
To add polynomials, combine like terms by adding their numerical coefficients.
To subtract a polynomial add the opposite (additive inverse) (p9)
Example Add: (2x2 - 4) + (x2 + 3x - 3)
1. Using a horizontal method to add like terms:
Remove parentheses. Identify like terms. Group the like terms together.
Add the like terms.
(2x2 - 4) + (x2 + 3x - 3) = 2x2 - 4 + x2 + 3x – 3 = 2x2 + x2 + 3x - 4 – 3
= 3x2 + 3x – 7
Using a vertical method to add like terms:
Arrange the like terms so that they are lined up under one another in vertical columns.
Add the like terms in each column following the rules for adding signed numbers.
2x2 + 0x - 4
+ x2 + 3x - 3
--------------------3x2 + 3x - 7
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Examples
Classwork and Homework
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Lesson #2 Aim: How do we divide polynomials?
Students will be able to
1. Divide a polynomial by a polynomial including polynomials with rational coefficients;
with and without remainders
2. Apply the operations of multiplication and addition of polynomials to check quotients
Do Now:
Homework:
Steps for Dividing a Polynomial by a Monomial:
1. Divide each term of the polynomial by the monomial.
a) Divide numbers (coefficients)
b) Subtract exponents
2.
Remember that numbers do not cancel and disappear! A number divided by itself
is 1. It reduces to the number 1.
3.
Remember to write the appropriate sign in between the terms.
Notice how the
numbers (the
coefficients) were
divided.
The polynomial on the
top has 3 terms and the
answer has 3 terms.
Answer:
Notice how the
exponents were
subtracted.
Notice how the last
term reduced to
one.
Steps for Dividing a Polynomial by a Binomial:
1. Remember that the terms in a binomial cannot be separated from one another
2.
when reducing. For example, in the binomial 2x + 3, the 2x can never be reduced
unless the entire expression 2x + 3 is reduced.
Factor completely both the numerator and denominator before reducing.
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3. Divide both the numerator and denominator by their greatest common factor.
Examples:
1.
How do we factor a
binomial?
2.
How do we factor the
difference of 2 perfect
squares?
3.
How do we factor a
trinomial?
4.
Notice that the -1 was
factored out of the
numerator to create a
binomial compatible with
the one in the
denominator.
2 - x = -1(x - 2)
Practice:
Lesson #3 Aim: How do we solve first degree equations and inequalities?
Students will be able to
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1. Apply the postulates of equality to solve first degree equations algebraically
2. Apply the postulates of inequality to solve first degree inequalities algebraically
3. Solve first degree equations and inequalities graphically
4. Graphically justify the solution of each linear equation and inequality found
algebraically
Do Now:
Homework:
Vocabulary:
An equation- is a sentence that uses the symbol = to state that two quantities are equal.
An inequality- is a sentence that uses one of the symbols of order, namely >,<,  , .
Solution Sets- Finite, infinite or empty. Subset of the domain that make the sentence true.
Open Sentence- One or more of the quantities in the relationship contains a variable. The
sentence is neither true nor false until the variable is replaced with one of the elements of
the domain.
Closed Sentence- Or a statement that can be judged to be true or false. No variables.
Properties of Equality:
Add/Subtract
For all numbers a, b, & c, if a = b, then a + c = b + c and a - c = b - c.
Property
Mult/Division
For all numbers a, b, and c, if a = b, then a * c = b * c, and if c not
Property
equal to zero, a ÷ c = b ÷ c.
Properties of Inequality
Add/Subtract Property If equals are added or subtracted from unequals the sums or
differences are unequal in the same order.
Mult/Division
If unequals are multiplied or divided by positive equals, the
Property
products or quotients are unequal in the same order. If they are
multiplied by negative equals the products or quotients are
unequal in the opposite order.
Distributive Property
For all numbers a, b, & c, a(b + c) = ab + ac.
Model Example Solve and check for an equality:
5(x-1)-3=2x-(3-x)
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5x-5-3=2x-3+x
5x-8=3x-3
2x=5
x=5/2
Simplify Each Side
Use additive inverse to have variable on one side
Use multiplicative inverse
Sample problems:
-3x+9=24
-3x=15,
x=-5
X=12
K=-39
Solving Inequalities
Model example for an inequality
2(5-x)>3+1
10-2x>4
-2x>-6
x<3
Steps: Clear parentheses, combine like terms, use additive inverse, use multiplicative
inverse, change direction of inequality.
Graphing inequalties on a number line.
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Lesson #4 Aim: How do we solve and graph compound linear inequalities involving the
conjunction and disjunction?
Students will be able to
1.Graph inequalities on a number line
2. Solve and graph conjunctions and disjunctions of two inequalities
3. Apply graphing compound inequalities to problems involving the conjunction or the
disjunction
Do Now: What is the definition of a compound sentence in English? What is a compound
in science? What is a compound equality in math? What is a compound inequality in
math?
Homework:
Vocabulary
Compound Inequality- Two inequalities that are joined by the word and or the word or.
The solution of one joined by and is any number that makes both inequalities true.
Two inequalities joined by or is any number that makes either inequality true.
A compound linear inequality is one that has two inequalities in one problem. For
example, 5 < x + 3 < 10 or -1 < 3x < 5. You solve them exactly the same way you solve
the linear inequalities shown above, except you do the steps to three "sides" (or parts)
instead of only two.
Example 9: Solve, write your answer in interval notation and graph the solution
set:
This is an example of a compound inequality
*Inv. of add. 2 is sub. by 2
*Apply steps to all three parts
Interval notation:
*All values between -6 and 8, with
a closed interval at -6 (including -6)
Graph:
*Visual showing all numbers
between -6 and 8, including -6 on
the number line.
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Interval notation:
This time we have a mixed interval since we are including where it is equal to -6, but not
equal to 8. x is between -6 and 8, including -6, so -6 is our smallest value of the interval
so it goes on the left and 8 goes on the right. The boxed end on -6 indicates a closed
interval on that side. A curved end on 8 indicates an open interval on that side.
Graph:
Again, we use the same type of notation on the endpoints as we did in the interval
notation, a boxed end on the left and a curved end on the right. Since we needed to
indicate all values between -6 and 8, including -6, the part of the number line that is in
between -6 and 8 was darkened.
Compound Inequalities – Conjunctions
 To solve a compound inequality, you must solve each part separately.
 A compound inequality containing and is true only if both parts of it are true.
 This means, the graph of a compound inequality containing and must be the
intersection of the graphs of the two solution parts.
 Where the graphs overlap or intersect determines the solution set.
 Example 1
7<2x+5<17
7<2x+5
7-5<2x
1<x
2x+5<17
2x<12
x<6
 Example 2
-2x-6>4 or
-2x>10
or
x<-5
x+5>4
x>-1
solution set is x<-5 or x>-1
 Model Example 1
4  r  5  1
4  5  r  5  5
1 r
r  5  1
r  5  5  1  5
r4
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 Model Example 3 4v+3<-5 or 2v+7<1
Collaborative Group Activity
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Lesson #5 Aim: How do we graph absolute value relations and functions?
Students will be able to
1. Explain how to use the graphing calculator to graph absolute value relations and
functions
2. Determine the appropriate window for each graph
3. Use the graphing calculator to graph absolute value relations and functions
4. Apply the transformations f(x+a), f(x) +a, -f(x), and af(x) to the absolute value function
Need Graphing Calculators and Overhead
Do Now:
Describe what the graph of the following 2 equations would look like.
y=-2x+3
Straight line, negative slope of -2, y intercept of 3.
2
y=x +2
Parabola, Opens upward
Homework:
Complete the following table: Describe each graph using math terminology.
Type
Equation
Description
Linear
Y=mx+b or
Graph is a straight line with m = slope and
y=-4x-3
Functions f(x) =mx+b
b= y intercept
Slope = -4 and
y intercept =-3
Constant y=b or f(x) =b Every x corresponds to a constant function
y=3
Function
y.
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Quadratic y=ax +bx+c
Graph is a parabola whose axis of symmetry y=x2-4x+3
Function f(x)= ax2+bx+c is a vertical line. Every x value corresponds
to one and only one y value the relation is a
function
Absolute Y=|x|
for every value of x there is one and only
Y=|2x-4|
Value
one value of y. A V shape
Function
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Graphing Absolute Value Equations
Example 1: Solve:
Enter left side in Y1. You can find abs( )
quickly under the CATALOG (above 0)
( or MATH → NUM, #1 abs( )
Enter right side in Y2
Use the Intersect Option (2nd CALC #5) to find
where the graphs intersect. Move the spider near
the point of intersection, press ENTER. Simply
hit ENTER twice more. You must repeat this
process to find the second point of intersection.
Answer: x = 4; x = -4
Example 2: Solve:

Answer: x = 2; x =
3.3333333
The x value is stored in the
calculator's memory. If you
wish to change 3.333333333 to
a fraction, simply return to the
home screen, hit x, hit Enter.
Now, change to fraction.
(MATH, #1►Frac )
Answer: x = 2; x = 10/3
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Lesson #6 Aim: How do we solve linear equations in one variable involving absolute
values?
Students will be able to
1. State the definition of the absolute value of x
2. Apply the definition of absolute value to solve linear equations involving absolute
values
3. Graph solutions to linear absolute value equations on the number line
4. Verify solutions to absolute value equations
5. Solve verbal problems resulting in linear equations involving absolute value
Do Now: What is the definition of absolute value?
Can the absolute number of a negative number ever be negative?
Homework:
To solve an absolute value equation, isolate the absolute value on one side of the equal
sign, and establish two cases:
Case 1:
Case 2:
Set the expression inside the
absolute value symbol equal to
the other given expression.
Set the expression inside the
absolute value symbol equal to
the negation of the other given
expression
. . . and always CHECK your answers.
The two cases create "derived" equations. These derived equations may not always be
true equivalents to the original equation. Consequently, the roots of the derived
equations MUST BE CHECKED in the original equation so that you do not list
extraneous roots as answers.
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Review the following procedures
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Lesson #7 Aim: How do we solve linear absolute value inequalities involving one
variable? (pg 81-82 Amsco)
Students will be able to
1. Solve inequalities of the form |x| < k and |x| > k
2. Solve inequalities of the form x ± a < k and x ± a > k
3. Graph solutions to linear absolute value inequalities on the number line
4. Verify solutions to absolute value inequalities
5. Solve verbal problems resulting in linear inequalities involving absolute value
Do Now:
3z=3
3z=-3 z=1 z=-1
Homework:
Procedures
 Solving an absolute value inequality problem is similar to solving an absolute value
equation.
 Isolate the absolute value on one side of the inequality symbol, then follow the rules
below:
If the symbol is > (or >) :
(or)
If a > 0, then the solutions to
are x < a and x > - a.
Also written: - a < x < a.
.
Think about it: absolute value is always
positive (or zero), so, of course, it is greater
than any negative number.
If a > 0, then the solutions to
are x > a or x < - a.
If the symbol is < (or <) :
If a < 0, all real numbers will satisfy
(and)
If a < 0, there is no
solution to
.
Think about it: absolute value is always
positive (or zero), so, of course, it cannot be
less than a negative number.
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R E M E M B E R: When working with any absolute value inequality, you must
create two cases.
To set up the two cases:
If <, the connecting word is "and".
x<a
Case 1: Write the problem without the
absolute value sign, and solve the
inequality.
Model Examples
If >, the connecting word is "or".
x > -a
Case 2: Write the problem without
the absolute value sign, reverse the
inequality, negate the value NOT
under the absolute value, and solve
the inequality
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|2x+3|<7
-7<2x+3<7
-10<2x<4
-5<x<2
Graph the solution
Write a derived equation
Solve
-6 -5 -4 -3 -2 -1 0 1 2 3
|3+y|-2 ≥0
|3+y|≥2
3+y≥2
3+y≤-2
Write derived equations
Y≥-1
y≤-5
Graph the solution
-6 -5 -4 -3 -2 -1 0 1 2 3
Lesson #8 Aim: How do we factor polynomials? (Page 22 Amsco)
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Students will be able to
1. Recognize when to factor out a greatest common factor
2. Factor by extracting the greatest common factor
3. Identify and factor quadratic trinomials, where a > 1
4. Factor quadratic trinomials that are perfect squares
5. Justify the procedures used to factor given polynomial expressions
Do Now:
Factor 120 to lowest prime factors 15*4,
(5*3*2*2)
3) Factor 15x2+30x 15x(x+2)
Homework:
Example 1
Example 2
2) Factor 48x3 3*2*2*2*2*x*x*x
4) Factor x2-49
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Example of factoring trinomials
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Factoring Trinomials with coefficients >1
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Lesson #9A Aim: How do we factor the difference of two perfect squares and factor
polynomials completely?
Students will be able to
1. Recognize and factor the difference of two perfect squares
2. Compare factoring the difference of two perfect squares to factoring a trinomial
3. Explain what is meant by factoring completely
4. Factor polynomials completely
5. Justify the procedures used to factor polynomials completely
6. Explain why it is more efficient to factor completely by first extracting the GCF
7. Factor cubic expressions of the form a3 - b3 or a3 + b3 (enrichment only)
Do Now: Factor Completely
1) 3x2-12x= 3x(x-4)
2) 9y3+3y2= 3y2(3y+1)
3) 4x2+4x+1= (2x+1)(2x+1)
Homework:
 What is a perfect square? Have students list perfect squares.
 What does it mean to factor completely?
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Lesson #9B Aim: How do we solve quadratic equations by factoring?
Students will be able to
1. Justify and explain the multiplication property of zero
2. Explain what is meant by the standard form of a quadratic equation
3. Transform a quadratic equation into standard form
4. Factor the resulting quadratic expression
5. Apply the multiplication property of zero to solving quadratic equations
6. Check the answers in the original equation
Do Now:
Homework:
Quadratic equations are normally expressed as
equal zero.
, where a does not
Many of the simpler quadratic equations with rational roots can be solved by factoring.
To solve a quadratic equation by factoring:
1
Start with the equation in the form
Be sure it is set equal to zero!
2
Factor the left hand side (assuming zero is on the right)
Set each factor equal to zero
3
4
Solve to determine the roots (the values of x)
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Some possible situations:
Factoring with GCF
(greatest common factor)
Factoring with DOTS
(difference of two squares)
Find the largest value that
Look carefully at this
can be factored from each of example to refresh this
the elements of the
expression.
Factoring Trinomials
In a quadratic equation in
descending order with a
leading coefficient of one,
look for the product of the
roots to be the constant tern
and the sum of the roots to
be the coefficient of the
middle
process:
Or Isolate the Variable
term.
Factoring Harder Trinomials
If the leading coefficient is not equal
to 1, you must think more carefully
about how to set up your factors.
Tricky One!!
Be sure to get the equation set equal to zero
before you factor.
Graph 2x2 - 5x + 2
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Quadratic equations are of the form
where a, b and c are real numbers and .
. Quadratic equations have two solutions. It is possible that one solution may repeat.
Solving by Factoring
Some quadratic equations can be solved by factoring. Set the equation equal to zero and
factor.
Example 1:
Example 2:
Example 2:
Solve by factoring:
Solve by factoring:
Solve by factoring:
Solve by Graphing
Some quadratic equations can be solved by graphing. Setting the equation equal to zero will
show the roots as locations on the x-axis.
Method 1: Set the equation equal to zero, if
Example 1: Solve by graphing:
necessary. Find the roots using the ZERO
command tool of the graphing calculator. For
help with the calculator, click here.
Example 2: Solve by graphing:
Method 2: Graph each side of the equation
separately. Use the INTERSECT command
tool to find when the graphs cross. Repeat this
process for both intersection points. For help
with the calculator, click here.
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Solving by Quadratic Formula
The solutions of some quadratic equations are not rational, and cannot be factored. For
such equations, the most common method of solution is the quadratic formula. The
quadratic formula can be used to solve ANY quadratic equation, even those that can be
factored.
Be sure you know this formula!!!
Note: The equation must be set equal to zero before using the formula.
Example:
As decimal values:
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Lesson # 10 Aim: How do we graph the parabola y = ax2 + bx + c?
Students will be able to
1. create a table of values and graph a parabola
2. explain the effect of a, b, and c on the graph of the parabola
3. identify the key elements of a parabola (axis of symmetry, turning point, intercepts,
opening up/down)
4. investigate and discover the effects of the transformations f(x+a), f(x) + a, af(x), f(-x)
and –f(x) on the graph of a parabola
5. produce a complete graph of a parabola using a graphing calculator
6. use the graphing calculator to identify the roots of a quadratic equation
Writing Exercise: Video games use life-like graphics to model 3-D effects on the 2-D
computer monitor. Video golf games show the path of a golf ball that is hit by each player.
Describe the shape of the path of the ball as shown on the screen and speculate on the type
of equation(s) the programmer needed to use in order to produce this graphic.
Homework: Graph the following equations by creating a table of values
Y=5x2
Y=1/2x2
y=-7x2
Development:/Motivation/Do Now:
Solve graphically: y = 2x + 1 by creating a table of values and graphing
2x+1
y
x
-2
2(-2)+1
-3
-1
2(-1)+1
-1
0
2(0)+1
1
1
2(1)+1
3
2
2(2)+1
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Pivotal Questions
What were the steps we followed to create the graph?
 Create a table of values (What goes in each column? How many rows?)
 Select values of x to substitute in the equation (How many values do we need? What
values of x should we use?)
 Solve for y
 Label x and y axis, Put in a scale for both axis
 Plot ordered pairs of x and y
 Connect the Points
What are the characteristics of this graph? (Linear, Straight Line, Slant Upwards)
Is it a positive or negative slope? (Positive)Why? (As we go left to right y goes up.)
What is the slope of the line? (2) How did you calculate the slope? (y2-y1/x2-x1)
(or Change in y over the change in x)
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Ok I want everyone to work with me in graphing a different type of equation
Y=3x2
What are the steps we are going to follow to graph this type of equation?
 Create a Table of Values
 Select values of x to substitute ( How many values do we need?)
 Solve for y (Do we multiply by 2 first or square the value of x?)
 Label x and y axis, Put in a scale for both axis
 Plot the x,y values on a graph
 Connect the points
Before we all sketch the graph lets look at the values of y and x
Is there any pattern to the numbers. Look at –2 and +2 (y has the same value)
As x increases by 1 what happens to the value of y? Does it change by a constant amount?
(No)
Lets all of us sketch the graph on the same chart that we used for a linear equation.
How would you describe the graph? (U shaped)
What happens as we move upward? (The opening gets wider)
What would happen if we fold the graph in half (Symmetry)
Where have we seen shapes like this in the real world.
 Satellite dish; Parabolic microphone used at football games; Solar Hot Dog cooker;
clothes line; archway; path of a basketball being shot, baseball being hit to the
outfield, football being kicked.
38
Similarities
Linear
Can be graphed on Coordinate
Plane
Same procedure to create graph
Lines go on forever
Solid line
Differences
Linear
Straight Line
No variable to the
second power.
Constant Slope
Need only 2 points to
graph
Quadratic
Curved Line
One of variables is to the
second power (squared)
Different slope
Turning Point
Need more than 2 points
to graph
Axis of Symmetry
Vocabulary:
Vertex- Highest or lowest point on a parabola.
Axis of symmetry- The line down the middle of the parabola.
Quadratic- it contains the variable squared, but not raised to any higher power. For
instance a quadratic equation in x contains x2 but not x3
Quadratic Equation- An equation which asks us to find where a particular quadratic
polynomial is equal to zero. It can always be written in the form: ax2+bx+c = 0.
Parabola - A particular shape of curve, given by an equation of the form y=ax2+bx+c (a, b,
c constants, a nonzero). The path of a thrown object is a parabola.
Group Activity- Graph the following equations:
Y=2x2+1
Y = x2 - 4
y = x2 – 2x+2
y
x X2-4
y
x x2 + 2x+2
x 2x2+1
-3 2(-3)2+1 19 -3 (-3)2-4
5
-3 -32+2(-3)+2
-2 2(-2)2+1 9
-2 (-2)2-4
0
-2 -22+2(-2)+2
-1 2(-1)2+1 3
-1 (-1)2-4
-3 -1 -12+2(-1)+2
0 2(-0)2+1 1
0 (-0)2-4
-4 0 -02+2(0)+2
1 2(1)2+1
3
1 (1)2-4
-3 1 12+2(1)+2
2 2(2)2+1
9
2 (2)2-4
0
2 22+2(2)+2
3 2(3)2+1
19 3 (3)2-4
5
3 32+2(3)+2
Summary:
 How are these equations different? How are they the same?
 How do we graph these type of equations?
 How many points do we need?
 What values should we use?
 What is the shape of the graph of a linear equation?
 What is the shape of the graph of a quadratic equation?
 Where do we see shapes like that in real life?
y
5
2
1
2
5
10
17
39
Lesson #11 Aim: How do we solve and graph a quadratic inequality algebraically?
Students will be able to
1. transform a quadratic inequality into standard form
2. solve a quadratic inequality algebraically and graph the solution on a number line
3. write the solution to a quadratic inequality as a compound inequality
Writing Exercise: How can the graphing calculator be used to verify the solution set of its
related quadratic inequality?
Do Now:
Homework:
A quadratic inequality is one that can be written in one of the following standard
forms:
or
or
or
Quadratic Inequalities involve variable expressions that look just like the expressions
found in quadratic equations. The only difference is an inequality symbol ( , , ,
instead of an equals sign.
For example:
or
The key is to make sure you have a 0 on one side of the inequality before you factor. We
will then use a "special number line" to keep track of our work.
For example, to solve the inequality
Step 1: Make sure you have a 0 on one side of the inequality.
Step 2: Factor (if possible).
Our goal is to determine when this product is positive or equal to zero.
Step 3: Construct a number line and identify the x values which make the individual
factors equal to 0.
x = 0 and x = 1 will make the factors equal zero. So we have the number line
Step 4: Use the number line and the factored form to determine where the product is
positive or equal to zero. We've already seen that the product equals zero when x = 0 and
when x = 1.
)
40
So we designate these two values as part of our solution by "shading them in" on the
number line.
Our number line is now divided into three pieces.
all numbers less than 0 all numbers between 0 and 1 all numbers greater than 1
Basic Strategy: Working with one piece at a time, we can choose a test value from each of
these pieces and determine whether that test value causes our product to be positive or
negative.
In this example our product consists of the factored form
Let's try an x value less than 0, say x = - 1.
Plugging in x = - 1 yields ( - 3)( - 2) = 6, a positive number. So our product is positive
when x < 0.
Let's try an x value between 0 and 1, say x =0 .5.
Plugging in x = .5 yields (1.5)( - 0.5) = -0 .75, a negative number. So our product is
negative when 0 < x < 1.
Let's try an x value greater than 1, say x = 2.
Plugging in x = 2 yields (6)(1) = 6, a positive number. So our product is positive when x
> 1.
We designate these results on the number line by "shading in" the pieces which
produce the desired outcome.
In this example, we shade in the pieces which make the product positive.
So the solution set for the inequality
is { all real numbers x such that
x 0 or x 1 }. We can also write this solution using interval notation
Note: If the inequality in this example were " " instead of " " our solution set would
not include the endpoints x = 0 and x = 1.
Remember, in this method, the factored form will tell you where to "break up the number
line," and which pieces to keep, simply by using test numbers to determine the sign of each
piece seperately.
Special Note: If the given quadratic has no real solutions, then the expression will either
41
always be positive or always be negative. In this case, simply plug x = 0 into the quadratic
and the resulting number value will allow you to answer the question.
Special Case: If the given quadratic does not factor conveniently, we must use the
quadratic formula to determine which numbers (if any) should be used to break up the
number line.
Example 1
Solve the given inequality and give your answer using interval notation as well as
inequality notation.
Solution
Step 1: Multiply out and make sure you have a 0 on one side of the inequality.
We now have a quadratic inequality.
Step 2: Factor (if possible).
In this example, our goal is to determine when this product is positive.
Step 3: Construct a number line and identify the x values which make the individual
factors equal to 0.
In this case, x = - 4 and x = - 3 will make the factors equal zero. So we have the number
line
Step 4: Use the number line and the factored form to determine where the product is
positive.
We've already seen that x = - 4 and x = - 3 make the product equal zero. Thus, these
values do not yield the desired outcome.
We designate this result with "open circles" on the number line at x = - 4 and x = - 3.
Our number line is now divided into three pieces.
Namely:
all numbers less than – 4, all numbers between - 4 and – 3, all numbers greater than
-3
Basic Strategy: Working with one piece at a time, we can choose a test value from each of
these pieces and determine whether that test value causes our product to be positive or
negative.
42
In this example our product consists of the factored form
Let's try an x value less than - 4, say x = - 5.
Plugging in x = - 5 yields ( - 1)( - 2) = 2, a positive number. So our product is positive
when x < - 4.
Let's try an x value between - 4 and - 3, say x = - 3.5.
Plugging in x = - 3.5 yields (.5)( - .5) = - 0.25, a negative number. So our product is
negative when - 4 < x < - 3.
Let's try an x value greater than - 3, say x = 0.
Plugging in x = 0 yields (4)(3) = 12, a positive number.
So our product is positive when x > - 3.
We designate these results on the number line by "shading in" the pieces which produce
the desired outcome. In this example, we shade in the pieces which make the product
positive.
So the solution set for the inequality
is { all real numbers x such that x
< - 4 or x > - 3 }. We can also write this solution using interval notation
43
Lesson #12 Aim: How can we use the graph of a parabola to solve quadratic inequalities in
two variables?
Students will be able to
1. identify the key elements (axis of symmetry, turning point, intercepts, opening up/down)
of a parabola
2. apply transformations to the graph of a parabola
3. produce a complete graph of a parabola using a graphing calculator
4. use the graphing calculator to calculate the roots of a quadratic equation
5. solve a quadratic inequality in two variables graphically
6. write the solution to a quadratic inequality as a compound inequality
Writing Exercise: In class, we plot 6 or 8 points to define the complete shape of a
parabola. How do we determine what x-values to use in order to create a complete graph of
the parabola?
Lesson #13 Aim: How do we solve more complex quadratic inequalities?
Students will be able to
1. transform a quadratic inequality into standard form
2. solve quadratic inequalities
3. graph the solution set of a quadratic inequality on the number line
4. apply graphing to the conjunction and the disjunction
5. graph the related parabola to identify the solution to a quadratic inequality
44
Lesson #14A Monday November 28
Aim: What are Roots and Radicals
Objectives: Students will be able to
 Understand the meaning of square roots and cube roots
 Understand the nth root of a number
 Evaluate simple expressions involving square roots
Homework: Page 401 # 6-12,19-22
Do Now:
Vocabulary:
Square Root- is one of two equal factors whose product is that number. Every positive real
number has two square roots.
Principal Square Root- of a positive number is its positive square root.
Cube Root- Is one of 3 equal factors whose product is that number.
Nth Root- Is one of n equal factors whose product is that number.
n k where k=radicand, n =the index
 The square root of 0 is 0.
 The square root of a negative number are not real numbers.
Find the following
144
100 x 4

3
8 y 4
 5 1
The square root of a product of positive numbers is equal to the product of the square roots
of the numbers.
a *b  a b
200  4 * 50 = 10 2
 The simplest form of a square root radical has the greatest perfect square factored out.
 The square root of a quotient of positive numbers is equal to the quotient of the square
roots of the numbers.
a
a
=
b
b
5
5
=
16
4
45
Lesson #14B Aim: How do we simplify radicals?
Objectives: Students will be able to
 simplify radicals with a numerical index of 2 or 3.
 simplify radicals involving literal radicands.
 explain the procedure for simplifying radical expressions.
 explain how to determine when a radical is in simplest form.
Homework:
Day 1 9-4 Worksheet Simplifying Radicals- Multiplication
Day 2 9-4 Worksheet Division with Radicals
Day 3 Page 406 # 5-12,32-36
Do Now:
Procedure To simplify (or reduce) a radical:
Find the largest perfect square which will divide evenly into the number under your
radical sign. This means that when you divide, you get no remainders, no decimals, no
fractions.
Reduce:
the largest perfect square that divides evenly into 48 is 16.
If the number under your radical cannot be divided evenly by any of the perfect squares,
your radical is already in simplest form and cannot be reduced further.
2. Write the number appearing under your radical as the product (multiplication) of the
perfect square and your answer from dividing.
3) Give each number in the product its own radical sign.
4. Reduce the "perfect" radical which you have now created.
5. You now have your answer.
What happens if I do not choose the largest perfect square to start the process?
If instead of choosing 16 as the largest perfect square to start this process, you choose 4,
look what happens.....
46
Unfortunately, this answer is not in simplest form.
The 12 can also be divided by a perfect square (4).
If you do not choose the largest perfect square to start the process, you will have to repeat
the process.
Example: Reduce
Don't let the number in front of the radical distract you. It is simply "along for the ride"
and will be multiplied times our final answer.
The largest perfect square dividing evenly into 50 is 25.
Reduce the "perfect" radical and multiply times the 3 (who is "along for the ride")
47
Lesson #15 Aim: How do we add and subtract radicals?
Students will be able
1. add and subtract like radicals with numerical or monomial radicands
2. add and subtract unlike radicals with numerical or monomial radicands
3. express the sum or difference of radicals in simplest form
4. explain how to combine radicals
Writing Exercise: Why is x6 a perfect square monomial and x9 not a perfect square
monomial?
 When adding or subtracting radicals, you must use the same concept as that of adding
or subtracting like variables.
In other words, the radicals must be the same before you add (or subtract) them.
Since the radicals are the same, simply add the numbers
in front of the radicals (do NOT add the numbers under
the radicals).
1. Are the radicals the same?
2. Can we simplify either
radical?
3. Simplify the radical.
NO
Yes,
can be simplified.
4. Now the radicals are the
same and we can add.
1. Simplify
Answer:
2. Simplify
Answer:
3. Since the radicals in steps 1 and 2 are now
the same, we can combine them.
4. You are left
with:
5. Can you combine these radicals?
Answer: NO
48
6. Therefore,
Answer:
----------------->
Lesson #16 Aim: How do we multiply and divide radicals?
49
Objectives: Students will be able to: multiply radical expressions, divide radical
expressions, express the products and quotients of radicals in simplest form, express
fractions with irrational monomial denominators as equivalent fractions with rational
denominators.
Do Now:
Homework:
When multiplying radicals, one must multiply the numbers OUTSIDE (O) the radicals
AND then multiply the numbers INSIDE (I) the radicals.
When dividing radicals, one must divide the numbers OUTSIDE (O) the radicals AND
then divide the numbers INSIDE (I) the radicals.
1. Multiply the outside numbers first
2. Multiply the inside numbers
3. Put steps 1 and 2 together and
simplify
4. Therefore, the answer is 72.
1. Divide the outside numbers first.
2. Divide the inside numbers.
3. Put steps 1 and 2 together and
simplify.
4. Therefore, the answer is:
2•3=6
50
51
Lesson #17 Aim: How do we rationalize a fraction with a radical denominator?
Objectives: Students will be able to: rationalize monomial denominators, rationalize
binomial denominators, express results in simplest form.
A fraction that contains a radical in its denominator can be written as an equivalent
fraction with a rational denominator. Never leave a radical in the denominator of a
fraction. Always rationalize the denominator.
1. When the denominator is a monomial (one term), multiply both the numerator and the
denominator by whatever makes the denominator an expression that can be simplified so
that it no longer contains a radical.
 Sometimes the value being multiplied happens to be exactly the same as the
denominator, as in Example 1:
Multiplying the top and bottom by will create the smallest
perfect square under the square root in the denominator.
Replacing
by 7 rationalizes the denominator.
 Sometimes you need to multiply by whatever makes the denominator a perfect square
or perfect cube or any other power that can be simplified, as in Examples 2 and 3.
Multiply by a value that will create the smallest perfect
square under the radical. This will prevent the need for
additional simplifications.
Choosing to multiply by (and not ) will create the
smallest perfect square under the radical in the denominator.
Multiplying by will create the smallest perfect cube
under the radical. Replacing
by 3, rationalizes the
denominator.
52
Make sure you multiply by whatever makes the radicand (the number under the radical
sign) the smallest possible value to be simplified. This will avoid having to further
simplify later on.
2. When there is more than one term in the denominator, the process is a little tricky.
You will need to multiply the denominator by its conjugate. The conjugate is the
same expression as the denominator but with the opposite sign in the middle.
Write the reciprocal of
53
Be sure to enclose expressions with multiple terms in ( ). This will help you to remember
to FOIL these expressions. Always reduce the root index (numbers outside radical) to
simplest form (lowest) for the final answer.
3. When working with the reciprocal of a rational expression , if there is a radical in the
denominator, you must rationalize the denominator.
An expression is in simplest radial form if:
1. The radicand of an nth root has no nth powers in it, (radical part is fully reduced)
2. the root index is as low as possible, and (fraction outside radical is fully reduced)
3. there are no radicals in the denominator (rationalize the denominator).
54
55
Lesson # 18 Aim: How do we complete the square?
Students will be able to
1. identify a perfect square trinomial
2. factor a perfect square trinomial
3. express a perfect square trinomial as the square of a binomial
4. state the relationship between the coefficient of the middle term and the constant term of
a perfect square trinomial
5. determine the constant term that needs to be added to a binomial or trinomial to make it
a perfect square trinomial
6. solve quadratic equations by completing the square
Factoring by Completing the Square

Find the x-intercepts of y = 4x2 – 2x – 5.
First off, remember that finding the x-intercepts means setting y equal to zero and solving
for the x-values, so this question is really asking you to "Solve 4x2 – 2x – 5 = 0".
This is the original problem.
Move the loose number over to the other
side.
Divide through by whatever is multiplied
on the square term.
Take half of the coefficient (don't forget
the sign!) of the x-term, and square it.
Add this square to both sides of the
equation.
Convert the left-hand side to squared
form, and simplify the right-hand side.
Square-root both sides, remembering the
"±" on the right-hand side. Simplify as
necessary.
Solve for "x =".
4x2 – 2x – 5 = 0
4x2 – 2x = 5
56
Remember that the "±" means that you
have two values for x.
When you complete the square, make sure that you are careful with the sign on the x-term
when you multiply by one-half. If you lose that sign, you can get the wrong answer in the
end. Also, don't be sloppy and wait to do the plus/minus sign until the end.
Here's another example:

Solve x2 + 6x – 7 = 0 by completing the square.
Do the same procedure as above, in exactly the same order. (Study tip:
Always working these problems in exactly the same way will help you
remember the steps when you're taking your tests.)
This is the original problem.
x2 + 6x – 7 = 0
Move the loose number over to the other side.
x2 + 6x
=7
Take half of the x-term (that is, divide it by two)
(and don't forget the sign!), and square it. Add
this square to both sides of the equation.
Convert the left-hand side to squared form.
Simplify the right-hand side.
Square-root both sides. Remember to do "±" on
the right-hand side.
Solve for "x =". Remember that the "±" gives you
two solutions. Simplify as necessary.
(x + 3)2 = 16
x+3=±4
x=–3±4
= – 3 – 4, –3 + 4
= –7, +1
If you are not consistent with remembering to put your plus/minus in as soon as
you square-root both sides, then this is an example of the type of problem where
you'll get yourself in trouble. You'll write your answer as "x = –3 + 4 = 1", and have
no idea how they got "x = –7", because you won't have a square root symbol
"reminding" you that you "meant" to put the plus/minus in. That is, if you're sloppy,
these easier problems will embarass you!
57
Here's another one:

Solve x2 + 6x + 10 = 0.
Apply the same procedure as before: Copyright © Elizabeth Stapel 20002005 All Rights Reserved
This is the original equation.
Move the loose number over to the other side.
x2 + 6x + 10 = 0
x2 + 6x = – 10
Take half of the coefficient on the x-term (that is,
divide it by two) (don't forget the sign!), and square it.
Add this square to both sides of the equation.
Convert the left-hand side to squared form. Simplify
the right-hand side.
(x + 3)2 = –1
Note: If you don't know about complex numbers yet,
then you have to stop at this step, because a square
can't equal a negative number! Otherwise, proceed...
Square-root both sides. Remember to put the "±" on
the right-hand side.
Note the square root of a negative number!
Solve for "x =", and simplify as necessary.
x = –3 ± i
If you don't yet know about complex numbers (the numbers with "i" in them), then you
would say that this quadratic has "no solution". If you do know about complexes, then you
would say that this quadratic has "no real solution" or that is has a "complex solution". In
either case, this quadratic had no "real" solution. Since solving "(quadratic) = 0" for x is
the same as finding the x-intercepts (assuming the solutions are real numbers), it stands to
reason that this quadratic should not intersect the x-axis.
This relationship is always true. If you come up with a real value on the right side (a zero
value is real, by the way; the square root of zero is just zero), then the quadratic will have
two x-intercepts (or only one, if you get plus/minus of zero on the right side); if you get a
negative on the right side, then the quadratic will not cross the x-axis.
58
Lesson #19
Aim: How do we apply the quadratic formula to solve quadratic equations
with rational roots?
Objectives: Students will be able to: explain how the method of completing the square
results in the quadratic formula, state the quadratic formula, use the quadratic formula to
solve quadratic equations, express rational roots in simplest form.
Do Now:
1) Solve by factoring x2+3x-4 (x+4)(x-1)
2) Solve by factoring x2-5x+6
(x-2)(x-3)
2
3) Solve by factoring x -10x+13=0 Cannot be factored See example 2
Homework: Page 426 # 4-6,10,11

Derive the Quadratic Formula by solving ax2 + bx + c = 0.
This is the original equation.
Move the loose number to the other side.
Divide through by whatever is multiplied
on the squared term.
Take half of the x-term, and square it.
Add the squared term to both sides.
Simplify on the right-hand side; in this
case, simplify by converting to a common
denominator.
Convert the left-hand side to square form
(and do a bit more simplifying on the
right).
Square-root both sides, remembering to
put the "±" on the right.
Solve for "x =", and simplify as necessary.
Discriminant:
ax2 + bx + c = 0
ax2 + bx = –c
59
1) B2-4ac is a positive number that is a perfect square, the quadratic equation has 2
rational roots.
2) B2-4ac =0, the quadratic equation has two equal rational roots or in effect one root.
3) B2-4ac, is a positive number that is not a perfect square, the quadratic equation has
irrational roots therefore the equation can not be solved by factoring.
4) B2-4ac is a negative number, the roots are imaginary.
Example 1
Here are some examples of how the Quadratic Formula works: Solve x2 + 3x – 4 = 0
This quadratic happens to factor: x2 + 3x – 4 = (x + 4)(x – 1) = 0.so x = –4 and x = 1. How
would this look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, it looks like this:
Example 2
Find the roots for 2x2+5x=12
Transform to 2x2+5x-12=0 a=2, b=5 and c=-12
B2-4ac= 25-4(2)(-12)= 121
Perfect square therefore it can be solved by factoring.
Example 3
Find the roots for x2-10x+13=0
60
Lesson #20 Aim: How do we apply the quadratic formula to solve quadratic
equations with irrational roots?
Objectives: Students will be able to
 state the quadratic formula.
 use the quadratic formula to solve quadratic equations.
 express irrational roots in simplest radical form.
 approximate irrational roots in decimal form.
When the roots of a quadratic equation are imaginary, they always occur in conjugate
pairs.
A root of an equation is a solution of that equation.
If a quadratic equation with real-number coefficients has a negative discriminant, then the
two solutions to the equation are complex conjugates of each other. (Remember that a
negative number under a radical sign yields a complex number.)
The discriminant is the b2- 4ac part of the quadratic formula (the part under
the radical sign). If the discriminant is negative, when you solve your
quadratic equation the number under the radical sign in the quadratic formula
is negative --- forming complex roots.
Quadratic equation:
Quadratic formula:
Example 1)
Find the solution set of the given equation over the set of complex numbers. Pick out the
coefficient values representing a, b, and c, and substitute into the quadratic formula, as you
would do in the solution to any normal quadratic equation. Remember, when there is no
number visible in front of the variable, the number 1 is there.
a = 1,
b = -10,
c = 34
a = 3, b = -4, c = 10
61
62
63
Lesson #21 Aim: How do we apply the quadratic formula to solve verbal problems?
Students will be able to
1. state the quadratic formula
2. create an appropriate quadratic equation that can be used to solve the verbal problem
3. use the quadratic formula to identify possible solutions (roots)
4. use the zero finder of a graphing calculator to verify the roots of a quadratic equation
5. verify each solution in the words of the problem
Writing Exercise: Tronty’s answer to his real world quadratic verbal problem gave him
two roots. He rejected the negative root. Deepak
Lesson #22 Aim: What are complex numbers and their properties?
Objectives: Students will be able to: define an imaginary number and a complex number;
 simplify powers of I; differentiate between complex and imaginary numbers; plot
points on the complex number plane; identify a complex number as a vector quantity;
compute the absolute value of a number.
Do Now:
Homework: Page 924 # 10-12,17-20,36,37
The Imaginary Unit is defined as i =
.
It is said that the term "imaginary" was coined by René Descartes in the seventeenth
century and was meant to be a derogatory reference since, obviously, such numbers did not
exist. Today, we find the imaginary unit being used in mathematics and science.
Electrical engineers use the imaginary unit (which they represent as j ) in the study of
electricity. Imaginary numbers occur when a quadratic equation has no roots in the set of
real numbers.
A pure imaginary number can be written in bi form where b is a real number and i is
Examples: pure imaginary numbers
*
Practice Page 924 # 1-5

 36  36  1  6i

 100  100  1  10i

2  49  2 49  1  14i
i=
or
-i=
-
64
Solve the following quadratic equation: x2 + 4 = 0 Remember: The Imaginary Unit is
defined as i =
.
A complex number is any number that can be written in the standard form a + bi, where
a and b are real numbers and i is the imaginary unit.
A complex number is a real number a, or a pure imaginary number bi, or the sum of both.
Note these examples of complex numbers written in standard a + bi form: 2 + 3i, -5 + 0i .
Complex Number
a
bi
7 + 2i
7
2i
1 - 5i
1
- 5i
8i
0
8i
a + bi
The set of real numbers and the set of imaginary numbers are subsets of the set of
complex numbers. Whenever there is a negative under the radical sign, it comes out from
underneath as i. When simplifying these radical numbers in terms of i, follow the usual
rules for simplifying radicals and treat the i with the rules for working with a variable.
Examples:
Number
bi Form
1
2
3
4
6i
2 x 7i
14i
65
Lesson #23 Aim: How do we add and subtract complex numbers?
Objectives: Students will be able to add and subtract complex numbers algebraically and
express answers in a+bi form; add and subtract complex numbers graphically and express
answers in a+bi form; find the additive inverse of complex numbers.
Do Now:
Homework:
Rule:
 Add like terms
 Find the sum of the real components.
 Find the sum of the imaginary components (the components with the i after them).
REMEMBER: Final answer must be in simplest
form.
Examples:
3) Express the sum of
and
in the form
.
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4) Add
and
.
Rules:
Subtracting Rule:
*Subtract like terms*
Find the difference of the real components. Find the difference of the imaginary
components (the components with the i after them).
REMEMBER: Final answer must be in simplest
form.
3) Subtract
4) Subtract
1.
2.
from
from
.
.
67
3.
4.
-1 + 6i
5.
6.
7.
8.
9.
10.
11. Add:
,
and
,
.
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12. Subtract
.
from
13. From the sum
, subtract
and
.
Lesson #24 Aim: How do we multiply complex numbers?
Objectives: Students will be able to multiply and combine expressions that involve
complex numbers; write the conjugate of a given complex number.
Multiplication:
Multiplying two complex numbers is accomplished in a manner similar to multiplying two
binomials. You can use the FOIL process of multiplication, the distributive property, or
your personal favorite means of multiplying.
Distributive Property:
(2 + 3i) • (4 + 5i) = 2(4 + 5i) + 3i(4 + 5i)
= 8 + 10i + 12i + 15i2
= 8 + 22i + 15(-1)
= 8 + 22i -15
= -7 + 22i Answer
Be sure to replace i2 with (-1) and proceed with
the simplification. Answer should be in a + bi
form.
The product of two complex numbers is a complex number.
(a+bi)(c+di) = a(c+di) + bi(c+di)
= ac + adi + bci + bdi2
= ac + adi + bci + bd(-1)
= ac + adi + bci - bd
= (ac - bd) + (adi + bci)
= (ac -bd) + (ad + bc)i answer in a+bi form
The conjugate of a complex number a + bi is the complex number a - bi. For example, the
conjugate of 4 + 2i is 4 - 2i. (Notice that only the sign of the bi term is changed.)
The product of a complex number and its conjugate is a real number, and is always
positive.
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(a + bi)(a - bi) = a2 + abi - abi - b2i2
= a2 - b2 (-1) (the middle terms drop out) = a2 + b2 Answer
This is a real number ( no i's ) and since both values are squared, the answer is positive.
1) Multiply: (3+5i)(3-5i) Answer (34)
2) Multiply: (8+9i)(7-3i) Answer (83+39i)
3) Multiply: (4-3i)(3-4i) Answer (-25i)
Lesson #25 Aim: How do we divide complex numbers and simplify fractions
Objectives: Students will be able to write the conjugate of a given complex number;
find the quotient of two complex numbers and express the result with a real denominator;
express the multiplicative inverse of a complex number in standard form.
Division:
When dividing two complex numbers:
 Write the problem in fractional form
 Rationalize the denominator by multiplying the numerator and the denominator by the
w
70
conjugate of the denominator.
 Remember that a complex number times its conjugate will give a real number.
This process will remove the i from the denominator.)
Example:
Dividing using the conjugate:
Answer
1) Simplify: (2+5i)2 Answer (-21+20i)
2) Simplify: 8+i(8-i) Answer (9+8i)
3) Simplify:
Answer (3+2i)
4) Simplify:
Answer 35/37 + (12/37)I
5) Simplify:
Answer -5-3i
6) Simplify:
Answer 2/15 + i/15
Lesson #26 Aim: How do we find complex roots of a quadratic equation using the
quadratic formula?
Objectives: Students will be able to
 Define the discriminant, determine the value of the discriminant, determine the nature
of the roots using the discriminant.
 Solve a quadratic equation that leads to complex roots, express in a + bi form.
Lesson 27: Aim: How do we use the discriminant to determine the nature of the roots of a
quadratic equation?
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Students will be able to
1. Define the discriminant
2. Determine the value of the discriminant
3. Determine the nature of the roots using the discriminant
Do Now:
Homework:
When the roots of a quadratic equation are imaginary, they always occur in conjugate
pairs. A root of an equation is a solution of that equation.
If a quadratic equation with real-number coefficients has a negative discriminant,
then the two solutions to the equation are complex conjugates of each other.
(Remember that a negative number under a radical sign yields a complex number.)
The discriminant is the b2- 4ac part of the quadratic formula (the part under the radical
sign). If the discriminant is negative, when you solve your quadratic equation the number
under the radical sign in the quadratic formula is negative --- forming complex roots.
Quadratic equation:
Quadratic formula:
B2-4ac>0 and a perfect square
B2-4ac>0 and not a perfect square
B2-4ac=0
B2-4ac<0
Roots of equation are real rational and unequal
Roots of equation are real, irrational and unequal
Roots are real, rational and equal
Roots are imaginary
Classwork:
Have students determine the discriminant for each equation and tell what the roots should
be. Then have them solve and prove if that is the case and what the answer is. Sketch a
graph of each equation. At how many places does the graph touch the x axis?
X2-2x-3=0 A=1 b=-2 c=-3 B2-4ac=(-2)2-4(1)(-3)=16
Real,rational and unequal
(3,-1)
2
2
2
X -6x+7=0 A=1 b=-6 c=7 B -4ac=(-6) -4(1)(7)=8
Real, irrational and unequal
2
2
2
Real, rational and equal
X -4x+4=0 A=1 b=-4 c=4 B -4ac=(-4) -4(1)(4)=0
2
2
2
Roots are imaginary
X -4x+5=0 A=1 b=-4 c=5 B -4ac=(-4) -4(1)(5)=-4
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Example 1) Find the value of the discriminants. Describe what the roots of the answer
should be. Then find the solution set of the given equation over the set of complex numbers.
Pick out the coefficient values representing a, b, and c, and substitute into the quadratic
formula.
HINT: When the directions say: Express over the set of complex numbers, look for a
negative value under the radical sign.
a = 1,
b = -10,
c = 34
a = 3, b = -4, c = 10
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 Be sure to set the quadratic equation equal to 0.
 Arrange the terms of the equation from the highest exponent to the lowest exponent.
Lesson #28 Aim: How do we find the sum and product of the roots of a quadratic
equation?
Objectives: Students will be able to find the sum and product of the roots of a quadratic
equation, find the missing coefficients given one root of the equation, check the solutions
to quadratic equations using the sum and product relationships.
Homework:
Do Now:
What are the sum of the roots? (1) What are the product of the roots? (-20)
We can find the sum and products of the roots through the following shortcut.
-b/a= 1
c/a= -20
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There is a definite relationship between the roots of a quadratic equation and the
coefficient of the second term and the constant term.
The sum of the roots of a quadratic equation is equal to the negation of the coefficient of
the second term divided by the leading coefficient.
The product of the roots of a quadratic equation is equal to the constant term divided by
the leading coefficient.
Lets try a few of these
Equation
Factors
2
X +5x-14
(x+7)(x-2)
2
X -4x+3
(x-3)(x-1)
2
2x +12x+10
(x+5)(2x+2)
2
6x +9x-15
(3x-3)(2x+5)
2
X -8x+16
(x-4)(x-4)
Solution
X=-7,2
X=3,1
X=-5,-1
X=1.-2.5
X=4,4
Sum
-5
4
-6
-1.5
8
Product
-14
3
5
-2.5
16
=-b/a
-5
4
-6
-9/6
8
c/a
-14
3
5
-15/6
16
Exercise: Take 2 Binomials and multiply using FOIL. Solve for the roots. Find their sum
and their product. Test that –b/a and c/a will give you the same 2 answers. Do this exercise
twice.
Lesson #29 Aim: How do we solve quadratic-linear systems of equations using the
graphing calculator?
.
Students will be able to
1. Explain what is meant by a system of equations and by the solution to a system of
equations
2. Explain how to use the graphing calculator to graph each equation
3. Explain how to use the graphing calculator to solve systems of equations
4. Determine the graphing window for a system of equations
5. Identify the graphs from their equations
6. Solve systems of equations using the graphing calculator
7. Check the solution to the system of equations
Lesson #30 Aim: How do we solve quadratic-linear systems of equations algebraically?
Note: This includes rational equations that can
be written as linear equations with restricted domain, which, if not carefully considered
might produce extraneous roots for the system. i.e. and y x x x
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y=1=2−.
Students will be able to
1. Explain how to solve for one variable in terms of the other
2. Explain how to substitute one equation into the other to create one equation in one
variable
3. Algebraically solve the system of equations for all possible solutions
4. Algebraically check the solutions to the system of equations
5. Graphically verify the solution of a quadratic-linear system found algebraically
A quadratic equation is defined as an equation in which one or more of the terms is squared
but raised to no higher power. The general form is ax2 + bx + c = 0, where a, b and c are
constants.
In Algebra and Geometry, we learned how to solve linear - quadratic systems algebraically
and graphically. With our new found knowledge of quadratics, we are now ready to attack
problems that cannot be solved by factoring, and problems with no real solutions.
The familiar linear-quadratic system:
(where the quadratic is in one variable)
Remember that linear-quadratic systems of this type can result in three graphical situations
such as:
The equations will intersect in two
locations. Two real solutions.
The equations will intersect in one
location. One real solution.
The equations will not intersect.
No real solutions.
Keep these images in mind as we proceed to solve these linear-quadratic systems
algebraically.
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Example 1:
When we studied these systems in Algebra, we encountered situations that could be
solved by factoring, such as this first example.
Solve this system of equations algebraically:
y = x2 - x - 6 (quadratic equation in one variable of form y = ax2 + bx + c )
y = 2x - 2
(linear equation of form y = mx + b)
Substitute from the linear
equation into the quadratic
equation and solve.
y = x2 - x - 6
2x - 2 = x2 - x - 6
2x = x2 - x - 4
0 = x2 - 3x - 4
Find the y-values by substituting There are 2 "possible" solutions
each value of x into the linear for the system: (4,6) and (-1,-4)
equation.
Check each in both equations.
y = 2(4) - 2 = 6
POINT (4,6)
y = 2(-1) - 2 = -4
POINT (-1,-4)
0 =(x - 4)(x + 1)
x-4=0
x=4
x + 1 =0
x = -1
See how to use your
TI-83+/84+ graphing
calculator with
quadratic-linear
systems.
Click calculator.
y = x2 - x - 6
6 = (4)2 - 4 - 6 = 6 checks
y = 2x - 2
6 = 2(4) - 2 = 6 checks
y = x2 - x - 6
-4 = (-1)2 - (-1) - 6 = -4 checks
y = 2x - 2
-4 = 2(-1) - 2 = -4 checks
Answer:
{(4, 6), (-1, -4)}
Example 2:
With our new found knowledge of quadratics, we are now ready to attack problems
that cannot be solved by factoring, and/or problems with no real solutions (such as
this second example).
Solve this system of equations algebraically:
y = x2 - 2x + 1 (quadratic equation in one variable of form y = ax2 + bx +
c)
y=x-3
(linear equation of form y = mx + b)
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Substitute from the linear
equation into the quadratic
equation.
y = x2 - 2x + 1
x - 3 = x2 - 2x + 1
0 = x2 - 3x + 4
Use quadratic
formula:
Find the y-values
There are 2 "possible" solutions.
by substituting each value
Check each in both equations.
of x into the linear
y = x2 - 2x + 1
equation.
POINT
No real solutions.
y=x-3
POINT
-------------------------------------
y = x2 - 2x + 1
Answer:
There are no real solutions.
The answers are complex numbers, which are not
graphed in the Cartesian coordinate plane.
y=x-3
Other linear - quadratic systems: (where the quadratic is in two variables)
Quadratics in two variables look like x2 + y2 = 16 where two variables are squared.
Example 3
Solve this system of equations algebraically:
x2 + y2 = 25 (quadratic equation of a circle center (0,0), radius 5)
4y = 3x
(linear equation)
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Substitute from the linear
equation into the quadratic
equation and solve.
Find the y-values
There are 2 "possible" solutions
by substituting each value of x for the system: (4,3) and (-4,-3)
into the linear
Check each in both equations.
equation.
x2 + y2 = 25
42 + 32 = 25
16 + 9 = 25
25 = 25 checks
4y = 3x
4(3)=3(4)
Answer:
12 = 12 checks
{(4, 3), (-4, -3)}
x2 + y2 = 25
(-4)2 + (-3)2 = 25
16 + 9 = 25
25 = 25 checks
4y = 3x
4(-3)=3(-4)
-12 = -12 checks
Solve this system of equations algebraically:
x2 + y2 = 26 (quadratic equation)
x-y=6
(linear equation)
Substitute from the linear
Find the y-values
There are 2 "possible" solutions
equation into the quadratic
by substituting each value of x for the system: (4,3) and (-4,-3)
equation and solve.
into the linear
Check each in both equations.
equation.
x2 + y2 = 26
52 + (-1)2 = 26
25 + 1 = 26
Answer:
26 = 26 checks
x-y=6
{(5, -1), (1, -5)}
5 - (-1) = 6
6 = 6 checks
x2 + y2 = 26
12 + (-5)2 = 26
1 + 25 = 26
26 = 26 checks
x-y=6
1 - (-5) = 6
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6 = 6 checks