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```					             Unit 10
• Learn the Three
Expressions
Equations
Lecture 1
Objectives
expression
• Simplify perfect
square factors under

Index
y

What number multiplied by itself
“y” times gives “x”?
Example
Simplify:

2
25  5
49  7
8  42 4       2 2 2
1. No perfect squares under the radical

2
24          46
4      6           2 6
Factor out perfect squares!
• Radicals with like indices can be multiplied.

3
5   3
7  35
3

3 18  54  9 6  3 6
Homework Set 10.1
• WS Simplifying
Lecture 2
Objectives
• Simplify perfect
Do you know
square factors under       what I am?
• Work with fractions    I am a Perfect Square!
2. No fractions under the radical

Convert a
2           2        fraction under
fraction of
3. No radicals in the denominator

2    3               6            6
  
 3                 
3                    9           3
Multiply numerator and denominator by the
Homework Set 10.2
• WS Simplifying
• Quiz next class
day
Lecture 3
Objectives
equation
equations by taking
square roots
A quadratic equation is like a linear equation
except that it contains a variable raised to
the second power.

ax  bx  c  0
2

Constant term
Solving a quadratic equation is much more
complicated than solving a linear equation. Try to
solve the following equation using the properties
of equality:

x  3x  4  0
2
While the properties of equality cannot
solve most quadratic equations, they do
work on some of them:

2x  8  0
2
Square Roots
Any quadratic equation the lacks a linear term
can be solved by:
1. Solving for the squared variable using the
properties of equality
2. Taking the square root of both sides of the
equation to solve for the variable
3. Make sure to include the “plus or minus” with
the square root step, because the square root of a

ax  bx  c  0
2
Example
Solve:   3x  24  0
2

3x  24
2

x 8
2

x  8
2

x  2 2
A few more complicated quadratics seem to
be all “set-up” for square roots:

( x  7)  162

( x  7)   16
2

x  7  4
x  7  4  3 or 11
Homework Set 10.3
Equations
Square Roots
10.3
Lecture 4
Objectives
equations by
factoring
Many quadratic equations can be solved by
factoring them into a product, and then using
something called the zero-product rule.

x  3x  0
2                   Factor out the “x”

x  x  3  0
Apply the zero-product
rule

x  0 or  3             Viola!
The Zero-Product Rule
The only way a set of numbers can multiply
to zero is if one of the numbers is zero.

If a b  0, then a  0 or b  0.
While not all quadratics can factor into
products, the zero-product rule is very
useful to solve those that do.
Examples
( x  7)  0
Solve:               2

x  7
(4 y  3)( y  2)  0
y  3       or 2
4
z (3z  2)( z  2)  0
z  0 or  2 or 2
3
Today we will focus on two kinds of factoring:

1. Factoring out a monomial.
2. Factoring a trinomial square.

ax  bx  c  0
2
Factoring Out a Monomial
When there is no constant term in a quadratic, you
can always factor out a common monomial
factor:

3x  4 x  0
2

x(3 x  4)  0
Then use the zero-product rule to get the solutions.
Trinomial Squares
Some trinomial quadratics (has all three terms)
factor into perfect squares. The only way they
do this, however, is if their coefficients have a
special relationship given by these forms:

a  2ab  b   a  b 
2                    2                    2

a  2ab  b   a  b 
2                    2                   2
Trinomial Squares
Notice that the first and last terms are perfect
squares, and the linear term is two times the
product of the square roots of the ends. Only
very special quadratic equations can be factored
this way.

a  2ab  b   a  b 
2                   2                  2

a  2ab  b   a  b 
2                   2                  2
Examples
Which of these are trinomial squares?

x  4x  4  0
2                                yes
9 x  24 x  16  0
2                                    yes
4x  6x  9  0
2              no
25x  70 x  49  0
2               yes
Solving Trinomial Squares
Trinomial squares are the easiest of all the
quadratics to solve. Simply factor them
according to their form, and then use the zero-
product rule.

a  2ab  b   a  b 
2                   2                   2

a  2ab  b   a  b 
2                   2                   2
Examples
Solve:   9 x  24 x  16  0
2

(3x  4)  0
2

x4
3
4 x  12 x  9  0
2

(2 x  3)  0
2

x  3
2
Homework Set 10.4
10.4
Lecture 5
Objectives
equations by
factoring
Today we will focus on another kind of
factoring. Many quadratic polynomials are the
result of multiplying two binomials together:

 x  3 x  2  x 2  5x  6
 x  7  x  3  x 2  4 x  21
 x 1 x  5  x 2  4 x  5
We can reverse this process and factor many
by doing the following:

x  bx  c   x  m x  n
2

where b  m  n and c  m n
All you have to do is find two numbers that
multiply to “c” and add to “b”!
Examples
Factor:

x  3x  2   x  1 x  2
2

y  y  20
2       y  5 y  4

r  5r  6   r  3 r  2
2

z  3z  70   z  10 z  7 
2
After these quadratics are factored, use the
zero-product rule to find the solutions:

x  5 x  24   x  8 x  3
2

x  8 or 3
Homework Set 10.5
10.5
Lecture 6
Objectives
equations by
factoring
Today we will focus on another kind of
factoring. Many quadratic polynomials are the
result of multiplying two binomials together:

 2 x  3 x  2  2 x 2  7 x  6
 3x  7  2x  3  6 x2  5x  21
 5x 13x  5  15x2  22 x  5
We can reverse this process and factor many
by doing the following:
1.   Multiply a x c and write it above b.
2.   Find two numbers “m” and “n” so that
m x n = a x c and m + n = b.
3.   Rewrite the quadratic but replace the linear term
with two terms with coefficients “m” and “n”.
4.   Factor by grouping.

ax  bx  c  0
2
Factor:
6                   Example
Rewrite the
original
2x  7 x  3
2                Multiply a x c and
write above b

equation
replacing b
m  6 and n  1    Find m x n = a x c
and m + n = b

2 x  6 x  1x  3
2                       Group and
factor

 2x  6x    1x  3
2

2x  x  3  1 x  3 Use the
Distributive

 2x 1 x  3
Property
Homework Set 10.6
10.6
• Quiz next class
day
Lecture 7
Objectives
• Learn the Factoring
Map
Quadratic equations contain a variable squared
and cannot be solved by using the properties of
equality. To solve a quadratic, several different
methods need to be used:
1. Take square roots
2. Factor:
1.   Simple monomial
2.   Trinomial square
3.   a=1 (short-cut)
4.   a1 (4-step)
Homework Set 10.7