2.1 Equations
Solving Equations with the
Variable on Both Sides
Objectives:
• to solve equations with the variable on both sides.
• to solve equations containing grouping symbols.
• Solve linear equations in one variable.
• Apply these skills to solve practical problems.
• Justify steps used in solving equations.
To solve these equations,
•Use the addition or subtraction
property to move all variables to one
side of the equal sign.
Let’s see a few examples:
1) 6x - 3 = 2x + 13
-2x
-2x
4x - 3 = 13
+3 +3
4x = 16
4
4
x=4
Be sure to check your
answer!
6(4) - 3 =? 2(4) + 13
24 - 3 =? 8 + 13
21 = 21
Let’s try another!
2) 3n + 1 = 7n - 5
-3n
-3n
1 = 4n - 5
+5
+5
6 = 4n
4
4
Reduce! 3 = n
2
Check:
3(1.5) + 1 =? 7(1.5) - 5
4.5 + 1 =? 10.5 - 5
5.5 = 5.5
Here’s a tricky one!
3) 5 + 2(y + 4) = 5(y - 3) + 10
• Distribute first.
5 + 2y + 8 = 5y - 15 + 10
• Next, combine like terms.
2y + 13 = 5y - 5
• Now solve. (Subtract 2y.)
13 = 3y - 5 (Add 5.)
18 = 3y
(Divide by 3.)
6=y
Check:
5 + 2(6 + 4) =? 5(6 - 3) + 10
5 + 2(10) =? 5(3) + 10
5 + 20 =? 15 + 10
25 = 25
Let’s try one with fractions!
4) 3
1
1
3
 x  x
8 4
2
4
Steps:
•Multiply each term
by the least common
3
1
1
3 denominator (8) to
(8)  (8) x  (8) x  (8)
8
4
2
4 eliminate fractions.
3 - 2x = 4x - 6
3 = 6x - 6
9 = 6x so x = 3/2
•Solve for x.
•Add 2x.
•Add 6.
•Divide by 6.
Two special cases:
6(4 + y) - 3 = 4(y - 3) + 2y
3(a + 1) - 5 = 3a - 2
24 + 6y - 3 = 4y - 12 + 2y
3a + 3 - 5 = 3a - 2
21 + 6y = 6y - 12
- 6y - 6y
21 = -12 Never true!
21 ≠ -12 NO SOLUTION!
3a - 2 = 3a - 2
-3a
-3a
-2 = -2 Always true!
We write IDENTITY.
2.1 Answers (4-64x 4)
(4-24 graphing)
(28-44 solve)
4. 0
8. 5
12. 10.2
16. .73
20. .5
24. 5.8
28. Null Set
32. -25/6
36. Null Set
40. All Reals, x ≠ ±5/2
44. Null Set
52. -29/4
56. Choose any a and b such that
b = -5/3a
60. r=C/2 
64. I = V/r
Understanding Algebra
Word Problems
2.2 Applied Problems
Word Problem Types
• Distance Problems
• Mixture Problems
• Work Problems
Distance Problems
Distance = (rate)(time)
Types
• Traveling in the Same Direction
– Overtaking
– Separating
• Traveling in Opposite Directions
– Traveling Toward Each Other
– Traveling Away From Each Other
• Going and Returning
Overtaking Problem
• A fishing boat leaves Tampa Bay at 4:00 a.m. and
travels at 12 knots. At 5:00 a.m. a second boat
leaves the same dock for the same destination and
travels at 14 knots. How long will it take the
second boat to catch the first?
• Let t be travel time of first boat
• Then t – 1 is the travel time of second boat
• 14(t – 1) = 12t
Separating in Same Direction
•
•
•
•
Key: Travel time same for both
Faster Vehicle = rft
Slower Vehicle = rst
Solution: rft - rst = Constant
– Constant is desired distance between
Separating Problem
• At the auto race one car travels 190 mph
while another travels 195 mph. How long
will it take the faster car to gain two laps on
the slower car if the speedway track is 2.5
miles long?
• Let t be the racing time
• 195t – 190t = (2)(2.5)
Coming Together Problem
• A freight train leaves Centralia for Chicago at the same
time a passenger train leaves Chicago for Centralia. The
freight train moves at a speed of 45 mph, and the passenger
train travels at a speed of 64 mph. If Chicago and Centralia
are 218 miles apart, how long will it take the two trains to
meet?
• Let t be time to meet
• 45t + 64t = 218
Going and Returning Problem
• Brandon and Shanda walk to Grandma’s house at a rate of
4 mph. They ride their bicycles back home at a rate of 8
mph over the same route that they walked. It takes one
hour longer to walk than to ride. How long did it take them
to walk to Grandma’s?
• Let h be time to walk
• Then h – 1 is time to ride
• 4h = 8(h – 1)
Mixture Problems
Coin Problem
• A coin bank contains four
more quarters than
nickels, twice as many
dimes as nickels, and five
more than three times as
many pennies as nickels.
If the bank contains
$22.25, how many of each
coin are in it?
• Let n be number of
nickels
• n + 4 = quarters
• 2n = dimes
• 3n + 5 = pennies
(3n + 5)(0.01) + n(0.05) + 2n(0.10) + (n + 4)(0.25) = 22.25
Interest Problem
• An investor has $500 more invested at 7% than he
does at 5%. If his annual interest is $515, how
much does he have invested at each rate?
• Let p be amount at 5%
• Then p + 500 is amount at 7%
• p(0.05)(1) + (p + 500)(0.07)(1) = 515
Solutions Problem
• How many gallons of cream that is 30%
butterfat must be mixed with milk that is 3%
butterfat to make 45 gallons that are 12%
butterfat?
• Let c be gallons of cream
• Then 45 – c is gallons of milk
• 0.30c + 0.03(45 – c) = 0.12(45)
Work Problems
work done = (rate of work)(time spent)
Work Problem
• Ron, Mike, and Tim are going to paint a house
together. Ron can paint one side of the house in 4
hours. To paint an equal area, Mike takes only 3
hours and Tim 2 hours. If the men work together,
how long will it take them to paint one side of the
house?
• Let t be time needed to paint the side.
• (1/4)t + (1/3)t + (1/2)t = 1
2-2 Word Problems Solutions
3. Gross Pay – deductions = Net (take home) pay ; X - .40x =
492
6. 40($10) + x($15) = 595; 13 hrs. overtime
9. x(2) + (600-x)(5) = 2400
12. x(1) + (15-x)10 = 15(2); 13.3 of 1% & 1.6 of 10%
15. (a) 1.5t + 2t = 224; (b) 64(1.5) = 96 m; 64(2) = 128 miles
18.1st – 1 + 4t miles; 2nd – 6t; r*t = d; 4(t+.25) + 6t = 2;
6 min or 1:21
21.Time (to target) + Time (from target) = Time (total);
x/3300 + x/1100 = 1.5
24.1st story = (8*30) = 240; 2nd story = (30*3) + ½(30)(h-3);
15h + 45 = 240; 13ft.
27. V = 2/3л*r3 + лr2h = 11250л
30. 1/8 + 1/5 = 1/x; 40/13 hrs.
33. GPA = total weighted honor pts. / total credit hours; 3.2 =
(4.8(2.75) + x(4.0)) / (48+x)
2.3 Quadratic Equations,
Solving a Quadratic Equation
• by factorization
• by graphical method
• by taking square roots
• by quadratic formula
• by using completing square
By factorization
x  7 x  10  0
2
( x  5)( x  2)  0
x  5  0 _ or _ x  2  0
x  5 _ or _ x  2
roots (solutions)
By graphical method
y
x  7 x  10  0
2
roots
O
x
By taking square roots
(2 x  3)  4
2
2x  3   4
2x  3  2
2 x  5 or 1
x  2.5 or 0.5
Solving a Quadratic Equation
by the quadratic Formula
By quadratic formula
If ax  bx  c  0, a  0
2
 b  b  4ac
x
2a
2
x  7 x  10  0
2
a= 1
b = -7
c = 10
 (7)  (7)  4(1)(10)
x
2(1)
2
x  5 or x  2
ax  bx  c  0
2
What are we going to do if we have non-zero values for
a, b and c but can't factor the left hand side?
x  6x  3  0
2
x  6 x  3
2
This will not factor so we will complete the
square and apply the square root method.
First get the constant term on the other side by
subtracting 3 from both sides.
9  3  ___
9
x  6 x  ___
2
x  6x  9  6
2
Let's add 9. Right now we'll see that it works and then we'll look at how
to find it.
We are now going to add a number to the left side so it will factor
into a perfect square. This means that it will factor into two
identical factors. If we add a number to one side of the equation,
we need to add it to the other to keep the equation true.
x  6x  9  6
2
Now factor the left hand side.
x  3x  3  6
x  3
 x  3
2
6
Now we'll get rid of the square by
square rooting both sides.
two identical factors
2
This can be written as:
Remember you need both the
positive and negative root!
 6
x3  6
Subtract 3 from both sides to get x alone.
x  3  6
These are the answers in exact form. We
can put them in a calculator to get two
approximate answers.
x  3  6  0.55
x  3  6  5.45
Let's solve another one by completing the square.
2 x  16 x  2  0
2
2
2
2
To complete the square we want the coefficient
of the x2 term to be 1.
2
x 2  8x  1  0
Divide everything by 2
16  1  ___
16
x 2  8x  ___
 8


 2 
Since it doesn't factor get the constant on the
other side ready to complete the square.
2
 16
So what do we add to both sides?
the middle term's coefficient divided by 2 and squared
x  4x  4  x  42  15
2
x  4   15
x  4   15
Factor the left hand side
Square root both sides (remember )
Add 4 to both sides to
get x alone
x  4  15
• http://www.youtube.com/watch?v=jGJrH49
Z2ZA&feature=related
In general, a quadratic equation may have :
(1) two distinct (unequal) real roots
(2) one double (repeated) real root
(3) no real roots
Two distinct (unequal) real roots
x-intercepts
One double (repeated) real roots
x-intercept
No real roots
no x-intercept
Nature of Roots
△=
2
b -
4ac
Since the expression b2 - 4ac can be
used to determine the nature of the
roots of a quadratic equation in the
form ax2 – bx + c = 0, it is called the
discriminant of the quadratic
equation.
Two distinct (unequal) real roots
△ = b2 - 4ac > 0
x-intercepts
One double (repeated) real roots
△ = b2 - 4ac = 0
x-intercept
No real roots
△ = b2 - 4ac < 0
no x-intercept
Homework pp. 90-91 (25-35 all, 35 graphing, 47, 55)
26.
a) 169
4
55. a)
16t 2  64t  48
64 2  4(16)( 48)
64 
2(16)
b) 9
c) ±10
d) ±9
28.
30.
4 5
5
 3
2
32. -3, 2/5
s  48 ft
1024
32
32
64 
 3 sec
32
64 
b)
and 1sec
s  0 ft
16t 2  64t  0
64 
64 2  4(16)( 0)
2(16)
4096
32
64
64 
 4 sec
32
64 
The Discoverers of Imaginary
Numbers
Complex numbers were first conceived and
defined by the Italian mathematician
Gerolamo Cardano, who called them
"fictitious", during his attempts to find
solutions to cubic equations. However,
Imaginary numbers were defined in 1572 by
Rafael Bombelli. At the time, such numbers
were regarded by some as fictitious or
useless, much as zero and the negative
numbers.
Imaginary Number Timeline
• http://www.google.com/search?q=history+o
f+imaginary+numbers&hl=en&safe=active
&tbs=tl:1&tbo=u&ei=lXW6SrvsKpGeMIT
ehOAP&sa=X&oi=timeline_result&ct=title
&resnum=11
2.4 Complex Numbers
OBJECTIVES
•Use the imaginary unit i to write complex numbers
•Add, subtract, and multiply complex numbers
•Use quadratic formula to find complex solutions
of quadratic equations
Consider the quadratic equation x2 + 1 = 0.
What is the discriminant ?
a = 1 , b = 0 , c = 1 therefore the discriminant is
02 – 4 (1)(1) = – 4
If the discriminant is negative, then the quadratic
equation has no real solution. (p. 114)
Solving for x , gives x2 = – 1
2
x  1
x  1
We make the following definition:
i  1
Note that squaring both sides yields i 2  1
Real numbers and imaginary numbers are
subsets of the set of complex numbers.
Real Numbers
Imaginary
Numbers
Complex Numbers
Definition of a Complex Number
If a and b are real numbers, the number a + bi is a
complex number written in standard form.
If b = 0, the number a + bi = a is a real number.
If b  0, the number a + bi is called an imaginary
number.
A number of the form bi, where b  0 , is called a
pure imaginary number.
Write the complex number in standard form
1   8  1  i 8  1  i 4  2  1  2i 2
Equality of Complex Numbers
Two complex numbers a + bi and c + di, are
equal to each other if and only if a = c and b = d
a  bi  c  di
Find real numbers a and b such that the equation
( a + 6 ) + 2bi = 6 –5i .
a+6=6
2b = – 5
a=0
b = –5/2
Addition and Subtraction of Complex Numbers,
p. 127
If a + bi and c +di are two complex numbers written
in standard form, their sum and difference are
defined as follows.
Sum:
(a  bi)  (c  di)  (a  c)  (b  d )i
Difference: (a  bi )  (c  di )  (a  c)  (b  d )i
Perform the subtraction and write the answer
in standard form.
( 3 + 2i ) – ( 6 + 13i )
(3 – 6) + (2 – 13)i
–3 – 11i
8   18  4  3i 2 
8  i 9  2  4  3i 2 
(8  4)  (3i 2  3i 2 )
4
Properties for Complex Numbers p.126
• Associative Properties of Addition and Multiplication
• Commutative Properties of Addition and Multiplication
• Distributive Property of Multiplication
Multiplying complex numbers is similar to multiplying
polynomials and combining like terms.
Perform the operation and write the result in standard form.
( 6 – 2i )( 2 – 3i )
F
O
I
L
12 – 18i – 4i + 6i2
12 – 22i + 6 ( -1 )
6 – 22i
Consider ( 3 + 2i )( 3 – 2i )
9 – 6i + 6i – 4i2
9 – 4( -1 )
9+4
13
This is a real number. The product of two
complex numbers can be a real number.
Complex Conjugates and Division p. 129
Complex conjugates-a pair of complex numbers of
the form a + bi and a – bi where a and b are
real numbers.
( a + bi )( a – bi )
a 2 – abi + abi – b 2 i 2
a 2 – b 2( -1 )
a2+b2
The product of a complex conjugate pair is a
positive real number.
To find the quotient of two complex numbers
multiply the numerator and denominator by
the conjugate of the denominator.
a  bi 
c  di 

a  bi  c  di 


c  di  c  di 
ac  adi  bci  bdi

2
2
c d
2
ac  bd  bc  ad i

2
2
c d
Perform the operation and write the result in
standard form.
(Try p.131 #45-54)
6  7i 
1  2i 

6  7i  1  2i 


1  2i  1  2i 
6  14  5i
6  12i  7i  14i


2
2
1 4
1 2
2
20  5i

5
20 5i


5
5
 4 i
Principle Square Root of a Negative Number,
If a is a positive number, the principle square
root of the negative number –a is defined as
a 
ai  i a .
Use the Quadratic Formula to solve the quadratic
equation.
9x2 – 6x + 37 = 0
a = 9 , b = - 6 , c = 37
What is the discriminant?
( - 6 ) 2 – 4 ( 9 )( 37 )
36 – 1332
-1296
Therefore, the equation has no real solution.
9x2 – 6x + 37 = 0
a = 9 , b = - 6 , c = 37
x
  6 
 6  4937 
29
2
6  36  1332
x
18
6   1296
6
1296
x


i
18
18
18
1 36i
x 
3 18
1
  2i
3
2.4 Answers: (3-42 x 3)
2.5 Answers (3-39 x 3)
6. -10+50i
12. 28-45i
18. -1
 19 4
24. 29  29 i
30. 6+58i
36. x=10; y = 3
42.  4  i
6. 2
12. -27, 0
18. 1
24. 9
30. 4
36. ±1, ±2
2.5 Other Types of Equations
Radical Equations; Absolute Value
Equations;
Thus 85 is NOT a solution.
The solution set is {5}.
The solution set is {7,-6}.
A constant function is a function of the form
f(x)=b
y
b
x
Identity function is a function of a form:
f(x)=x
(1,1)
(0,0)
The square function
yx
2
Cube Function
(1,1)
(-1,-1)
Square Root Function
Rational Function
6. 2
2.5 Answers (3-39 x 3)
12. -27, 0
18. 1
24. 9
30. 4
36. ±1, ±2
2.6 Inequalities
Solving Absolute Value
Equations & Inequalities

<
>




>

>
<

<


There are two kinds of notation
for graphs
Remember---these
mean
theofsame
inequalities: open circle or filled in circle notation
thing---just
twobrackets.
different
notations.
and interval notation
You should
be
familiar with both.
x  1
[
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
circle filled in
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
squared end bracket
Both of these number lines show the inequality
above. They are just using two different notations.
Because the inequality is "greater than or equal to"
the solution can equal the endpoint. That is why the
circle is filled in. With interval notation brackets, a
square bracket means it can equal the endpoint.
Remember---these
Let's look at the two differentmean
notationsthe
with same
a
different inequality
sign.different notations.
thing---just
two
x  1
)
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
circle not filled in
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
rounded end bracket
Since this says "less than" we make the arrow go the
other way. Since it doesn't say "or equal to" the
solution cannot equal the endpoint. That is why the
circle is not filled in. With interval notation brackets, a
rounded bracket means it cannot equal the endpoint.
Compound Inequalities
Let's consider a "double inequality"
(having two inequality signs).
2 x3
(
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
]
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
I think of these as the "inbetweeners".
x is inbetween the two numbers. This is an "and"
inequality which means both parts must be true. It
says that x is greater than –2 and x is less than or
equal to 3.
Compound Inequalities
Now let's look at another form of a "double inequality"
(having two inequality signs).
x  2 or x  3
)
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
[
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Instead of "and", these are "or" problems. One part
or the other part must be true (but not necessarily
both). Either x is less than –2 or x is greater than or
equal to 3. In this case both parts cannot be true at
the same time since a number can't be less than –2
and also greater than 3.
Just like graphically there are two different
notations, when you write your answers you can
use inequality notation or interval notation.
Again you should be familiar with both.
[
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
x  1
Inequality notation
for graphs shown
above.
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
[1, )
Interval notation
for graphs shown
above.
Let's have a look at the interval notation.
[1, )
[
unbounded
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
For interval notation you list the smallest x can
be, a comma, and then the largest x can be so
solutions are anything that falls between the
smallest and largest.
The bracket before the –1 is square because this is greater than
"or equal to" (solution can equal the endpoint).
The bracket after the infinity sign is rounded because the
interval goes on forever (unbounded) and since infinity is not
a number, it doesn't equal the endpoint (there is no endpoint).
Let's try another one.
(2,4]
Rounded bracket means
cannot equal -2
Squared bracket means
can equal 4
The brackets used in the interval notation above
are the same ones used when you graph this.
(
]
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
This means everything between –2 and 4 but not including -2
Let's look at another one
(,4)
)
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Notice how the bracket notation for graphing
corresponds to the brackets in interval notation.
Remember that square is "or equal to" and round is up
to but not equal. By the infinity sign it will always be
round because it can't equal infinity (that is not a
number).
Now let's look at an "or" compound inequality
x  2 or x  3
)
[
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
(,2)
 [3, )
When the solution consists of more than one interval,
we join them with a union sign.
There are two intervals to list when you list in
interval notation.
Properties of Inequalities
Essentially, all of the properties that you learned to
solve linear equations apply to solving linear
inequalities with the exception that if you multiply or
divide by a negative you must reverse the
inequality sign.
So to solve an inequality just do the same steps as
with an equality to get the variable alone but if in the
process you multiply or divide by a negative let it ring
an alarm in your brain that says "Oh yeah, I have to
turn the sign the other way to keep it true".
Example:
2x  6  4x  8
- 4x
- 4x
 2x  6  8
+ 6 +6
 2 x  14
-2
We turned the sign!
-2
x  7
Ring the alarm!
We divided by a
negative!
2.6 Inequalities Involving Absolute Value
Absolute Value (of x)
•
•
•
•
Symbol |x|
The distance x is from 0 on the number line.
Always positive
Ex: |-3|=3
-4
-3
-2
-1
0
1
2
Ex: x = 5
• What are the possible values of x?
x=5
or
x = -5
To solve an absolute value equation:
ax+b = c, where c>0
To solve, set up 2 new equations, then solve
each equation.
ax+b = c or ax+b = -c
** make sure the absolute value is by itself
before you split to solve.
Ex: Solve 6x-3 = 15
6x-3 = 15 or 6x-3 = -15
6x = 18 or 6x = -12
x = 3 or x = -2
* Plug in answers to check your solutions!
Ex: Solve 2x + 7 -3 = 8
Get the abs. value part by itself first!
2x+7 = 11
Now split into 2 parts.
2x+7 = 11 or 2x+7 = -11
2x = 4 or 2x = -18
x = 2 or x = -9
Check the solutions.
Solving Absolute Value Inequalities
1. ax+b < c, where c>0
Becomes an “and” problem
Changes to: –c<ax+b<c
2. ax+b > c, where c>0
Becomes an “or” problem
Changes to: ax+b>c or ax+b<-c
Ex: Solve & graph.
4 x  9  21
• Becomes an “and” problem
 21  4x  9  21
12  4x  30
15
3 x 
2
-3
7
8
Solve & graph.
3x  2  3  11
• Get absolute value by itself first.
3x  2  8
• Becomes an “or” problem
3x  2  8 or 3x  2  8
3x  10 or
3x  6
10
x
or x  2
3
-2
3
4
Answers to 2.6
pp. 117-118 (3-33 x3, 41)
6. (-3, ∞)
12. (-5, -3]
18. x > -3
24.
30.
−8
( , ∞)
5
−1
(-2, ]
3
2.7 More on Inequalities
Inequalities with Quadratic
Functions
Quadratic inequalities
ax2+bx+c>0
…means “for what values of x is
this quadratic above the x axis”
e.g. x2+ x - 20 >0
ax2+bx+c<0 …means “for what values of x is
this quadratic below the x axis”
e.g. x2+ x - 20 < 0
Quadratic inequalities (2)
e.g. x2+ x - 20 >0
Factorises to (x-4)(x+5) >0
Numbers that multiply together to give more than 0?
A) both greater than zero
B) both less than zero
So, (x-4)>0
and (x+5)>0
x > 4 and x > -5
Only possible if x > 4
(then it must be >-5)
Either:
x>4
So, (x-4)<0
and (x+5)<0
x < 4 and x < -5
Only possible if x < -5
(then it must be <4)
or
x < -5
Quadratic inequalities (3)
e.g. x2 + x - 20 <0
Factorises to (x-4)(x+5) < 0
Numbers that multiply together to give less than 0?
A) first >0, second <0
B) first <0, second >0
So, (x-4)>0
and (x+5)<0
x > 4 and x < -5
IMPOSSIBLE
So, (x-4)<0
and (x+5)>0
x < 4 and x > -5
DO-ABLE
x < 4 and x > -5
-5 < x < 4
Quadratic inequalities (4)
- may be easier just looking at a graph
e.g. (x-4)(x+5) >0
…means “for what values of x is
this quadratic above the x axis”
Crosses at
x=4
and x=-5
When x>4
and x<-5
e.g. (x-4)(x+5) < 0
When x<4
and x>-5
…means “for what values of x is
this quadratic below the x axis”
Try this one
For what values of x is x2 - 3x - 18 > 0
[ Factorises to (x+3)(x-6) >0 ]
Numbers that multiply together to give more than 0
A) both greater than zero
B) both less than zero
So, (x+3)>0
and (x-6)>0
x > -3 and x > 6
Only possible if x > 6
Either:
x>6
So, (x+3)<0
and (x-6)<0
x < -3 and x < 6
Only possible if x < -3
or
x < -3
Quadratic with linear
Solve: x2 – 8x + 16 > 2x +7
y = 2x + 10
Estimate ?
x<1
x>9
y = x2 – 8x +16
Quadratic with linear (2)
Solve: x2 – 8x + 16 > 2x +7
Algebraically:
1. Rearrange first
2. Solve like the others
x2 – 8x + 16 > 2x +7 (-2x)
x2 – 10x + 16 > 7
(-7)
x2 – 10x + 9 > 0 Like the ones we did
(x-9)(x-1) > 0
x>9 or x<1
Try this one
Solve: x2 + x + 4 > 4x +14
First: try a sketch
Algebraically:
1. Rearrange first
2. Solve like the others
x2 + x + 4 > 4x +14
(-4x)
x2 – 3x + 4 > 14
(-14)
x2 – 3x - 10 > 0
(x+2)(x-5) > 0
x<-2 or x>5
Answers to 2.7
pp. 125-126 (3-30 x 3, 43, 44)
6. (-∞, −3] ∪ [−1, ∞)
12.(- ∞, -3]∪ [4, ∞)
18.(- ∞,
−3
3
)∪ ( ,
4
4
∞)
24. (-3, 3)∪ (3, ∞)
30. (-4, 0) ∪ (4, ∞)
44. 8 ≤ 𝑡 ≤ 12
Answers to Ch. 2 Review Packet
1. x = 9
17. (0, 14]
2. y = 2
18. ( , )
3. 9 mi./hr.
19. x < 2 or x > 5
4.
5.
6.
7.
8.
9.
10.
4
11
mi./hr.
h = 5.32 ft.
x = -14
width = 6 ft.
21i – 20
74
10+24𝑖
13
11. 65 + 142i
12.
2𝑖
−247
13. x = 2 ± 5i
14. x = ± 3, x = ± 3i
15. x = 6, x = -12
−11 −4
, ]
7
7
16. (
−8 6
7 7