Homework Log
Thurs & Learning Objective:
Fri
To remember everything in Ch 2
10/22
Lesson
Rev
Hw: #216 Pg. 155 #1 – 85 odd
Homework Log
Mon
Learning Objective:
To remember everything in Ch 2
10/24
Hw: Extra Credit Test Review
Lesson
Rev
10/22/15 Chapter 2 Review
Advanced Math/Trig
Learning Objective

To remember everything in
Chapter 2!
Solve a Rational Equation
𝑥(1−𝑥) 1
1−𝑥
1
1.
+
3 𝑥(1−𝑥)
𝑥
1
=
−1 𝑥(1−𝑥)
𝑥
1
LCD:
x(1 – x)
x + 3(1 – x) = –1(1 – x)
x + 3 – 3x = –1 + x
3 – 2x = –1 + x
–3x = –4
x=
4
3
4
3
Absolute Value Equations
2.
−4 2𝑥 − 3 + 7 = −9
–7 –7
−4 2𝑥 − 3 = −16
–4
–4
2𝑥 − 3 = 4
2x – 3 = 4
or
2x – 3 = – 4
2x = 7
2x = –1
7
−1
−1 7
x=
or x =
,
2
2
2
2
Warm–up #2 Solutions
3. Solve 𝑥 + 5 = 3𝑥 − 2
x + 5 = 3x – 2
or x + 5 = –(3x – 2)
–2x = –7
or x + 5 = –3x + 2
4x = –3
x=
7
2
or
x=
3
−
4
Check for extraneous solutions!
Nope!
Solving for a Variable
(S + F) P =
4. Solve for S:
PS + PF = S
– PS
– PS
PF = S – PS
PF = S(1 – P)
𝑃𝐹
=𝑆
1−𝑃
𝑆
(S + F)
𝑆+𝐹
Simple Interest
5. Part of $20,000 is to be invested at 15% and the
remainder at 9%. How much should be invested at
each rate to yield an annual interest income of $2520.
Principal
Inv 1
Inv 2
Total
x
20000 – x
20000
•
rate
.15
.09
•
time
=
1
1
.15x + .09(20000 – x) = 2520
Interest
.15x
.09(20000 – x)
2520
equation!
Simple Interest #5 cont’d
.15x + .09(20000 – x) = 2520
.15x + 1800 – .09x = 2520
.06x = 720
x = 12000
20000 – 12000 = 8000
$12,000 at 15%
$8,000 at 9%
Investment
6. If $9000 is invested at 7% per year, how
much additional money needs to be invested at
14% per year so that the total annual interest
income from the investments is $1330?
Principal
Inv 1
Inv 2
Total
9000
x
9000 + x
•
rate
.07
.14
630 + .14x = 1330
•
time
1
1
=
Interest
9000(.07) = 630
.14x
1330
equation!
Investment Cont’d
630 + .14x = 1330
.14x = 700
x = 5000
$5,000 at 14%
Mixture
7. I want to dilute 40 L of a solution that is
80% acid to one that is 50% acid. How much
water should be added to the acid solution?
Amount
Solution 1
Solution 2
Mix
40
x
40 + x
•
%
80
0
50
3200 = 50(40 + x)
=
Total
40(80) = 3200
0
50(40 + x)
24 L
Distance Problem
8. Laura & Luke left school at the same time and
went in opposite directions. Laura was driving 40
mph faster than Luke. After 3 hours, they were 330
miles apart. How fast was Laura driving?
Rate
Laura
Luke
•
Time
x + 40
x
Total
3(x + 40) + 3x = 330
3x + 120 + 3x = 330
x = 35
3
3
= Distance
3(x + 40)
3x
330
75 mph
Drain/Work Problem
9. An Olympic sized pool can be filled by pipe A in
12 hours and by pipe B in 10 hours. There is also a
drain pipe that drains the entire pool in 6 hours. If
the valves of pipe A, pipe B and the drain pipe are
open, how long will it take to fill the pool?
Alone
Rate
Pipe A
Pipe B
Drain Pipe
𝟏
𝟏𝟐
𝟏
𝟏𝟎
𝟏
−
𝟔
•
Time
Together
x
x
x
=
Part of Job
Completed
𝒙
𝟏𝟐
𝒙
𝟏𝟎
𝒙
−
𝟔
Drain/Work Problem #9
Cont’d
Alone
Rate
Pipe A
Pipe B
Drain Pipe
𝟏
𝟏𝟐
𝟏
𝟏𝟎
𝟏
−
𝟔
•
Time
Together
x
x
x
=
Part of Job
Completed
𝒙
𝟏𝟐
𝒙
𝟏𝟎
𝒙
−
𝟔
Pipe A’s part + Pipe B’s part + Drain’s part= 1 Whole Job
𝑥 (60) 𝑥
Completed
(60) 𝑥 (60)
+
− = 1(60)
12
10
6
5x + 6x – 10x = 60
x = 60 hours
Work Problem
10. Working together, Scott and Jenna can sweep
a porch in 10 minutes. If Jenna worked alone, it
would have taken her 15 minutes. How long does
it take Scott to sweep the porch alone?
Alone
Rate
Scott
Jenna
𝟏
𝒙
𝟏
𝟏𝟓
•
Time
Together
10
10
=
Part of Job
Completed
𝟏𝟎
𝒙
𝟏𝟎
𝟏𝟓
Scott’s part + Jenna’s part = 1 Whole Job Completed
Work Problem #10 Cont’d
Alone
Rate
Scott
Jenna
𝟏
𝒙
𝟏
𝟏𝟓
•
Time
Together
10
10
=
Part of Job
Completed
𝟏𝟎
𝒙
𝟏𝟎
𝟏𝟓
Scott’s part + Jenna’s part = 1 Whole Job Completed
(15𝑥)
(15𝑥)10
10
+
= 1 (15𝑥)
𝑥
15
150 + 10x = 15x
150 = 5x
30 minutes
Solve Absolute Value
Inequalities
11.
2𝑥 − 5 < 3
Less ThAND
2x – 5 < 3
and
2x – 5 > – 3
+5 +5
+5 +5
2x < 8
2x > 2
2
2
2
2
x < 4 and x > 1
1<x<4
(1, 4)
Solve Absolute Value
Inequalities
12.
5𝑥 + 3 > 7
GreatOR
5x + 3 > 7
or
–3 –3
5x > 4
5 5
x > 4/5 or x < – 2
5x + 3 < – 7
–3 –3
5x < – 10
5
5
4
(–∞, –2) ∪ ( , ∞)
5
Solve Absolute Value
Inequalities
13.
−2𝑥 + 3 > – 5
All Real Numbers
(–∞, ∞)
Absolute Value is always positive & will
ALWAYS be greater than a negative
number!!
Solve Absolute Value
Inequalities
14.
−3𝑥 + 4 < −20
No Solution
∅
Absolute Value is always positive & will
NEVER be less than a negative number!!
Solve by Factoring
15. 2𝑥 3 + 5𝑥 2 = 3𝑥
2𝑥 3 + 5𝑥 2 − 3𝑥 = 0
–6
𝑥 2𝑥 2 + 5𝑥 − 3 = 0
2
𝑥 2𝑥 + 6𝑥 − 1𝑥 − 3 = 0
x(2x – 1)(x + 3) = 0
x=0
{–3, 0,
x+3=0
1
}
2
2 #s that mult to
6
–1
5
& add to
2x – 1 = 0
Solve by Completing the
Square
2
16. 𝑥 − 8𝑥 + 20 = 0
– 20
– 20
𝑥 2 − 8𝑥 + 16 = – 20 + 16
−8 2
2
= (−4)2 = 16
(𝑥 − 4)2 = −4
x – 4 = ±2𝑖
4 ± 2𝑖
Solve by Completing the
Square
17. 4𝑥 2 − 10𝑥 = −3
4
𝑥2 −
5
−
2
4
25
16
5
𝑥+
2
÷2
2
=
5 2
25
= −
=
4
16
5 2 13
(𝑥 − ) =
4
16
4
3
=− +
4
5 1 2
− ∙
2 2
25
16
5
4
=±
±
13
4
𝑥−
5
4
Or
5± 13
4
13
4
Solve by Factoring
2
18. (𝑥 + 2) −5 𝑥 + 2 = 14 Let 𝑢 = 𝑥 + 2
(𝑥 + 2)2 −5 𝑥 + 2 − 14 = 0
𝑢2 − 5𝑢 − 14 = 0
2
𝑢 − 7𝑢 + 2𝑢 − 14 = 0
u(u – 7) + 2(u – 7) = 0
(u – 7)(u + 2) = 0
–14
–7 2
2 #s that mult to
–5
& add to
Solve by Factoring
18. (u – 7)(u + 2) = 0
((x +2) – 7)((x +2) + 2) = 0
Now replace u
with 𝑥 + 2
(x – 5)(x + 4) = 0
x–5=0
{– 4, 5}
x+4=0
2 Answers!!
Highest Power is 2!!
Sum – Product Rule
19. Find a monic quadratic eq’n whose
roots are 5 & –3
x2 – (sum)x + product = 0
sum:
5 + –3 = 2
product: (5)(–3) = –15
x2 – (2)(x) + (–15) = 0
x2 – 2x – 15 = 0
Quadratic Formula
Given: ax 2  bx  c  0
b  b  4ac
x
2a
2
Xavier is a negative boy who couldn’t decide (yes or no)
whether to go to a radical party.
It turns out that this boy is a total square because he
missed out on 4 awesome chicks.
And the party was all over at 2 AM.
Discriminant
Discriminant – tells the nature of the roots
𝑏 2 − 4𝑎𝑐
(Part under the radical)
Discriminant
Roots
Zero
1 real double root
Positive
2 real roots
Negative
2 imaginary roots