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Unit 9 ~ Contents
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
9.15
9.16
Algebra Beauty and Awe ~ Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Dividing Larger Polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Polynomial Division With a Binomial Remainder . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Quadratic Equations:Taking the Square Root of Both Sides. . . . . . . . . . . . . . . . . . . . 5
Dividing Radicals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Adding and Subtracting Rationals With Unlike Denominators . . . . . . . . . . . . . . . . . 11
Quiz 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
How Attractive! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Permutations and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Rationalizing Denominators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Quadratic Equations: Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Dividing Polynomials With Missing Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Quiz 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
In Perfect Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Complex Rationals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
Quadratic Equations: Solving by Completing the Square . . . . . . . . . . . . . . . . . . . . . 34
The Domain of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Review for Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Under the Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
9 . 11
Complex Rationals
A complex rational is a fraction over another fraction. Below are several examples.
Figure 2
Figure 1
5
3
6 – 5
a
b3
2
3
5
6
Figure 4
Figure 3
a2
2b
2
1+ x
5
1+ y
2
3
Two methods can be used to simplify complex rationals. The best method to use depends on the
nature of the expression.
Method 1. This method works best with a single fraction over another single fraction, such as in
Figures 1 and 2 above. Because a fraction bar indicates division, rewrite the expression horizontally
using the division sign. Multiply by the reciprocal to complete the division.
Example 1.
Division changed to multiplication by the reciprocal.
2
3
5
6
2
3
=
÷ 65
=
2 6
•
3 5
=
12
15
4
5
=
Simplified answer.
Division rewritten horizontally.
Example 2.
a
b3
a
2b
2
30 ~ Algebra I Unit 9
=
a
b3
a2
÷ 2b
=
a 2b
•
b3 a2
=
2ab
a 2b 3
=
2
ab2
Example 3.
Division changed to multiplication by the reciprocal.
x2 – 9
x
x–3
x3
x2 – 9
x
=
÷
x–3
x3
=
x2 – 9
x3
•
x
x–3
=
2
x
(x + 3)(x – 3)
• x–3
x
3
Division rewritten horizontally.
=
x 2 (x + 3)
Simplified answer.
Method 2. This method works best when the numerator and/or denominator of a complex
rational contains more than one term, as is the case in Figures 3 and 4 on the previous page. To
simplify such expressions, find the LCD of all the terms in the numerator and denominator. Then
eliminate the fractions by multiplying the numerator and the denominator by the LCD.
Example 4.
5 3
–
6 5
2
3
=
Example 5.
2
1+ x
5
1+ y
=
( 56 – 35 )
( 23 )
( 11 + 2x )
( 11
5
+ y
)
•
•
(30)
(30)
(xy)
(xy)
=
25 – 18
20
=
7
20
30 is the LCD of the three terms 6, 5,
and 3. Multiplying the numerator and the
denominator by 30 eliminates the fractions.
=
xy + 2y
xy + 5x
=
y (x + 2)
x (y + 5)
xy is the LCD of the four terms 1, x, 1,
and y. Multiplying the numerator and the
denominator by xy eliminates the fractions.
9.11 Complex Rationals ~ 31
Example 6.
x2
6 –6
x
+3
2
=
Simplify. 9.11
3
1. 2
2
3
m 2n 2
4.
p
mn
p2
7.
x2 – x – 6
x2 – x – 2
x 2 + 2x – 15
x 2 – 4x + 4
Review
Divide. 9.9
10. (x 3 + 8) ÷ (x + 2)
( x6
2
–
6
1
)•
( 2x + 31 )
(6)
=
(6)
2.
x 2 – 36
3x + 18
32 ~ Algebra I Unit 9
(x + 6)(x – 6)
3(x + 6)
=
x–6
3
6 is the LCD of the four terms 6, 1, 2,
and 1. Multiplying the numerator and the
denominator by 6 eliminates the fractions.
14
27
3.
1
9
7x 3y
z2
21y 3
z
x2 – 1
5.
2
6.
1
2
+
3
8. 7
9. 1 +
x+1
8
2
Think of 2 as
2
.
1
2x + 1
x2
10x + 5
3x
1+
1
x
1
y
11. (6x 3 + 5x 2 – 4) ÷ (3x – 2)
Complete the square in the quadratic equations. 9.8
12. x 2 + 2x – 6 = 0
=
13. x 2 + 10x + 25 = 0
14. 4x 2 + 12x – 4 = 0
Factor polynomials completely. 8.2
15. 2k 3 – 12k 2 + 18k
16. 2x 4y – 162y
$17. 54xy 2 – 18xy + 9y 2 – 3y
Follow the directions. 9.6
$ 18.
$ 19.
The Family Friendly lunch buffet has 6 choices of meat, 8 choices of vegetables, 13 choices of
fruits/salads, and 7 choices of desserts. If a temperate customer took only one of each choice, how
many possible varieties of meal are there?
Nine women are in the Good Samaritan Sewing Circle. How many permutations exist for choosing a
President, Secretary, and Treasurer? What is the probability that Caleb’s mother will be Secretary,
Kezia’s mother will be President, and Guillermo’s mother will be Treasurer? Round the answer to the
nearest hundredth of a percent.
Factor the trinomials. If any is not a perfect square, write not a perfect square. 6.3
20. 16y 2 – 24y + 9
21. 9x 2 + 6x + 1
Divide the rational expressions. 7.7
23. (x 2 – 9) ÷
x–3
x+3
24.
x 2 + 4x + 4
x+2
÷
x 2 + 6x + 9
x+3
$ 22. 4x 2 – 2x + 0.25
25.
6x 2y + 15xy 6x 2y – 6xy
÷ 3xy – 12y
x–4
Solve the systems of equations, using the method of your choice. 6.4, 6.12, 7.2
26. 2x + 3y = 17
3x – 18y = 3
27. 8x – 2y = – 20
16x + 2y = – 28
Simplify. 9.11
29.
1
x– 3
1
3– x
30.
1
y– y
1
1+ y
28. 2x – 7y = – 64
2x – 3y = – 24
31.
x2
6 –6
x
2 +3
9.11 Complex Rationals ~ 33
9.12
Quadratic Equations: Solving by
Completing the Square
Lesson 9.8 stated that completing the square could be used to solve any quadratic equation.
After completing the square, three more steps are needed to solve quadratic equations. First, the
perfect square trinomial created by completing the square is factored into a binomial squared. Then
the square root of both sides of the equation is taken. Finally, the value(s) for x are found by solving
the remaining equation.
Example 1. Solve the quadratic equation x 2 + 14x + 13 = 0.
x 2 + 14x + 13 = 0
Original equation.
x 2 + 14x = –13
Constant moved to the right side.
(x + 7)2 = 36
Perfect square trinomial factored.
x 2 + 14x + 49 = –13 + 49
(x + 7) = ±6
x = –7 ±6
The solutions are: x = –1 and x = –13.
49 added to both sides to complete the square.
Square root taken of both sides.
7 subtracted from both sides.
Example 2. Solve the quadratic equation 3x 2 + 18x – 15 = 0.
3x 2 + 18x – 15 = 0
3x + 18x – 15 0
=
3
3
2
x 2 + 6x – 5 = 0
x 2 + 6x = 5
x 2 + 6x + 9 = 5 + 9
(x + 3)2 = 14
(x + 3) = ± √14
x = – 3 ± √14
Original equation.
Both sides divided by the coefficient of the x 2 term (3).
Simplified.
Constant term “moved” to the right side.
9 added to both sides to complete the square.
Perfect square trinomial factored.
Square root taken of both sides.
3 subtracted from both sides.
The solutions are: x = – 3 + √14 and x = –3 – √14 .
34 ~ Algebra I Unit 9
Example 3. Complete the square of the quadratic equation 2x 2 + 5x – 3 = 0.
2x 2 + 5x – 3 = 0
Original equation.
2x 2 + 5x – 3 0
=
2
2
x2 +
5
3
x– =0
2
2
x2 +
x2 +
Both sides divided by the coefficient of the x 2 term (2).
Simplified.
5
3
x=
2
2
Constant term “moved” to the right side.
5
25 3 25
x+
= +
2
16 2 16
25
added to both sides to complete the square.
16
(x +
Perfect square trinomial factored.
5 2 49
) =
4
16
x+
5
7
=±
4
4
x=–
The solutions are x =
Square root taken of both sides.
5 7
±
4 4
1
and x = – 3.
2
5
subtracted from both sides.
4
Complete the square and solve the quadratic equations. 9.12
1. x 2 – 4x – 21 = 0
2. x 2 + 16x + 55 = 0
3. x 2 + 5x + 4 = 0
7. x 2 + 4x – 12 = 0
8. 3x 2 – 6x – 18 = 0
9. x 2 – x – 2 = 0
4. x 2 – 8x + 9 = 0
Review
Simplify. 9.11
x
y
10.
y
x
5. x 2 + 22x + 21 = 0
11.
3r 2
s
6. 3x 2 + x – 1 = 0
12.
4s 3
r
1
2– x
1
x+ 2
Solve the quadratic equations for x by taking the square root of both sides. 9.2
13. (x + 2)2 = 16
14. x 2 – 5 = 0
15. 25x 2 = 5
16. x 2 – 20x + 100 = 100
9.12 Quadratic Equations: Solving by Completing the Square ~ 35
Simplify by rationalizing denominators. 9.7
1
17. √y
18.
19. √r
√s
√19
√xy
Simplify by dividing radicals. 9.3
20. 3 √50
√2
22. √14 – √28
3 √7
21. √98x
4 √2
Divide. 9.9
24. (x 3 + 1) ÷ (x + 1)
23. (9x 3 + 14x + 5) ÷ (3x + 1)
Solve by using direct variation. 7.14
25. If n varies directly as m, and n is 22 when m is 7, what is n when m is 3?
26. If y varies directly as the square of x, and y is 6.3 when x is 3, what is y when x is 5?
27. The distance away you are from a hunter is directly proportional to the time it takes before you hear
the sound of the rifle. If someone 510 meters away hears the shot in 1.5 seconds, how far away is
someone who hears it in only 1 second after the shot?
$ 28.
Think about the last problem. What is the constant of proportionality and what does it represent?
Solve the quadratic equations by factoring. 8.11
29. x 2 – 9 = 0
30. 16x 2 + 24x + 9 = 0
Graph these systems of inequalities. 8.7
32. 7y ≤ x – 7, and 8y ≥ 3x – 40
Simplify the expressions. 6.11
34.
x4 – 1
x+1
35.
33. y ≥ x, and y ≥ – x, and y ≤ 2
8x 2y – 6xy 2
4x 2y 2
Add or subtract. Leave your answers in factored form. 9.4
1
5
37. 3x 2y + 2x 3y 2
36 ~ Algebra I Unit 9
31. 6x 2 – 7x – 5 = 0
6e + 3
4
38. 2e + 1 – (2e + 1)2
36.
x2 – 9
x+3
2x
4
39. (2x – 3) – (x + 4)
Use a system of equations to solve the problems. 8.13
40. An isoceles triangle whose base is half the length of a side has a perimeter of 22.5 centimeters.
How long are the sides?
41. Adam drove 180 more miles than Jared on an 1,100 mile trip. How many miles did each drive?
Solve by using inverse variation. 8.3
42. If s varies inversely as the square of r, and s is 0.005 when r is 20, what is s when r is 10?
43. An empty space shuttle orbiter weighs 151,205 pounds at sea level. How much less does it weigh
at the edge of the stratosphere, 31 miles above the earth? (Remember that weight is inversely
proportional to the square of the distance from the center of the earth which is approximately 4,000
miles in radius.)
44. If n varies inversely as m, and n is 90 when m is 5, what is n when m is 3?
Solve for the requested information. 7.12
45. Gene began to divide the profits from his business into two savings accounts: 75% for his personal
needs, and the other 25% to an account to be donated to a mission hospital in 18 months when it is
scheduled to begin. Both accounts earn the same interest rate of 2%. At the end of 18 months the
total interest for the two accounts was $270. What were the principals in each account (these
would in reality be the “average” principal amounts over 18 months, but treat them as lump sum
investments)? (Hint: The principal in the personal account was 3 times the principal in the mission
account.)
Complete the square and solve the quadratic equations. 9.12
46. x 2 + 4x – 5 = 0
47. 4x 2 + 8x – 12 = 0
48. 5x 2 + 2x – 3 = 0
9.12 Quadratic Equations: Solving by Completing the Square ~ 37
9.13
The Domain of a Function
One way to think of functions is as machines which accept input value and produce output
values. The input value is called the independent variable (x) and the output value is called the
dependent variable (y). The function tells what will happen with the input to produce the output.
input (x )
function
x=4
output (y )
5x + 3
23
The set of possible input values (x) is the domain of the function. The domain can be unlimited
such as the set of all real numbers, as is the case for the function above: y = 5x + 3. Then any
positive or negative real number can be used for x. At other times the domain has limits such as the
function y = 1 in which x cannot be 0:
x
domain = {real numbers, x ≠ 0}.
The domain of a function is usually determined from a given list of ordered pairs, an equation,
or its graph. When a list of ordered pairs is given, the domain is the x-values of the ordered pairs. If
an equation or a graph is given, the domain is all the values of the x-axis that the graph covers.
Example 1. Give the domain for each of the following.
a. (6, 7) (10, 8) (5, 9) (4, 7) (–1, 5)
b. y = 6x – 7
1
c. y =
x–3
d. y = 2x 2 – 3x + 4
e. y = √2x
38 ~ Algebra I Unit 9
domain = {6, 10, 5, 4, –1}
domain = {all real numbers}
domain = {real numbers, x ≠ 3}
domain = {all real numbers}
domain = {real numbers, x ≥ 0}
Example 2. Give the domain for each graph.
a.
domain = {real numbers, x ≥ 0}
b.
domain = {real numbers, –1 < x ≤ 3}
y
y
x
c.
domain = {real numbers}
y
x
x
Specify the domain and tell whether or not it represents a function. 9.13
1. (2, 8) (3, 27) (4, 64) (5, 125)
3.
2. y = 4x + 7
4.
y
y
x
x
5.
6.
y
x
y
x
9.13 The Domain of a Function ~ 39
7.
8.
y
y
x
x
Review
Complete the square and solve the quadratic equations. 9.12
9. x 2 – 6x + 7 = 0
10. x 2 – 2x – 15 = 0
11. x 2 + 7x + 10 = 0
Combine like radicals. If they cannot be combined, write cannot be combined. 7.9
12. x √3 + x √3
3
3
13. √16 – √27
Divide. 9.1
14. 3 √x – x √x + √x
$ 16. (3x 3 – 18x 2 + x – 6) ÷ (x – 6)
15. (8h3 + 8h2 + 12h – 6) ÷ (4h2 + 6h + 8)
Add or subtract as indicated. 8.4
17.
5x√2
√8
+
3x√2
√8
Simplify. 9.11
x+2
x–1
20.
x–2
x–1
Factor by grouping. 7.3
23. 12x + 3xy – 4y – y 2
40 ~ Algebra I Unit 9
3x 2 – x
–12 – 2x 2
18. x 2 – x – 12 + x 2 – x – 12
21.
1
1
+
x
y
y
2
24. 4r 2 + 4rs – 7r – 7s
19.
22.
3j 2 + j + 3
2j 2 + j + 4
–
j–1
j–1
4xy – 8x
y2
3x 2y – 6x 2
y
25. 10x 2 + 4x – 5xy – 2y
Specify the domain and tell whether or not it represents a function. 9.13
4
26. y = x
27. y = √ x
28. y = x 2 – 5x + 3
29.
30.
y
x
y
x
9.13 The Domain of a Function ~ 41