Download Intermediate Algebra Notes Contents Joseph Lee Fall 2016

Intermediate Algebra Notes
Joseph Lee
Fall 2016
Contents
Introduction
2
Unit 1
Slope and y-intercept . . . . . . . . . .
Linear Equations . . . . . . . . . . . .
Functions . . . . . . . . . . . . . . . .
Two Variable Systems of Equations . .
Three Variable Systems of Equations .
Applications of Systems of Equations .
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Unit 2
Factoring the Greatest Common Factor
Factoring Trinomials . . . . . . . . . . .
Factoring Using Special Formulas . . . .
Solving Equations by Factoring . . . . .
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22
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33
Unit 3
Simplifying Rational Expressions . . . . . . .
Adding and Subtracting Rational Expressions
Complex Rational Expressions . . . . . . . .
Solving Equations with Rational Expressions
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Unit 4
Radicals . . . . . . . . . . . . . . . . . . . . . .
Rational Exponents . . . . . . . . . . . . . . .
Simplifying Radicals . . . . . . . . . . . . . . .
Adding, Subtracting, and Multiplying Radicals
Dividing Radicals . . . . . . . . . . . . . . . . .
Solving Equations with Radicals . . . . . . . .
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50
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55
62
69
74
78
Unit 5
Complex Numbers . . . . . . .
Completing the Square . . . . .
The Quadratic Formula . . . .
Solving Equations . . . . . . .
Graphing Quadratic Functions
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1
Joseph Lee
Introduction
The question I receive most often, regardless of the course, is, “When am I ever going to use this?” I think
the question misses the point entirely. While I do not determine which classes students need to get their
degree, I do think it is a good policy that students are required to take my course – for more reasons than
just my continued employment, which I support as well.
If a student asked an English instructor why he or she had to read Willa Cather’s My Ántonia, the
instructor would not argue that understanding nineteenth century prairie life was essential to becoming a
competent tax specialist or licensed nurse. The instructor would not argue that reading My Ántonia would
benefit the student directly through a future application. Instead, the benefit of reading this beautiful piece
of American literature is entirely intrinsic. The mere enjoyment and appreciation is enough to justify its
place in a post-secondary education. Moreover, the result arrived to at the end of my course is as beautiful as any prose or poetry a student will encounter in his or her studies here at Metro or any other college.
2
Tuesday, September 6, 2016
Joseph Lee
Unit 1
Slope and y-intercept
Definition of Slope
The slope, m, of a line is the ratio of the vertical change to the horizontal change.
m=
vertical change
horizontal change
Example 1. Determine the slope of the line passing through points (1, 1) and (4, 3), as shown below.
y
(4, 3)
(1, 1)
x
Slope Formula
The slope of any line passing through the points (x1 , y1 ) and (x2 , y2 ) is given by the following formula:
m=
y2 − y1
.
x2 − x1
Example 2. Find the slope of the line passing through (1, 6) and (4, 2).
Example 3. Find the slope of the line passing through (−2, −4) and (6, 12).
3
Tuesday, September 6, 2016
Joseph Lee
Example 4. Find the slope of the line passing through (−3, 2) and (4, 2).
Example 5. Find the slope of the line passing through (3, −2) and (3, −8).
Definition of y-intercept
The y-intercept of a line is the point (0, b) where the line intersects the y-axis.
Slope-Intercept Form
The equation of a line is in slope-intercept form if it is written as
y = mx + b,
where m is the slope of the line and (0, b) is the y-intercept.
Example 6. Determine the slope and y-intercept for the following equation.
y = 2x + 3
Example 7. Determine the slope and y-intercept for the following equation.
1
y = x−2
4
4
Tuesday, September 6, 2016
Joseph Lee
Example 8. Determine the slope and y-intercept for the following equation.
4x − 3y = 9
5
Tuesday, September 6, 2016
Joseph Lee
Linear Equations
Example 1. Write the equation of the line that has a slope of − 23 and a y-intercept of (0, 4).
Example 2. Write the equation of the following line.
y
(2, 1)
x
(0, −2)
Point-Slope Form
An equation of the line with slope m passing through the point (x1 , y1 ) is given by
y − y1 = m(x − x1 ).
Example 3. Write the equation of the line that passes through the point (2, 1) with a slope of 32 .
Example 4. Write the equation of the line that passes through the point (−4, 3) with a slope of − 12 .
6
Tuesday, September 6, 2016
Joseph Lee
Standard Form
An equation in standard form is written as
Ax + By = C,
where A, B, and C are integers and A > 0.
Example 5. Write the equation of the line that passes through the point (3, −8) with a slope of −2.
Slope-Intercept Form:
Standard Form:
Example 6. Write the equation of the line that passes through the points (2, −3) and (−2, 1).
Slope-Intercept Form:
Standard Form:
7
Tuesday, September 6, 2016
Joseph Lee
Example 7. Write the equation of the line that passes through the points (5, −2) and (−1, 1).
Slope-Intercept Form:
Standard Form:
Parallel and Perpendicular Lines
Two lines are parallel if they have the same slope.
opposite reciprocals1 . Additionally:
Two lines are perpendicular if the slopes are
• Any two horizontal lines are parallel.
• Any two vertical lines are parallel.
• A horizontal line and a vertical line are perpendicular.
Example 8. Determine whether the following lines are parallel, perpendicular, or neither.
2x − 6y = 5
3x + y = 5
Example 9. Determine whether the following lines are parallel, perpendicular, or neither.
4x − 2y = 8
−6x + 3y = 4
1
Two numbers are opposite reciprocals if their product is −1.
8
Tuesday, September 6, 2016
Joseph Lee
Example 10. Write the equation of the line passes though the point (3, 1) and is parallel to the line
2x + y = 6.
Slope-Intercept Form:
Standard Form:
Example 11. Write the equation of the line passes though the point (0, 3) and is perpendicular to
the line 2x + 3y = 5.
Slope-Intercept Form:
Standard Form:
9
Thursday, September 8, 2016
Joseph Lee
Functions
Definition: Relation
A relation is a correspondence between two sets.
Elements of the second set are called the range.
Elements of the first set are called the domain.
Definition: Function
A function is a specific type of a relation where each element in the the domain corresponds to
exactly one element in the range.
Example 1. Determine the domain and range of the following relation2 . Does the relation define a
function?
{(Joseph, turkey), (Joseph, roast beef), (Michael, ham)}
Domain:
Range:
Function?
Example 2. Determine the domain and range of the following relation. Does the relation define a
function?
{(1, 3), (2, 4), (−1, 1)}
Domain:
Range:
Function?
Example 3. Determine the domain and range of the following relation. Does the relation define a
function?
{(3, 5), (4, 5), (5, 5)}
Domain:
Range:
Function?
2
This relation relates math instructors and the sandwiches they enjoy.
10
Thursday, September 8, 2016
Joseph Lee
Example 4. Determine the domain and range of the following relation. Does the relation define a function?
y
(2, 3)
(2, 1)
x
(−2, −1)
(0, −2)
Domain:
Range:
Function?
Example 5. Determine the domain and range of the following relation. Does the relation define a
function?
y
(2, 0)
(−1, −1)
(−2, −2)
Domain:
Range:
Function?
11
(1, −1)
x
Thursday, September 8, 2016
Joseph Lee
Example 6. Determine the domain and range of the following relation. Does the relation define a function?
y
(−2, 4)
(2, 4)
(−1, 1)
(0, 0)
(1, 1)
x
Domain:
Range:
Function?
2
Example 7. Evaluate the function f (x) = − x + 2 for the following values.
3
f (−3)
f (0)
f (1)
f (a)
12
Thursday, September 8, 2016
Joseph Lee
Example 8. Evaluate the function below.
f (x)
(2, 0)
(−1, −1)
x
(1, −1)
(−2, −2)
f (−2)
f (−1)
f (0)
f (1)
f (2)
Example 9. Evaluate the function below.
y
(−2, 4)
(2, 4)
(−1, 1)
(0, 0)
f (−2)
f (1)
f (−1)
f (2)
f (0)
13
(1, 1)
x
Thursday, September 8, 2016
Joseph Lee
Two Variable Systems of Equations
Example 1. Solve the following system of equations by graphing.
3x + y = 1
−x + y = −3
y
x
Solution:
Example 2. Solve the following system of equations by graphing.
2x + 3y = 12
y = − 23 x + 4
y
x
Solution:
14
Thursday, September 8, 2016
Joseph Lee
Example 3. Solve the following system of equations by graphing.
y = −3x + 4
3x + y = 1
y
x
Solution:
Dependent, Independent, Inconsistent Systems
A system of equations is called inconsistent if it has no solutions.
A system of equations is called dependent if it has infinitely many solutions.
A system of equations is called independent if it has a single solution.
Example 4. Solve the following system of equations by substitution.
x − 2y = 8
4x + y = 5
Solution:
15
Thursday, September 8, 2016
Joseph Lee
Example 5. Solve the following system of equations by substitution.
5x + 3y = 6
4x + 2y = 2
Solution:
Example 6. Solve the following system of equations by elimination.
x − 2y = 8
4x + y = 5
Solution:
Example 7. Solve the following system of equations by elimination.
5x + 3y = 6
4x + 2y = 2
Solution:
16
Thursday, September 8, 2016
Joseph Lee
Example 8. Solve the following system of equations by elimination.
4x + 6y = 24
y = − 23 x + 4
Solution:
Example 9. Solve the following system of equations by elimination.
y = −3x + 4
6x + 2y = −2
Solution:
17
Tuesday, September 13, 2016
Joseph Lee
Three Variable Systems of Equations
Example 1. Solve the following system of equations.
x+y+z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution:
Example 2. Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution:
18
Tuesday, September 13, 2016
Joseph Lee
Example 3. Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution:
Example 4. Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution:
19
Tuesday, September 13, 2016
Joseph Lee
Example 5. Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution:
20
Tuesday, September 13, 2016
Joseph Lee
Applications of Systems of Equations
Example 1. Joseph has a collection of quarters and dimes. A friend counts his change and determines he
has $3.70. Joseph knows he has 25 coins. How many quarters and dimes does Joseph have?
Quarters:
Dimes:
Example 2. Ten liters of a 12% HCl solution is mixed with a 20% HCl solution to make a mixture
that is 15% HCl. How many liters of the 20% HCl solution were used to make the mixture?
Liters of the 20% HCl solution:
Example 3. Against the wind, a plane can fly 2880 miles in 5 hours. Flying with the same wind at
its tail, it only takes 4.5 hours. Determine the speed of the wind and the speed of the plane in still air.
Speed of the wind:
Speed of the plane in still air:
21
Thursday, September 22, 2016
Joseph Lee
Unit 2
Factoring the Greatest Common Factor
Example 1. Factor.
15x3 − 25x2 y
Example 2. Factor.
14x2 + 35x − 21
Example 3. Factor.
12p2 q 4 r2 − 6p3 q 2 r2 + 20pq 4 r3
Example 4. Factor.
4x(3x − 1) − y(3x − 1)
Example 5. Factor.
4x2 (x2 + 2) − 3x(x2 + 2)
Example 6. Factor.
(3x − 2)(x + 3) + y(x + 3)
22
Thursday, September 22, 2016
Joseph Lee
Example 7. Factor.
12x3 − 9x2 + 8x − 6
Example 8. Factor.
xy + 3y − 5x − 15
Example 9. Factor.
7x2 y − 14xy + 28xy − 56y
Example 10. Factor.
8mn − 12m − 12n + 18
23
Thursday, September 22, 2016
Joseph Lee
Factoring Trinomials
Example 1. Multiply.
(3x − 7)(2x + 5)
Example 2. Factor.
6x2 + x − 35
Example 3. Factor.
2x2 + 3x − 5
Example 4. Factor.
3x2 + 11x − 20
Thursday, September 22, 2016
Joseph Lee
Example 5. Factor.
5x2 − 13x + 6
Example 6. Factor.
4x2 + 12x − 7
Example 7. Factor.
6x2 − 27x + 12
Example 8. Factor.
9m4 n − 15m3 n2 − 6m2 n3
Thursday, September 22, 2016
Joseph Lee
Example 9. Factor.
(x + 7)2 − 6(x + 7) − 16
Substition:
Example 10. Factor.
4x6 − 4x3 − 15
Substition:
Example 11. Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Substition:
Tuesday, September 27, 2016
Joseph Lee
Factoring Using Special Formulas
Example 1. Multiply.
(a + b)2
Example 2. Multiply.
(a − b)2
Example 3. Multiply.
(a + b)(a − b)
Factoring Squares and the Difference of Squares
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
a2 − b2 = (a + b)(a − b)
27
Tuesday, September 27, 2016
Joseph Lee
Example 4. Factor.
4x2 + 28x + 49
Example 5. Factor.
9x2 − 30x + 25
Example 6. Factor.
100x4 y 2 + 60x2 yz 3 + 9z 6
Example 7. Factor.
32x2 − 80xy + 50y 2
Example 8. Factor.
x2 − 16
28
Tuesday, September 27, 2016
Joseph Lee
Example 9. Factor.
4m8 − 9n2
Example 10. Factor.
12x3 − 3x
Example 11. Factor.
x4 − 16
29
Tuesday, September 27, 2016
Joseph Lee
Example 12. Factor.
(x + y)2 + 2(x + y) + 1
Substitution:
Example 13. Factor.
25 − (3z + 4)2
Substitution:
Example 14. Multiply.
(a + b)(a2 − ab + b2 )
30
Tuesday, September 27, 2016
Joseph Lee
Factoring the Sum of Cubes and the Difference of Cubes
a3 + b3 = (a + b)(a2 − ab + b2 )
a3 − b3 = (a − b)(a2 + ab + b2 )
Example 15. Factor.
8x3 + 27
Example 16. Factor.
1 − 64x3
Example 17. Factor.
125x6 − 27y 3 z 9
31
Tuesday, September 27, 2016
Joseph Lee
Example 18. Factor.
216 − (a + b)3
Subsitution:
Example 19. Factor.
(3x − 2y)3 + (x + 1)3
Subsitution:
32
Tuesday, September 27, 2016
Joseph Lee
Solving Equations by Factoring
Example 1. Solve.
x2 − 5x + 6 = 0
Example 2. Solve.
3x2 − 10x − 8 = 0
Example 3. Solve.
3x3 − 3x2 − 6x = 0
33
Tuesday, September 27, 2016
Joseph Lee
Example 4. Solve.
2x2 = 5x + 3
Example 5. Solve.
(x − 3)(x − 5) = −1
Example 6. Solve.
3x(x − 2) = 4(x + 1) + 4
34
Tuesday, September 27, 2016
Joseph Lee
Pythagorean Theorem
If a right triangle has legs of lengths a and b and hypotenuse of length c, then
a2 + b2 = c2 .
c
b
a
Example 7. Solve for x.
x+2
x
x+1
Example 8. Solve for x.
x
x−8
x−1
35
Tuesday, October 4, 2016
Joseph Lee
Unit 3
Simplifying Rational Expressions
36
Example 1. Simplify.
10
25
Example 2. Simplify.
x2 − 10x + 25
x2 − 25
Example 3. Simplify.
3x2 y 2
6xy 3
Example 4. Simplify.
x3 − y 3
x−y
Tuesday, October 4, 2016
Joseph Lee
Example 5. Multiply.
x2 − 2x − 8 x + 1
·
x2 + 2x + 1 x2 − 4x
Example 6. Multiply.
x2 − 2x − 8 x + 1
·
x2 + 2x + 1 x2 − 4x
Example 7. Multiply.
2x
3x2 − 5x − 2
·
3x + 1
4x2 + 8x
Tuesday, October 4, 2016
Joseph Lee
Example 8. Multiply.
x2 + x − 6 x2 + 4x + 4
· 2
4 − x2
x + 4x + 3
Example 9. Divide.
2x2 − 9x − 5
4x2 − 1
÷ 2
2
2x − 13x + 15 4x − 8x + 3
Example 10. State the domain of each rational expression.
x2
x−4
+ 5x + 6
Tuesday, October 4, 2016
Joseph Lee
Adding and Subtracting Rational Expressions
Example 1. Add.
3x − 3
x2 − 5x
+ 2
2
x − 7x + 12 x − 7x + 12
Example 2. Subtract.
7x − 3 4x − 9
−
4x + 8 4x + 8
Example 3. Add.
7
13
+
48 60
39
Tuesday, October 4, 2016
Joseph Lee
Example 4. Rewrite the following fractions with the least common denominator.
3
2
,
2
4x y 3xy 3
Example 5. Rewrite the following fractions with the least common denominator.
x2
x−1
3x
,
− 5x + 6 4x − 12
Example 6. Add.
2
5
+ 2 4
3
2
3x y
4x y
40
Tuesday, October 4, 2016
Example 7. Add.
Example 8. Subtract.
Joseph Lee
x−1
x
+
4x + 8 6x + 12
x − 26
4
− 2
x + 2 x − 3x − 10
41
Tuesday, October 4, 2016
Example 9. Add.
Joseph Lee
3x − 2
x−5
+ 2
− 5x − 3 2x + 5x + 2
2x2
Example 10. Subtract.
x−1
3x
−
x−4 4−x
42
Thursday, October 6, 2016
Joseph Lee
Complex Rational Expressions
Example 1. Simplify the complex rational expression.
x2
y
x3
y4
Example 2. Simplify the complex rational expression.
3
4
5
6
43
Thursday, October 6, 2016
Joseph Lee
Example 3. Simplify the complex rational expression.
x−2
x
x+1
x−4
Example 4. Simplify the complex rational expression.
2
x
5
1+
x
3−
44
Thursday, October 6, 2016
Joseph Lee
Example 5. Simplify the complex rational expression.
4
x2
2
1+
x
1−
Example 6. Simplify the complex rational expression.
7 10
+
x x2
2 15
1− − 2
x x
1−
45
Thursday, October 6, 2016
Joseph Lee
Example 7. Simplify the complex rational expression.
2
x+3
x
3+
x+3
x−
Example 8. Simplify the complex rational expression.
3
4
−
x+2 x+1
5
1
+
x+1 x+2
46
Tuesday, October 11, 2016
Joseph Lee
Solving Equations with Rational Expressions
Example 1. Solve.
2 1
7
+ =
x 2
6
Example 2. Solve.
4
3x
=
−5
x+4
x+4
Example 3. Solve.
3
3x
+7=
x−1
x−1
47
Tuesday, October 11, 2016
Joseph Lee
Example 4. Solve.
x−
8
2x
=−
x−4
x−4
Example 5. Solve.
1+
10
25
=
2
x
x
48
Tuesday, October 11, 2016
Joseph Lee
Example 6. Solve.
18
x
=
−3
−9
x+3
x2
Example 7. Solve.
4x
3
3x
−
= 2
x − 2 2x − 1
2x − 5x + 2
49
Tuesday, October 18, 2016
Joseph Lee
Unit 4
Radicals
Definition: Principal nth Root
The principal nth root of a is denoted
If
If
√
n
√
√
n
a. The principal square root of a is denoted by
a = b, then bn = a.
a = b, then b is positive and b2 = a.
Example 1. Evaluate.
Example 2. Evaluate.
√
√
81
169
Example 3. Evaluate.
r
Example 4. Evaluate.
Example 5. Evaluate.
1
25
√
−36
√
3
64
50
√
a.
Tuesday, October 18, 2016
Example 6. Evaluate.
Example 7. Evaluate.
Example 8. Evaluate.
Example 9. Evaluate.
Example 10. Evaluate.
Example 11. Simplify.
Example 12. Simplify.
Joseph Lee
√
3
−64
√
4
√
4
16
−16
√
15
√
3
√
√
31
x2
x10
51
Tuesday, October 18, 2016
Example 13. Simplify.
Example 14. Simplify.
Example 15. Simplify.
Example 16. Simplify.
Example 17. Simplify.
Joseph Lee
√
16x6
√
x4
√
3
√
4
√
4
x3
x4
81x8
Example 18. Simplify. Assume all variables represent nonnegative values.
√
x2
52
Tuesday, October 18, 2016
Joseph Lee
Example 19. Simplify. Assume all variables represent nonnegative values.
√
9x6
Example 20. Simplify. Assume all variables represent nonnegative values.
p
25x6 y 2
Example 21. Simplify. Assume all variables represent nonnegative values.
s
4x2
y4
Example 22. Simplify. Assume all variables represent nonnegative values.
√
3
8x6
Example 23. Simplify. Assume all variables represent nonnegative values.
p
4
625x8 y 12
Example 24. Simplify. Assume all variables represent nonnegative values.
p
(x + 1)2
53
Tuesday, October 18, 2016
Example 25. Let f (x) =
Example 26. Let f (x) =
Example 27. Let f (x) =
Example 28. Let f (x) =
Joseph Lee
√
√
√
√
x + 7. Find f (2).
x + 7. Find f (18).
x + 7. Find f (−7).
x + 7. Find f (−11).
Example 29. State the domain of f (x) =
Example 30. State the domain of g(x) =
√
x + 7.
√
7x − 42.
54
Tuesday, October 18, 2016
Joseph Lee
Rational Exponents
First Definition of a Rational Exponent
x1/n =
√
n
x
Example 1. Simplify.
91/2
Example 2. Simplify.
641/3
Example 3. Simplify.
−251/2
Example 4. Simplify.
(−25)1/2
Example 5. Simplify. Assume all variables represent positive values.
(81x2 )1/2
55
Tuesday, October 18, 2016
Joseph Lee
Example 6. Simplify. Assume all variables represent positive values.
4x1/2
Example 7. Simplify. Assume all variables represent positive values.
(4x)1/2
Example 8. Simplify. Assume all variables represent positive values.
1/2
144x8
169y 6
Second Definition of a Rational Exponent
xm/n =
m
√
n
x
Example 9. Simplify. Assume all variables represent nonnegative values.
813/4
Example 10. Simplify. Assume all variables represent nonnegative values.
642/3
56
Tuesday, October 18, 2016
Joseph Lee
Example 11. Simplify. Assume all variables represent nonnegative values.
323/5
Example 12. Simplify. Assume all variables represent nonnegative values.
(8x3 )2/3
Third Definition of a Rational Exponent
xm/n =
√
n
xm
Example 13. Simplify. Assume all variables represent nonnegative values.
x2/3
Example 14. Simplify. Assume all variables represent nonnegative values.
2
(x2 y 3 ) 11
Example 15. Simplify. Assume all variables represent nonnegative values.
(x + 4)3/5
57
Tuesday, October 18, 2016
Joseph Lee
Negative Exponent Property
x−m =
1
xm
Example 16. Simplify.
16−1/2
Example 17. Simplify.
−49−1/2
Example 18. Simplify.
(−49)−1/2
Example 19. Simplify.
27−2/3
Example 20. Simplify.
1
125
−2/3
Example 21. Simplify.
−64−2/3
58
Tuesday, October 18, 2016
Joseph Lee
Basic Properties for Exponents
xm · xn = xm+n
xm
= xm−n
xn
(xm )n = xmn
ax
by
m
=
am xm
bm y m
Example 22. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
x1/2 · x2/3
Example 23. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
x3/4
x2/3
Example 24. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
x3/4
1/2
Example 25. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
(x2 y 3 )2/3
59
Tuesday, October 18, 2016
Joseph Lee
Example 26. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
3 1/2
x
y4
Example 27. Simplify. Write the answer with positive exponents. Assume all variables represent positive
values.
6 1/2
9x
y8
Example 28. Simplify by writing as a root with a smaller index.
√
4
25
Example 29. Simplify by writing as a root with a smaller index.
√
6
8
Example 30. Simplify by writing as a root with a smaller index.
√
8
x6
Example 31. Simplify by writing as a root with a smaller index.
p
9
x3 y 6
60
Tuesday, October 18, 2016
Joseph Lee
Example 32. Simplify. Write the answer as a radical. Assume all variables represent positive values.
√ √
x3x
Example 33. Simplify. Write the answer as a radical. Assume all variables represent positive values.
√
√
5
4
x3 x2
Example 34. Simplify. Write the answer as a radical. Assume all variables represent positive values.
√
3
x2
√
x
Example 35. Simplify. Write the answer as a radical. Assume all variables represent positive values.
√ √
3
5· 7
Example 36. Simplify. Write the answer as a single radical. Assume all variables represent positive
values.
q
3 √
x
Example 37. Simplify. Write the answer as a single radical. Assume all variables represent positive
values.
q
3 √
4
x
61
Thursday, October 20, 2016
Joseph Lee
Simplifying Radicals
Multiplying Radicals with the Same Index
√
n
Example 1. Simplify.
Example 2. Simplify.
Example 3. Simplify.
Example 4. Simplify.
x·
√
√
√
n
3·
5·
√
3
y=
√
√
2·
√
4
4·
√
n
xy
27
125
√
3
32
√
4
4
Example 5. Simplify. Assume all variables represent nonnegative values.
√
√
x3 · x5
62
Thursday, October 20, 2016
Joseph Lee
Observation
If x represents a nonnegative value, then
Example 6. Simplify.
√
x·
√
√
7·
x = x.
√
7
Example 7. Simplify. Assume all variables represent nonnegative values.
p
p
3x2 y · 3x2 y
Dividing Radicals with the Same Index
r
n
√
n
x
x
= √
n
y
y
Example 8. Simplify.
r
Example 9. Simplify.
5
9
√
14
√
2
63
Thursday, October 20, 2016
Joseph Lee
Example 10. Simplify.
r
3
Example 11. Simplify.
25
27
√
4
48
√
4
3
Rules for Simplifying Radicals
A radical expression is simplified if:
1. there are no factors inside any radical raised to a power greater than or equal to the index,
2. there are no fractions inside any radicals, and
3. there are no radicals in any denominator of any fraction.
Example 12. Simplify.
Example 13. Simplify.
√
√
8
48
64
Thursday, October 20, 2016
Example 14. Simplify.
Joseph Lee
√
72
Example 15. Simplify.
√
4 12
Example 16. Simplify.
√
−3 54
Example 17. Simplify.
√
252
65
Thursday, October 20, 2016
Example 18. Simplify.
Example 19. Simplify.
Example 20. Simplify.
Example 21. Simplify.
Joseph Lee
√
720
√
3
√
3
√
4
32
162
208
66
Thursday, October 20, 2016
Joseph Lee
Example 22. Simplify. Assume all variables represent nonnegative values.
√
x3
Example 23. Simplify. Assume all variables represent nonnegative values.
√
x11
Example 24. Simplify. Assume all variables represent nonnegative values.
√
3
x11
Example 25. Simplify. Assume all variables represent nonnegative values.
p
32x5 y 4
Example 26. Simplify. Assume all variables represent nonnegative values.
p
3
54x8 y 7
67
Thursday, October 20, 2016
Joseph Lee
Example 27. Simplify. Assume all variables represent nonnegative values.
p
4
8x8 y 7
Example 28. Simplify. Assume all variables represent nonnegative values.
p
p
5x3 y 2 · 10xy 3
Example 29. Simplify. Assume all variables represent nonnegative values.
p
p
4x2 y 2x2 y 3xy 4 6xy
Example 30. Simplify.
√
540
√
12
68
Tuesday, October 25, 2016
Joseph Lee
Adding, Subtracting, and Multiplying Radicals
Example 1. Add.
√
√
4 3+7 3
Example 2. Subtract.
√
√
3
3
2 7−6 7
Example 3. Add.
√
Example 4. Subtract.
18 +
√
32
√
√
6 20 − 2 80
69
Tuesday, October 25, 2016
Example 5. Subtract.
Joseph Lee
√
3
√
3
54 − 5 2
Example 6. Subtract. Assume all variables represent nonnegative values.
√
√
2 48x5 − 7x 12x3
Example 7. Simplify.
Example 8. Simplify.
√
√
√
4
4
2x 16x − 4 4 x + 7 x5
√ √
√ 6
5− 3
70
Tuesday, October 25, 2016
Example 9. Simplify.
Joseph Lee
√ √ √
6 5 2 8 − 3 10
Example 10. Simplify.
Example 11. Simplify.
4+
√ √ 3 6− 7
√
√ √
√ 2 6−3 5 5 6−9 5
Example 12. Simplify. Assume all variables represent nonnegative values.
√
√
x+3
x+7
71
Tuesday, October 25, 2016
Joseph Lee
Example 13. Simplify.
Example 14. Simplify.
√ 2
2+5 3
√
√ 2
3 2−6 7
Example 15. Simplify. Assume all variables represent nonnegative values.
2
√
x+4
Example 16. Simplify.
8+
√ √ 3 8− 3
72
Tuesday, October 25, 2016
Example 17. Simplify.
Example 18. Simplify.
Example 19. Simplify.
Joseph Lee
√
√ √
√ 3 7−4 5 3 7+4 5
√
8·
√
6−
√
5·
√
15
√
3
1500 √
3
√
+ 108
3
3
73
Tuesday, October 25, 2016
Joseph Lee
Dividing Radicals
Example 1. Simplify.
1
√
2
Example 2. Simplify.
4
√
3
Example 3. Simplify.
r
9
5
Example 4. Simplify. Assume all variables represent positive values.
r
2
x
74
Tuesday, October 25, 2016
Joseph Lee
Example 5. Simplify.
1
√
3
2
Example 6. Simplify.
1
√
3
4
Example 7. Simplify. Assume all variables represent positive values.
r
3x
3
4y 2
Example 8. Simplify.
1
√
4
8
75
Tuesday, October 25, 2016
Joseph Lee
Example 9. Simplify.
Example 10. Multiply.
√
3 8
√
15
√
x+
√ √
√ y
x− y
Example 11. Simplify.
3
√
4− 2
Example 12. Simplify.
√
√
3+
6
√
76
18
Tuesday, October 25, 2016
Example 13. Simplify.
Joseph Lee
√
2− 3
√
2+ 3
Example 14. Simplify. Assume all variables represent positive values.
√
x
√
√
x + 2y
Example 15. Rationalize the numerator.
√
2
3
Example 16. Rationalize the numerator.
√
3− x
√
2
77
Thursday, October 27, 2016
Joseph Lee
Solving Equations with Radicals
Example 1. Solve.
Example 2. Solve.
Example 3. Solve.
Example 4. Solve.
√
√
√
√
x=4
x = 13
x = −5
x−3=2
78
Thursday, October 27, 2016
Example 5. Solve.
Example 6. Solve.
Example 7. Solve.
Joseph Lee
√
√
3
√
4
4x + 1 = 5
x + 8 = −3
2x + 1 = −2
79
Thursday, October 27, 2016
Example 8. Solve.
Example 9. Solve.
Joseph Lee
√
√
3
2x + 8 =
7x − 4 =
Example 10. Solve.
x+1=
√
√
3
√
80
3x + 2
1 − 3x
x + 13
Thursday, October 27, 2016
Joseph Lee
Example 11. Solve.
p
x2 − x + 3 − 1 = 2x
Example 12. Solve.
√
x−1=
√
81
2x + 2
Thursday, November 3, 2016
Joseph Lee
Unit 5
Complex Numbers
Definition: Complex Number
The imaginary unit, i, can be defined in the following two ways.
• i2 = −1
√
• i = −1
The sum of any real number a and imaginary number bi,
a + bi,
is called a complex number.
Example 1. Simplify.
√
−36
Example 2. Simplify.
√
−12
Example 3. Simplify.
√
−98
82
Thursday, November 3, 2016
Joseph Lee
Example 4. Add the complex numbers.
(−2 + 3i) + (6 − 9i)
Example 5. Subtract the complex numbers.
(5 − 12i) − (8 − 3i)
Example 6. Multiply.
(4i)(5i)
Example 7. Multiply.
(−2i)(i)
83
Thursday, November 3, 2016
Joseph Lee
Example 8. Multiply.
4i(5 − 2i)
Example 9. Multiply.
(3 + 4i)(5 − i)
Example 10. Multiply.
(1 − 3i)(3 − 2i)
Example 11. Multiply.
(4 + i)2
84
Thursday, November 3, 2016
Joseph Lee
Example 12. Multiply.
(2 − 5i)2
Example 13. Multiply.
(7 − 2i)(7 + 2i)
Example 14. Multiply.
(3 + 4i)(3 − 4i)
Example 15. Divide.
3
4i
85
Thursday, November 3, 2016
Joseph Lee
Example 16. Divide.
4 − 5i
3i
Example 17. Divide.
7 + 4i
2 − 3i
Example 18. Divide.
5−i
4 + 5i
86
Thursday, November 3, 2016
Joseph Lee
Completing the Square
Definition: Quadratic Equation
A quadratic equation is an equation that can be written as
ax2 + bx + c = 0
where a, b, and c are real numbers and a 6= 0.
Example 1. Solve.
x2 = 4
Theorem: Square Root Principal
If x2 = a, then x =
√
√
a or x = − a.
Example 2. Solve.
x2 = 9
Example 3. Solve.
x2 = 12
87
Thursday, November 3, 2016
Joseph Lee
Example 4. Solve.
x2 = −25
Example 5. Solve.
x2 − 4 = 32
Example 6. Solve.
3x2 + 4 = 58
Example 7. Solve.
(x − 3)2 = 4
88
Thursday, November 3, 2016
Joseph Lee
Example 8. Solve.
(2x − 1)2 = −5
Completing the Square
The polynomial
x2
+ bx can be turned into a square by adding
2
x + bx +
1 2
b .
2
1 2
1 2
b = x+ b
2
2
Example 9. Solve by completing the square.
x2 + 8x + 2 = −5
89
Thursday, November 3, 2016
Joseph Lee
Example 10. Solve by completing the square.
x2 − 6x + 1 = 0
Example 11. Solve by completing the square.
x2 + x − 2 = 5
90
Thursday, November 3, 2016
Joseph Lee
Example 12. Solve by completing the square.
x2 − 5x +
41
=0
4
Example 13. Solve by completing the square.
9x2 + 12x − 5 = 0
91
Thursday, November 3, 2016
Joseph Lee
Example 14. Solve by completing the square.
4x2 − 20x + 33 = 0
92
Tuesday, November 8, 2016
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax2 + bx + c = 0, where a, b, and c are real numbers, and a 6= 0,
√
−b ± b2 − 4ac
x=
.
2a
Proof.
ax2 + bx + c = 0
ax2 + bx
= −c
ax2 + bx
a
=
−c
a
c
b
=−
x2 + x
a
a
2
2
b
b
c
b
x2 + x +
=− +
a
2a
a
2a
b2
c
b2
b
x2 + x + 2 = − + 2
a
4a
a 4a
b 2
4ac
b2
x+
=− 2 + 2
2a
4a
4a
s
b
x+
2a
2
b
x+
2a
2
b
x+
2a
2
=
b2
4ac
− 2
2
4a
4a
=
b2 − 4ac
4a2
r
=±
b2 − 4ac
4a2
√
b
b2 − 4ac
x+
=± √
2a
4a2
√
b
b2 − 4ac
x+
=±
2a
2a
√
b
b2 − 4ac
x =− ±
2a
2a
√
−b ± b2 − 4ac
x =
2a
93
Tuesday, November 8, 2016
Joseph Lee
Example 1. Solve using the quadratic formula.
x2 + 2x − 8 = 0
Example 2. Solve using the quadratic formula.
3x2 − 5x − 2 = 0
Example 3. Solve using the quadratic formula.
x2 − 3x − 7 = 0
94
Tuesday, November 8, 2016
Joseph Lee
Example 4. Solve using the quadratic formula.
2x2 + x + 1 = 0
Example 5. Solve using the quadratic formula.
4x2 = 7x − 3
Example 6. Solve using the quadratic formula.
3x − x2 = 1
95
Tuesday, November 8, 2016
Joseph Lee
Solving Equations
Example 1. Solve.
3x
14
30
−
= 2
x−2 x+3
x +x−6
Example 2. Solve.
1+
4
2
= 2
x
x
96
Tuesday, November 8, 2016
Example 3. Solve.
Joseph Lee
√
Example 4. Solve.
x+
x+3=x−3
√
2x + 12 = −1
97
Tuesday, November 8, 2016
Joseph Lee
Example 5. Solve.
(x + 7)2 − (x + 7) − 12 = 0
Substitution:
Example 6. Solve.
x4 − 5x2 − 36 = 0
Substitution:
98
Tuesday, November 8, 2016
Joseph Lee
Graphing Quadratic Functions
Definition: Quadratic Function
A quadratic function is a function that can be written as
f (x) = ax2 + bx + c
where a, b, and c are real numbers and a 6= 0. The graph of a quadratic function is called a parabola.
Example 1. Graph f (x) = x2 .
f (x)
x
Vertex:
Axis of Symmetry:
Example 2. Graph f (x) = −x2 .
f (x)
x
Vertex:
Axis of Symmetry:
99
Tuesday, November 8, 2016
Joseph Lee
Example 3. Graph f (x) = 2x2 .
f (x)
x
Vertex:
Axis of Symmetry:
1
Example 4. Graph f (x) = x2 .
2
f (x)
x
Vertex:
Axis of Symmetry:
Standard Form for a Quadratic Function
A quadratic function is in standard form if it is written as
f (x) = a(x − h)2 + k.
The vertex of the parabola is located at (h, k). The axis of symmetry is given by x = h.
100
Tuesday, November 8, 2016
Joseph Lee
Example 5. Graph f (x) = 2(x + 2)2 − 3.
f (x)
x
Vertex:
Axis of Symmetry:
Example 6. Graph f (x) = −3(x − 1)2 + 4.
f (x)
x
Vertex:
Axis of Symmetry:
101