Download Intermediate Algebra Final Exam Review Sheet

Int. Alg. Final Exam Review Sheet
December 2007
Page 1 of 24
Section R.3: Operations on Signed Numbers
To find the least common denominator:
1. Factor each denominator.
2. Write down the common factor(s), and then copy any additional factors.
3. Multiply those numbers; the product is the least common denominator.
4. Convert each rational number to an equivalent fraction with the LCD as denominator.
The Distributive Property of Real Numbers:
If a, b, and c are real numbers, then
a  b  c   a  b  a  c
 a  b  c  a  c  b  c
Section R.4: Order of Operations
Exponents
Repeated multiplication can be written more efficiently using the notation of exponents:
3  3  3  3  34
Many calculators and computer languages use the “caret” symbol to denote exponents:
3  3  3  3  3^ 4
Order of Operations:
1. Evaluate expressions within parentheses first. Begin with the innermost parentheses and work outward.
2. Evaluate expressions involving exponents next, working from left to right.
3. Evaluate expressions involving multiplication and division next, working from left to right.
4. Evaluate expressions involving addition and subtraction next, working from left to right.
Section R.5: Algebraic Expressions
Simplifying Algebraic Expressions:

by Combining Like Terms: 2 x  3 y  9  4 x 2  2 y  5x  19  4 x 2  7 x  5 y  28
2 x  x  y    xy  x   2 x 2  2 xy  xy  x

by removing parentheses using the distributive property:
 2 x 2  xy  x
Int. Alg. Final Exam Review Sheet
December 2007
Page 2 of 24
Section 1.1: Linear Equations
We write an equation when we know what we want the answer to a multi-step to calculation to turn out
to be.
1. Key fact that allows us to solve equations: Properties of Equality (i.e., the equality remains true if you
do the same thing to both sides of the equation.): An equation can be transformed into an equivalent
equation by adding or subtracting the same quantity on both sides of the equation, or by multiplying or
dividing both sides of the equation by the same nonzero quantity.
2. The goal in manipulating equations when we solve them is to isolate the variable on one side of the
equation.
3. When solving equations, proceed opposite the order of operations:
a. Simplify both sides of the equation as much as possible by combining like terms and removing
parentheses suing the distributive property.
b. Look for something to add or subtract to both sides.
c. Look for something to multiply or divide both sides.
d. Look for an exponent to raise both sides to.
Section 1.2:An Introduction to Problem Solving
English
Math
is, was, are, yields, equals, gives,
results in, is equal to, is equivalent
to
=
product of, of (with a fraction or %) 
sum of
+
difference
-
x more than y
x+y
x less than y
y-x
increase
+
decrease
-
twice x or 2 times x
2x
n times x
nx
Two consecutive integers
n,
n+1
Area of a rectangle: A = Width  Length
Perimeter = sum of the lengths of the sides of a shape
Rate  Time = Amount
Distance = rate  time
Mixture Problems: Portion from Item A + Portion from Item B = Total
part
Percentage problems: percentage 
whole
Int. Alg. Final Exam Review Sheet
December 2007
Page 3 of 24
Problem Solving with Mathematical Models
1. Identify what you are looking for.
2. Give names to the unknowns.
3. Translate the problem into the language of mathematics. Use pictures to help you when possible.
Solve the equation.
4. Check the reasonableness of your answer.
5. Answer the original question.
Section 1.3: Using Formulas to Solve Problems
Shape
Square
Rectangle
Triangle
Trapezoid
Parallelogram
Circle
Cube
Rectangular Solid
Sphere
Right Circular Cylinder
Cone
Some Geometric Formulas
Formula
Area A  s 2
Perimeter P  4s
Area A  lw
Perimeter P  2l  2w
1
Area A  bh
2
Perimeter P  a  b  c
1
Area A  h  B  b 
2
Perimeter P  a  b  c  B
Area A  ah
Perimeter P  2a  2b
Area A   r 2
Circumference C  2 r   d
Volume V  s 3
Surface Area S  6 s 2
Volume V  lwh
Surface Area S  2lw  2lh  2wh
4
Volume V   r 3
3
Surface Area S  4 r 2
Volume V   r 2 h
Surface Area S  2 r 2  2 rh
1
Volume V   r 2 h
3
Int. Alg. Final Exam Review Sheet
December 2007
Page 4 of 24
Section 1.4: Linear Inequalities
Properties of Inequalities: An inequality can be transformed into an equivalent inequality by adding or
subtracting any quantity to both sides, or multiplying or dividing by any positive quantity. If both sides are
multiplied or divided by a negative quantity, then the inequality symbol gets reversed.
Section 1.5: Compound Inequalities
Intersection of two sets: all the elements in both sets.
Union of two sets: only the elements common to both sets.
Compound Inequalities Using and: the solution is the intersection of the two sets.
Compound Inequalities Using or: the solution is the union of the two sets.
Section 1.6: Absolute Value Equations and Inequalities
To solve absolute value equations, use the facts that:
 u  a is equivalent to u = a or u = -a.
 u  v is equivalent to u = v or u = -v.
To solve absolute value inequalities, use the facts that:
 x  c is equivalent to  c  x  c
 x  c is equivalent to  c  x  c
 x  c is equivalent to x  c or x  c
 x  c is equivalent to x  c or x  c
Section 2.1: Rectangular Coordinates and Graphs of Equations
To graph a linear equation by plotting points, make a table with two rows of values, plot the points, then draw
the line.
Graph y  2 x  5
Table:
Plot points:
Draw line:
y
y
x
y
0
-5
2
-1
x
x
Int. Alg. Final Exam Review Sheet
December 2007
Page 5 of 24
To graph a nonlinear equation by plotting points, make a table with many rows of values, plot the points, then
draw a smooth curve connecting the points.
Graph y = 0.5x2 + 1.
y
x
-3
-2
-1
0
1
2
3

y
5.5
3.0
1.5
1.0
1.5
3.0
5.5













An x-intercept of a graph is the x-coordinate of a point on the graph that crosses or touches the x-axis.
A y-intercept of a graph is the y-coordinate of a point on the graph that crosses or touches the y-axis.
y




      







x


y = x^4 - 5x^2 + 4

This graph has x-intercepts at x = -2, x = -1, x = +1, and x = +2, and a y-intercept at y = 4.
Section 2.2: Relations
A relation is a “link” from elements of one set to elements of another set.
If x and y are two elements in these set, and if a relation exists between x and y, then we say:
x corresponds to y,
or y depends on x,
and we write x  y.
We may also write a relation where y depends on x as an ordered pair (x, y).
The domain of a relation is the set of all inputs to the relation.
The range of a relation is the set of all outputs of the relation.
Section 2.3: An Introduction to Functions
Section 2.4: Functions and Their Graphs

x

Int. Alg. Final Exam Review Sheet
December 2007
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Section 3.1: Linear Equations and Linear Functions
Definition: Linear Equation
A linear equation (in two variables) is an equation of the form
Ax + By = C
Where A, B, and C are real numbers and A and B both cannot be zero. This is called the standard form for the
equation of a line.
To graph a linear equation using intercepts:
1. Let y = 0 and solve for x (for the x-intercept)
2. Let x = 0 and solve for y (for the y-intercept)
Equation of a Vertical Line
A vertical line is given by an equation of the form
x=a
Where a is the x-intercept.
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y=b
Where b is the y-intercept.
Section 3.2: Slope and Equations of Lines
Definition: Slope
The slope m between two points with coordinates (x1, y1) and (x2, y2) is defined by the formula
y y2  y1
m

.
x x2  x1
If x1 = x2, then the line between the two points is a vertical line, and the slope m is undefined.
Properties of Slope:
m > 0  the line slants upward from left to right
m < 0  the line slants downward from left to right
m = 0  the line slants is horizontal
m = undefined  the line is vertical
Point-Slope Form of a Line
If you know the slope of a line (m) and one point on that line (x1, y1), then the equation of the line is given by
the equation
y  y1  m  x  x1  (Point-Slope Form of a Line)
An equation of a line L with slope m and y-intercept b has a slope-intercept form of
y  mx  b .
Int. Alg. Final Exam Review Sheet
December 2007
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Equation of a line from two points:
1) Find the slope.
2) Use the point-slope form.
Section 3.3: Parallel and Perpendicular Lines
Definition: Relationship between the slopes of parallel lines
Two nonvertical lines are parallel if and only if their slopes are equal and they have different y-intercepts.
Vertical lines are parallel if they have different x-intercepts.
Definition: Relationship between the slopes of perpendicular lines
Two nonvertical lines are perpendicular if and only if the product of their slopes is -1. Alternatively, their
slopes are negative reciprocals of one another. Vertical lines are perpendicular to horizontal lines.
Section 3.4: Linear Inequalities in Two Variables
 Step 1: Write the inequality as an equality, then graph the equation using a dashed line if it is a strict
inequality (< or >), or using a solid line if it is not a strict inequality ( or ).
 Step 2: Pick a test point and see if the ordered pair satisfies the inequality. If it does satisfy, shade the half of
the plane on the side of the line containing the point. If it does not satisfy, shade the other half.
Section 3.5: Building Linear Models
You should be able to write equations for linear models given a verbal description, a direct variation, or a set of
data.
Section 4.1: Systems of Linear Equations in Two Variables
Solution by substitution:
1. Solve one equation for one of the variables.
2. Substitute that expression for that variable into the other equation.
3. Solve for the remaining variable.
4. Back-substitute into the first equation to get the value for the other variable.
Section 4.2: Problem Solving: Systems of Two Linear Equations Containing Two Unknowns
You should be able to write out a model for a real-world situation involving two equations in two unknowns,
then solve that system to get an answer.
Int. Alg. Final Exam Review Sheet
December 2007
Page 8 of 24
Section 4.3: Systems of Linear Equations in Three Variables
Example of a system of three equations in three unknowns that is in triangular form:
x  2 y  z  1

y  2z  5


z  3

Notice: the name triangular form comes from the “blank” triangular space in the lower left corner due to no x
or y variables. Also, this system is really easy to solve using back-substitution:
y  23  5
y65
y  1
x  2y  z 1
x  2   1  3  1
x 1  1
x2
Steps for Solving a System of Three Linear Equations in Three Unknowns Using Elimination
1. Eliminate the same one variable from two of the equations using elimination.
2. Use elimination to remove a second variable from those two equations.
3. Solve for the remaining variable.
4. Substitute the answer for that variable into the remaining equations.
Section 4.4: Using Matrices to Solve Systems
Example of a system of three equations in three unknowns that is in triangular form:
x  2 y  z  1
 1 2 1 1 

y  2 z  5   0 1 2 5 


 0 0 1 3 
z  3

Use row operations to convert the matrix into triangular form.
 x  2y  z  3
1


2 x  5 y  6 z  7   2
2 x  4 y  z  5
 2

2
1
5
6
4 1
3
7  
5 
1
0

 2
2
1
1
4
4 1
3
1  
5 
1
0

 0
2
3
1 4 1 
0 3 1
1
Int. Alg. Final Exam Review Sheet
December 2007
Page 9 of 24
Section 4.5: Determinants and Cramer's Rule
Definition: Determinant of a 2  2 matrix
a b
a b 
Suppose that a, b, c, and d are real numbers. The determinant of the 2  2 matrix 
, written as
,

c d
c d 
a b
is
 ad  bc .
c d
Cramer’s Rule for solving a system of two linear equations in two unknowns
ax  by  s
The solution to the system of equations 
is given by
 cx  dy  t
s b
a s
Dy
t d
c t
D
x
 x and y 

a b
a b
D
D
c
d
c
Provided that D 
a b
c d
d
 ad  bc  0
Definition: Determinant of a 3  3 matrix
 a1,1 a1,2

The determinant of the 3  3 matrix  a2,1 a2,2
 a3,1 a3,2
a1,3 
a1,1

a2,3  , written as a2,1
a3,3 
a3,1
a1,2
a1,3
a2,2
a2,3 , is calculated using the
a3,2
a3,3
determinants of 2  2 matrices as follows:
a1,1
a1,2
a1,3
a2,1
a2,2
a2,3  a1,1
a3,1
a3,2
a3,3
a2,2
a2,3
a3,2
a3,3
 a1,2
a2,1
a2,3
a3,1
a3,3
 a1,3
a2,1
a2,2
a3,1
a3,2
.
Cramer’s Rule for solving a system of three linear equations in three unknowns
 a1 x  b1 y  c1 z  d1

The solution to the system of equations a2 x  b2 y  c2 z  d 2
a x  b y  c z  d
3
3
3
 3
with
a1 b1 c1
d1 b1 c1
a1 d1 c1
a1 b1 d1
D  a2 b2 c2  0 , Dx  d 2 b2 c2 , Dy  a2 d 2 c2 , and Dz  a2 b2 d 2
a3 b3 c3
d3 b3 c3
a3 d3 c3
a3 b3 d3
is given by
D
D
D
x  x , y  y , and z  z .
D
D
D
Int. Alg. Final Exam Review Sheet
December 2007
Page 10 of 24
Section 4.6: Systems of Linear Inequalities
2x
Graph the system 
 x

3y

6

4y

4
.
Section 5.1: Adding and Subtracting Polynomials
To add polynomials, combine like terms.
To subtract polynomials, combine like terms.
A polynomial function is a function whose rule is a polynomial. The domain of a polynomial function is all
real numbers. The degree of a polynomial function is the value of the largest exponent on the variable.
Section 5.2: Multiplying Polynomials
Extended form of the Distributive Property:
a  b1  b2  b3   bn   ab1  ab2  ab3   abn
To multiply polynomials, use distribution, or multiply each term in the first polynomial by each term in the
second polynomial.
Example: (x2 + 3x + 9)(x2 – 2x +3) = x4 –2x3 +3x2
+3x3 -6x2 +9x
+9x2 -18x +27
4
3
=x
+x
+6x2 -9x
+27
Section 5.3: Dividing Polynomials; Synthetic Division
Dividing a polynomial by a monomial: Divide the monomial into each term of the polynomial, and cancel
when possible.
Dividing a polynomial by a polynomial using long division:
Long division of polynomials is a lot like long division of numbers:
a. Arrange divisor and dividend around the dividing symbol, and be sure to write them in
descending order of powers with all terms explicitly stated.
b. Divide leading terms, then multiply and subtract.
c. Repeat until a remainder of order less than the divisor is obtained.
Example:
2 x3  x  18
?
x3
Int. Alg. Final Exam Review Sheet
December 2007
Page 11 of 24
2 x 2  6 x  17
x  3 2 x  0 x 2  x  18
3

 2 x3  6 x 2

6x2  x

 6 x 2  18 x

 17 x  18
  17 x  51
33

2 x3  x  18
33
 2 x 2  6 x  17 
x3
x3
Dividing a polynomial by a polynomial using synthetic division: THIS IS A SHORTCUT THAT ONLY
WORKS WHEN THE DEGREE OF THE DIVISOR IS 1 (I.E., THE DIVISOR IS x – c) !!!
Synthetic division is a shorthand way to divide a polynomial by a linear factor.:
a. Write c outside the bar and the coefficients of the dividend inside the bar.
b. Bring the leading coefficient straight down.
c. Compute c times the number in the bottom row, and write the answer in the middle row to the
right.
d. Add and repeat until all coefficients are used up.
The Remainder Theorem
If the polynomial P(x) is divided by x – c, then the remainder is the value P(c).
The Factor Theorem
If P is a polynomial function, then x – c is a factor of P if and only if P(c) = 0.
(This can be used to see if a divisor divides evenly into a dividend quickly)
Section 5.4: Greatest Common Factor; Factoring by Grouping
Factoring the greatest common factor:
1. Identify the greatest common factor (GCF) in each term.
2. Rewrite each term as the product of the GCF and the remaining factor.
3. Use the Distributive Property to factor out the GCF.
4. Use the Distributive Property to verify that the factorization is correct.
Example:
10 x 2 y 2  15 xy 3  25 x3 y 4  5 xy  2 xy  5 xy  y 2  5 xy  x 2 y 3
 5 xy   2 xy  y 2  5 x 2 y 3 
Int. Alg. Final Exam Review Sheet
December 2007
Page 12 of 24
Factoring by grouping:
1. Group the terms with common factor. You may need to rearrange the terms.
2. In each grouping, factor out the common factor.
3. Factor out the common factor that remains.
4. Check your work.   .
Example:
5 y  5 z  ay  az   5 y  5 z    ay  az 
 5 y  z   a  y  z 
  5  a  y  z 
Section 5.5: Factoring Trinomials
1. Factoring x2 + qx + p:
a. Look for factors of p that sum to q
b. Example: x2 – 4x – 12 =
i. factors of 12 are: 1, 2, 3, 4, 6, 12
ii. the factors that add up to –4 are: -4 = –6 and +2
iii.  x2 – 4x – 12 = (x – 6)(x + 2)
2. Factoring Ax2 + Bx + C:
a. List factors of AC that add to the coefficient B, then write the middle term as a sum of those
factors
b. Factor by grouping
c. Example: 2x2 – 9x – 18 =
i. (2)(-18) = -36
ii. factors of -36 that add to -9 are: -9 = -12 and + 3
iii. 2x2 – 9x – 18 = 2x2 – 12x + 3x – 18 = 2x(x – 6) + 3(x – 6) = (2x + 3)(x – 6)
Section 5.6: Factoring Special Products
1. Difference of Two Squares: a2 – b2 = (a + b)(a – b)
a. Example: 9x2 – 81 = (3x + 3)(3x – 3)
2. Square of Sum: a2 + 2ab + b2 = (a + b)2
a. Example: x2 + 12x + 36 = (x + 6)2
3. Square of Difference: a2 – 2ab + b2 = (a – b)2
a. Example: x2 - 2x + 1 = (x – 1)2
4. Difference of Cubes: a3 – b3 = (a – b)(a2 + ab + b2)
a. Example: x3 – 8 = (x – 2)(x2 + 2x + 4)
5. Sum of Cubes: a3 + b3 = (a + b)(a2 – ab + b2)
a. Example: 27x3 + 64 = (3x + 4)(9x2 – 12x + 16)
Section 5.7: Factoring: A General Strategy
See the book…
Int. Alg. Final Exam Review Sheet
December 2007
Page 13 of 24
Section 5.8: Polynomial Equations
The Zero-Product Property
If the product of two numbers is zero, then at least one of the numbers is zero. That is,
if ab = 0, then a = 0 or b = 0, or both a and b are 0.
Steps for solving any polynomial equation (quadratic or higher):
1. Expand the polynomial equation (if needed), and collect all terms on one side; combine like terms.
2. Factor the polynomial on the one side.
3. Set each factor from step 2 equal to zero (this is justified by the zero-product property rule).
4. Solve each first degree equation for the variable.
5. Check answers in original equation.
Example: x3 – 4x2
= 12x
3
2
x – 4x – 12x = 0
x(x3 – 4x2 – 12)= 0
x(x – 6)(x + 2) = 0
x = 0 or
x–6=0
or
x+2=0
x=0
x=6
x = -2
Section 6.1: Multiplying and Dividing Rational Expressions
A rational expression is simplified when all common factors in the numerator and denominator have been
cancelled.
Example:
 x  5   x  3 x  5
x 2  2 x  15  x  5 x  3



2 x 2  3x  9  2 x  3 x  3  2 x  3  x  3 2 x  3
a c ac
 
b d bd
x 3
( x  3)
4 2 x  1 4( x  3)(2 x  1)
Example:
 (8 x  4) 
 

4
2
2 x  5x  3
(2 x  1)( x  3) 1
1
( x  3)(2 x  1)
a c a d ad
To divide rational expressions, use the arithmetic rule    
(i.e., invert and multiply)
b d b c bc
x3
( x  3)
1
1
Example:
 x2  9 


2
2 x  5x  3
 2 x  1 x  3  x  3 x  3  2 x  1 x  3 x  3
To multiply rational expressions, use the arithmetic rule


Int. Alg. Final Exam Review Sheet
December 2007
Page 14 of 24
Section 6.2: Adding and Subtracting Rational Expressions
When rational expressions have common denominators, they can be added or subtracted by simply adding
or subtracting the numerators (like with arithmetic fractions).
To find the least common denominator of a rational expression, factor each denominator, then list all the
common factors and all the uncommon factors.
To add or subtract rational expressions, find the LCD, then write each rational expression with the common
denominator, then add or subtract numerators.
Example:
3 x  1 2 x  1  3 x  1 x  1  2 x  1 x  1



x 1
x 1
 x  1 x  1  x  1 x  1

(3 x 2  2 x  1)  (2 x 2  3 x  1)
 x  1 x  1

x2  5x  2
 x  1 x  1
Section 6.3: Complex Rational Expressions
Keys:
1. Method #1: Simplify numerator and denominator to a single rational expression, then invert and
multiply.
2. Method #2: Multiply numerator and denominator by the L.C.D. of all fractions…
Section 6.4: Rational Equations
Steps for Solving a Rational Equation
1. Determine the domain of the rational equation.
2. Determine the LCD of all the denominators.
3. Multiply both sides of the equation by the LCD and simplify each side of the equation into polynomials.
4. Solve the resulting polynomial equation (using factoring and the zero-product property if needed).
5. Verify your solutions using the original equation. Make sure they are in the domain of the equation (i.e.,
that they are not extraneous solutions).
Int. Alg. Final Exam Review Sheet
December 2007
Page 15 of 24
3x
3
5 x  17

 2
x7 x2
x  5 x  14
3x
3
5 x  17


x7 x2
 x  7  x  2 
 x  7  x  2 
3 
5 x  17
 3x



  x  7  x  2   
1
 x  7  x  2 
 x7 x2
3 x  x  2   3  x  7   5 x  17
3 x 2  6 x  3 x  21  5 x  17
3x 2  4 x  4  0
3x 2  6 x  2 x  4  0
3x  x  2   2  x  2   0
 3x  2  x  2   0
x  2 / 3 or
x2
Section 6.5: Rational Inequalities
Steps for Solving a Rational Inequality
1. Write the inequality so it is a single rational expression on one side of the inequality and zero on the other
side. Completely factor the numerator and denominator of the rational expression.
2. Determine all numbers that make the factors of the rational expression zero.
3. Use those zeros to separate the real number line into intervals.
4. Choose a test point in each interval, and determine the sign of the rational expression at that test point. If
the sign for a test value matches the inequality, then the entire interval containing the test point is a solution.
Section 6.6: Models Involving Rational Expressions
Int. Alg. Final Exam Review Sheet
December 2007
Page 16 of 24
Section 7.1: nth Roots and Rational Exponents
Definition: The principal nth root of a number
n
a , where n is an integer greater than or equal to 2, computes to a number b such that if
n
a  b , then b n  a .
If n is an even number, then a and b must be positive.
If n is an odd number, then a and b can be any real number.
n
2
3
4
5
6
7
8
9
10
2^n
4
8
16
32
64
128
256
512
1,024
3^n
4^n
5^n
6^n
7^n
8^n
9^n
10^n
9
16
25
36
49
64
81
100
27
64
125
216
343
512
729
1,000
81
256
625
1,296
2,401
4,096
6,561
10,000
243
1,024
3,125
7,776
16,807
32,768
59,049
100,000
729
4,096
15,625
46,656
117,649
262,144
531,441 1,000,000
2,187
16,384
78,125
279,936
823,543 2,097,152 4,782,969
6,561
65,536
390,625 1,679,616 5,764,801
19,683
262,144 1,953,125
59,049 1,048,576 9,765,625
n
2
3
4
5
6
7
8
9
10
(-2)^n
4
-8
16
-32
64
-128
256
-512
1,024
(-3)^n
(-4)^n
(-5)^n
(-6)^n
(-7)^n
(-8)^n
(-9)^n
(-10)^n
9
16
25
36
49
64
81
100
-27
-64
-125
-216
-343
-512
-729
-1,000
81
256
625
1,296
2,401
4,096
6,561
10,000
-243
-1,024
-3,125
-7,776
-16,807
-32,768
-59,049 -100,000
729
4,096
15,625
46,656
117,649
262,144
531,441 1,000,000
-2,187
-16,384
-78,125 -279,936 -823,543 -2,097,152 -4,782,969
6,561
65,536
390,625 1,679,616 5,764,801
-19,683 -262,144 -1,953,125
59,049 1,048,576 9,765,625
a
1
n
n a
Section 7.2: Simplifying Expressions Using the Laws of Exponents
a a  a
m
a 
m n
n
mn
 a mn
Formulas for integer exponents:
am
1
 a mn  nm
n
a
a
0
a  1 for a  0
ab n
 a nb n
an
a
   n
b
b
a n 
1
for a  0
an
a
 
b
n
n
b
 
a
n
Int. Alg. Final Exam Review Sheet
December 2007
Page 17 of 24
Section 7.3: Simplifying Radical Expressions
Big Idea: A radical is in simplest form when:
As many powers as possible are pulled out of the radical.
The order of the radical is as low as possible.
There are no radicals in the denominator.
n
an 
m n
 a
n
a  mn a
n
a
 a  b  
n
n
n
a
n
b
n
n
ab
a
b
Steps to simplify a radical expression:
Write each factor of the radicand as the product of two factors, one of which is a perfect power of the
radicand.
Write the radicand as the product of two radicals (using the product property of radicals).
Take the nth root of the perfect power.
Section 7.4: Adding, Subtracting, and Multiplying Radical Expressions
Big Idea: Radicals can only be added or subtracted when they are similar. Similar radicals have the same
order and same radicand.
If the radicals are not similar, then they can not be combined by addition or subtraction.
Example of similar radicals that can be added:
43 7  53 7  93 7
Example of radicals that are not similar because of different orders and thus can not be added: 3 7  5 7
Example of radicals that are not similar because of different radicands and thus can not be added:
11  17
Radicals that do not look similar may in fact be similar if they are simplified first:
8  2  42  2  2 2  2  3 2
Int. Alg. Final Exam Review Sheet
December 2007
Page 18 of 24
Sections 7.5: Rationalizing Radical Expressions
Big Idea: A radical is in simplest form when:
As many powers as possible are pulled out of the radical.
The order of the radical is as low as possible.
There are no radicals in the denominator. (This is simply because before calculators, it was a pain to
divide by a radical long-hand).
To eliminate radical factors in the denominator, multiply numerator and denominator by an appropriate power
of the denominator to make the denominator a perfect power of the order of the radical. This is called
rationalizing the denominator.
To eliminate a sum of square roots in the denominator, multiplying numerator and denominator by the
conjugate of the denominator. This trick is based on the difference of squares: (a + b)(a – b) = a2 – b2.
Section 7.6: Functions Involving Radicals
Domain of a radical function:
If the index of the radical is even, then the domain is all numbers that make the radicand non-negative.
If the index of the radical is odd, then the domain is all real numbers.
Section 7.8: The Complex Number System
Definition: The Imaginary Unit
The imaginary unit, denoted by the symbol i, is the number whose square root is –1. That is, i2 = –1.
Taking the square root of both sides, we can see that i  1
Evaluating Square Roots of negative Numbers
If N is a positive real number, then the principal square root of –N is calculated as:
 N  N  1  N
 1 
Ni
Definition: Complex Numbers
Complex numbers are numbers of the form a + bi, where a and b are real numbers, and i is the imaginary unit.
The number a is called the real part of the complex number, and
the number b is called the imaginary part of the complex number.
Sum of Complex Numbers
To add complex numbers, add reals to reals, and add imaginaries to imaginaries:
(a + bi) + (c + di) = (a + c) + (b + d)i
Difference of Complex Numbers
To subtract complex numbers, subtract reals from reals, and subtract imaginaries from imaginaries:
(a + bi) – (c + di) = (a – c) + (b – d)i
Note: Adding or subtracting complex numbers is just like combining like terms.
Product of Complex Numbers
To multiply complex numbers, use the Distributive Property (or the FOIL method), then collect like terms:
(a + bi)(c + di) = (ac – bd) + (ad + bc)i
Int. Alg. Final Exam Review Sheet
December 2007
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Definition: Complex Conjugate:
For a complex number a + bi, its conjugate is defined as a – bi.
Quotient of Complex Numbers
To divide complex numbers, multiply top and bottom by the complex conjugate of the denominator. Note that
the denominator can be simplified using the difference of squares formula.
a  bi a  bi c  di  ac  bd    bc  ad  i



c  di c  di c  di
c2  d 2
Sections 8.1: Solving Quadratic Equations by Completing the Square
The Square Root Property:
If x2 = p, then x   p or x   p .
Completing the Square
To obtain a perfect square trinomial:
Identify the coefficient of the first-degree term. Multiply this coefficient by
1
, then square the result. That is,
2
2
b
if you have ax2 + bx + c, a value of   will complete the square.
 2
Solving a Quadratic Equation by Completing the Square:
Write the quadratic equation as x2 + bx = -c.
Complete the square on x2 + bx by making it a perfect square trinomial. Don’t forget to add the same
amount to the right hand side.
Factor the perfect square trinomial.
Solve using the Square Root Property.
Section 8.2: Solving Quadratic Equations by the Quadratic Formula
The Quadratic Formula:
If ax2 + bx + c = 0, then x 
b  b 2  4ac
2a
The Discriminant
The discriminant for a quadratic equation ax 2  bx  c  0 is the quantity b 2  4ac . Its value tells us the number
and type of solution to expect from the quadratic formula.
Discriminant
Positive and a
perfect square
Positive and not a
perfect square
Zero
Negative
Number of
Solutions
2
Type of Solutions
Practice Example
Rational
1
2
Irrational
2
1 (a repeated root)
2
Rational
Complex
3
4
Int. Alg. Final Exam Review Sheet
December 2007
Page 20 of 24
Section 8.3: Solving Equations Quadratic in Form
To solve x4 – 4x2 – 12 = 0, let u = x2. Then the equation becomes u2 – 4u – 12 = 0, so
u
4
 4 
4  64
2
48
u
2
u
2
 4 1 12 
2 1
which means that
x2  2  4
x   24
Section 8.4: Graphing Quadratic Equations Using Transformations
Adding or subtracting a constant from every occurrence of the variable x shifts the graph left or right.
The graph of y = f(x – h) simply shifts the graph of y = f(x) by h units to the right.
The graph of y = f(x + h) simply shifts the graph of y = f(x) by h units to the left.
Multiplying a function by a constant makes the graph steeper or even reflects it about the x-axis.
Int. Alg. Final Exam Review Sheet
December 2007
Page 21 of 24
Adding or subtracting a constant from the entire function shifts the graph up or down.
The graph of y = f(x) + k simply shifts the graph of y = f(x) by k units up.
The graph of y = f(x) – k simply shifts the graph of y = f(x) by k units down.
To quickly sketch the graph of a quadratic equation of the form y = a(x – h)2 + k:
Locate the point (h, k). This is the vertex.
If a is positive, the parabola opens up; if a is negative, the parabola opens down.
If |a| > 1, the parabola is “steep”. If |a| < 1, the parabola is “squashed.”
Draw the parabola with the correct open up/down and steepness or squashedness.
Section 8.5: Graphing Quadratic Equations Using Properties
f  x   ax 2  bx  c
f  x   missing steps...
2
  b   4ac  b 2
f  x  a  x     
4a
  2a  
To quickly and ACCURATELY sketch the graph of a quadratic equation of the form y = a(x – h)2 + k:
 b 4ac  b2 
Locate the point   ,
 . This is the vertex.
4a 
 2a
b
Draw a dashed line at x  
. This is the symmetry axis of the parabola.
2a
Calculate and plot the zeros of the function (if there are any).
Compute a couple points on either side of the symmetry axis, then sketch the curve.
Int. Alg. Final Exam Review Sheet
December 2007
Page 22 of 24
Section 9.2: Exponential Functions
The exponential function is a constant number raised to an exponent that is a variable:
y = bx
Facts about the exponential function:
The constant number b is called the base.
The base must be greater than zero (b > 0).
Some common bases are 2, 10, and e.
The base can’t be equal to one (b  1).
The variable x, which is the exponent, can be any real number (i.e., the domain of the function is all real
numbers).
The dependent variable y will take on all values greater than zero (i.e., the range is y > 0).
The graph always passes through the point (0, 1), because any number raised to the power of zero equals
one.
The graph also always contains the points (1, b) and (-1, 1/b).
The graph of the function increases if b > 1, while the graph decreases if 0 < b < 1, as shown below:
This is for b > 1
This is for 0 < b < 1
Definition of e:
n
 1
The number e is defined as the number that the expression 1   approaches as n gets bigger without bound.
 n
Solving exponential equations:
If au = av, then u = v.
Compound Interest Formula:
The amount A in a bank account for a principal P compounded at an annual interest rate r n times per year for t
 r
years is: A  P  1  
 n
nt
Int. Alg. Final Exam Review Sheet
December 2007
Page 23 of 24
Section 9.3: Logarithmic Functions
Big Idea: The logarithmic function “undoes” the exponential function.
x = by is equivalent to y = logb x.
The first formula is in “exponential form,” while the second equation is in “logarithmic form.”
The logarithm is just an exponent. In fact, it helps sometimes to not to say the word logarithm, but instead “the
exponent on b that gives x for an answer.”
Facts about the logarithmic function for b > 1:
The constant number b is called the base.
Some common bases are 2, 10, and e.
a. When the base is 10, we write using shorthand: log10 100 = log 100
b. When the base is e, we write using shorthand: loge 7 = ln 7
The answer generated by the logarithmic function can be any real number (i.e., the range of the function is
all real numbers - < y < ).
The number you plug into the logarithmic function can only be positive (i.e., the domain is x > 0).
The graph always passes through the point (1, 0), because any number raised to the power of zero equals
one.
There are no
There are no y-intercepts.
The logarithm of a fraction between zero and one is negative.
The logarithm of a number greater than one is positive.
The graph of the function increases, as shown below:
Solving logarithmic equations:
Convert the equation to exponential form, and go from there.
Int. Alg. Final Exam Review Sheet
December 2007
Page 24 of 24
Definition:
The loudness L, measured in decibels, of a sound intensity x measured in units of watts per square meter is:
x
L  x   10 log 12
10
Section 9.4: Properties of Logarithms
Exponent Rule
b0  1
Corresponding Logarithm Rule
logb 1  0
b1  b
logb b  1
b b
log b b n  n
n
n
bu b v  bu  v
logb xy  logb x  logb y
bu
 bu  v
v
b
 x
logb    logb x  logb y
 y
log b  x n   n log b x
b 
m n
 b mn
blogb M  M
Change of Base Formula: log a M 
log b M
log b a
Determining the Domain of a variable:
Assume the domain is all real numbers.
Rule out any numbers that would make the denominator equal to zero.
Rule out any numbers that would make the radicand negative for even radicals.
Rule out any numbers that would make the argument of a logarithm negative.
For word problems, rule out any numbers that do not make sense relative to the word problem (like negative
costs or distances or weights, etc.)