College Algebra Exam 1 Material.

College Algebra Exam 1 Material

Special Binomial Products to Memorize
When a binomial is squared, the result is always a “perfect square trinomial” (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 Both of these can be summarized as a formula: Square the first term Multiply 2 times first term times second term Square the last term

Homework Problems Section: R.3 Page: 33 Problems: 49 – 52
MyMathLab Homework Assignment 1

Raising a Binomial to a Power Other Than Two
You should recall that you CAN NOT distribute an exponent over addition or subtraction: We have just seen that: (a + b)2 is NOT equal to a2 + b2 (a + b)2 = (a + b)m is NOT equal to am + bm (a – b)m is NOT equal to am – bm

Patterns in Binomials Raised to Whole Number Powers

Binomial Theorem Patterns observed on previous slide are the basis for the Binomial Theorem that gives a short cut method for raising any binomial to any whole number power:

Using Binomial Theorem
To raise any binomial to nth power: Write expansion of (a + b)n using patterns Use this as a formula for the desired binomial by substituting for “a” and “b” Simplify the result

Homework Problems Section: Page: Problems: Binomial Worksheet
There is no MyMathLab Homework Assignment that corresponds with these problems

Binomial Expansion Worksheet
Use the Binomial Theorem to expand and simplify each of the following: 1. 2. 3. 4. 5. 6. 7. 8.

Equation a statement that two algebraic expressions are equal
Many different types with different names Examples: Many other types of equations – (we will learn names as we go)

Solutions to Equations
Since an equation is a statement, it may be “true” or “false” All values of a variable that make an equation “true” are called “solutions” of the equation Example consider this statement: x + 3 = 7 Is there a value of x that makes this true? “x = 4” is the only solution to this equation

All Types of Equations Classified in One of Three Categories on the Basis of Its Solutions:
Conditional Equation Identity Contradiction

Conditional Equation An equation that is true for only certain values of the variable, but not for all Previous example: x + 3 = 7 Is true only under the “condition” that x = 4, and not all values of x make it true Conditional Equation

Identity An equation that is true for all possible values of the variable Example: 2(x + 3) = 2x + 6 What values of x make this true? All values, because this is just a statement of the distributive property Identity

Contradiction An equation that is false for every possible value of the variable Example: x = x + 5 Why is it impossible for any value of x to make this true? It says that a number is the same as five added to the number – impossible Contradiction

Classifying Equations as Conditional, Identity, Contradiction
Classification normally becomes possible only as an attempt is made to “solve” the equation We will examine classifying equations as we begin to solve them

Equivalent Equations Equations with exactly the same solution sets
Example: Why are each of the following equivalent? 2x – 3 = 7 2x = 10 x = 5 They all have exactly the same solution set: {5}

Finding Solutions to Equations
One way to find the solutions to an equation is to write it as a simpler equivalent equation for which the solution is obvious Example which of these equivalent equations has an obvious solution? 3(x – 5) + 2x = x + 1 x = 4 Both have only the solution “4” obvious for the second equation, but not for first

Procedures that Convert an Equation to an Equivalent Equation
Addition Property of Equality: When the same value is added (or subtracted) on both sides of an equation, the result is an equivalent equation Multiplication Property of Equality When the same non-zero value is multiplied (or divided) on both sides of an equation, the result is an equivalent equation .

Linear Equations in One Variable
Technical Definition: An equation where, after parentheses are gone, every term is a constant or a constant times a variable to the first power. Shorter Definition: A polynomial equation in one variable of degree 1. Examples:

Solving Linear Equations
Get rid of parentheses Get rid of fractions and decimals by multiplying both sides by LCD Collect like terms Decide which side will keep variable terms and get rid of variable terms on other side Get rid of non-variable terms on variable side Divide both sides by the coefficient of variable

Solve the Equation Identify the type of equation:
Get rid of parentheses: Get rid of fractions and decimals by multiplying both sides by LCD:

Example Continued Collect like terms:
Decide which side will keep variable terms and get rid of variable terms on other side: Get rid of non-variable terms on variable side: Divide both sides by coefficient of variable:

Homework Problems Section: 1.1 Page: 90 Problems: Odd: 9 – 27

Contradiction Solve: 2x – (x – 3) = x + 7 What type of equation?
Solve by linear steps: 2x – x + 3 = x + 7 x + 3 = x + 7 What’s wrong? This says that 3 added to a number is the same as 7 added to the number - No matter what type of equation, when we reach an obvious impossibility, the equation is a classified as a “contradiction” and has “no solution”

Identity Solve: x – (2 – 7x) = 2x – 2(1 – 3x) What type of equation?
Solve by linear steps: x – 2 + 7x = 2x – 2 + 6x 8x – 2 = 8x – 2 What looks strange? Both sides are identical In any type of equation when this happens we classify the equation as an “identity” and say that “all real numbers are solutions”

Formulas Any equation in two or more variables can be called a “formula” Familiar Examples: A = LW Area of rectangle P = 2L + 2W Perimeter of rectangle I = PRT Simple Interest D = RT Distance In all these examples each formula has a variable isolated and we say the formula is “solved for that variable”

Formulas Continued Some formulas may not be solved for a particular variable: In cases like this we need to be able to solve for a specified variable (A or B) In other cases, when an equation is solved for one variable, we may need to solve it for another variable P = 2L + 2W is solved for P, but can be solved for L or W

Solving Formulas for a Specified Variable
When solving for a specified variable, pretend all other variables are just numbers (their degree is “zero”) Ask yourself “Considering only the variable I am solving for, what type of equation is this?” If it is “linear” we can solve it using linear techniques already learned, otherwise we will have to use techniques appropriate for the type of equation

Solving a Formula for a Specified Variable
Solve for “t”: Is this equation “linear in t” ? No, it is second degree in “t” – not first degree! Since it’s not linear in t we can’t solve by using linear equation techniques. (Later we can solve this for t, but not with linear techniques.)

Solve for “g”: Is this equation “linear in g” ? Yes, so we can solve like any other linear equation: Get rid of parentheses: (Not necessary for this formula) Get rid of fractions: What is LCD?

Example Two Solve for y:

MyMathLab Homework Assignment 4 MyMathLab Homework Quiz 1 will be due for a grade on the date of our next class meeting!!!

Linear Applications General methods for solving an applied (word) problem: Read problem carefully taking notes, drawing pictures, thinking about formulas that apply, making charts, etc. Read problem again to make a “word list” of everything that is unknown Give a variable name, such as “x” to the “most basic unknown” in the list (the thing that you know the least about) Give all other unknowns in you word list and algebraic expression name that includes the variable, “x” Read the problem one last time to determine what information has been given, or implied by the problem, that has not been used in giving an algebra name to the unknowns and use this information to write an equation about the unknowns Solve the equation and answer the original question

Solve the Application Problem
A 31 inch pipe needs to be cut into three pieces in such a way that the second piece is 5 inches longer than the first piece and the third piece is twice as long as the second piece. How long should the third piece be? Read the problem carefully taking notes, drawing pictures, thinking about formulas that apply, making charts, etc. Perhaps draw a picture of a pipe that is labeled as 31 inches with two cut marks dividing it into 3 pieces labeled first, second and third

Example Continued 2. Read problem again to make a “word list” of everything that is unknown What things are unknown in this problem? The length of all three pieces (even though the problem only asked for the length of the third). Word List of Unknowns: Length of first Length of second Length of third

Example Continued Give a variable name, such as “x” to the “most basic unknown” in the list (the thing that you know the least about) What is the most basic unknown in this list? Length of first piece is most basic, because problem describes the second in terms of the first, and the third in terms of the second, but says nothing about the first Give the name “x” to the length of first

Example Continued Give all other unknowns in the word list an algebraic expression name that includes the variable, “x” The second is 5 inches more than the first. How would the length of the second be named? x + 5 The third is twice as long as the second. How would the length of the third be named? 2(x + 5) Word List of Unknowns: Algebra Names: Length of first x Length of second x + 5 Length of third 2(x + 5)

Example Continued Read the problem one last time to determine what information has been given, or implied by the problem, that has not been used in giving an algebra name to the unknowns and use this information to write an equation about the unknowns What other information is given in the problem that has not been used? Total length of pipe is 31 inches How do we say, by using the algebra names, that the total length of the three pieces is 31? x + (x + 5) + 2(x + 5) = 31

Example Continued 6. Solve the equation and answer the original question This is a linear equation so solve using the appropriate steps: x + (x + 5) + 2(x + 5) = 31 x + x x + 10 = 31 4x + 15 = 31 4x = 16 x = 4 Is this the answer to the original question? No, this is the length of the first piece. How do we find the length of the third piece? The length of the third piece is 2(x + 5): 2(4 + 5) = (2)(9) = 18 inches = length of third piece

Solve this Application Problem
The length of a rectangle is 4 cm more than its width. When the length is decreased by 2 and the width increased by 1, the new rectangle has a perimeter of 18 cm. What were the dimensions of the original rectangle? Draw of picture of two rectangles and label them as “original” and “new”. Also write notes about relationships between the widths and lengths. Write the formula for perimeter of rectangle: P = 2L + 2W

Example Continued Make a word list of all unknowns: length of original
width of original length of new width of new Give the name “x” to the most basic unknown: width of original = x

Example Continued Read problem again to give algebra names to all other unknowns: length of original: width of original: length of new: width of new: Read problem one more time to determine what other information is given that has not been used and use it to write an equation: Perimeter of new rectangle is 18 cm Use formula: P = 2L + 2W 18 = 2(x + 2) + 2(x + 1)

Example Continued Solve equation and answer the original question:
18 = 2(x + 2) + 2(x + 1) 18 = 2x x + 2 18 = 4x + 6 12 = 4x 3 = x length of original: x + 4 = = 7 width of original: x = 3

Solving Application Problems with Formulas & Charts
There are four types of problems that can easily be solved by means of formulas and charts: 1. Motion problems: D = RT (Distance equals Rate multiplied by Time) 2. Work problems: F = RT (Fraction of job completed equals Rate of work multiplied by Time worked) 3. Mixture problems: IA = (IP)(SA) (Ingredient Amount equals Ingredient Percent multiplied by Substance Amount) 4. Simple Interest problems: I = PRT (Interest equals Principle multiplied by Rate (%) multiplied by Time (in years or part of a year)

Formula: D = RT Given R and T this formula can be used as is to find D
Example: If R = 5 mph and T = 3 hr, what is D? D = (5)(3) = 15 miles Given any two of the three variables in the formula, the other one can always be found: How would you find T if D and R were given? T = D / R How would you find R if D and T were given? R = D / T

Solving Motion Problems with Formula and Chart
Immediately write formula: D = RT as a heading on a chart Make and label one line in the chart for everything “moving” Write “x” in the box for the most basic unknown Fill out the remainder of that column based on information given in the problem Fill out one more column based on the most specific information given in the problem Fill out the final column by using the formula at the top Read problem one more time and write an equation about D, R, or T, based on other information given in the problem that was not used in completing the chart Solve the equation and answer the original question

Solving a Motion Problem
Lisa and Dionne are traveling to a meeting. It takes Lisa 2 hours to reach the meeting site and 2.5 hours for Dionne, since she lives 40 miles farther away. Dionne travels 5 mph faster than Lisa. Find their average speeds. Immediately write formula: D = RT as a heading on a chart: Make and label one line in the chart for everything “moving”: D = R T Lisa Dionne

Example Continued 3. Write “x” in the box for the most basic unknown:
What is it? Lisa’s speed, because if we find it, we can calculate Dionne’s speed by adding 5 mph D = R T Lisa Dionne

Example Continued 4. Fill out the remainder of that column based on information given in the problem What is the other item in that column? Dionne’s speed. How would we describe it with an algebra description? x + 5 D = R T Lisa Dionne

Example Continued 5. Fill out one more column based on the most specific information given in the problem Is the most specific information given about how far each one traveled, or about how much time each one took? The time each took: Lisa’s time was 2 hours, and Dionne’s 2.5 hours D = R T Lisa Dionne

Example Continued 6. Fill out the final column by using the formula at the top Formula says that D = RT, so the final column is: D = R T Lisa Dionne 7. Read problem one more time and write an equation about D, R, or T, based on other information given in the problem that was not used in completing the chart What other information is given in the problem that was not used in making the chart? Dionne’s distance was 40 miles more than Lisa’s distance: 2.5(x + 5) = 2x + 40

Example Continued 2.5(x + 5) = 2x + 40 2.5x + 12.5 = 2x + 40
x = 55 mph (Lisa’s speed) x + 5 = 60 mph (Dionne’s speed)

Homework Problems Section: 1.2 Page: 102 Problems: 19 – 24, 27 – 28

Solving Work Problems With Formula: F = RT
This formula says that the fraction of a job completed equals the rate of work (portion of the job done per unit time) multiplied by the amount of time worked. Example: If the rate tells us that 1/8 of the job is being done per hour and work is done for 3 hours, then R = 1/8 and T = 3. What is the fraction of the job completed? F = 3(1/8) = 3/8 (3/8 of the job is completed) What fraction of the job remains? 5/8 Why? Because anytime a whole job is done, the fraction completed will be “1”

Other Forms of Formula: F = RT
Given F and R, find T: T = F / R Given F and T, find R: R = F / T General note about formula F = RT when one thing works alone to complete a job, it’s fraction done is “1”, but when two or more things work to finish a job, then the sum of their fractions must be “1” Example: If your friend and you work together to finish a job and you do 2/3 of the job, then your friend must do: 1/3 of the job

Solving Work Problems with Formula and Chart
1. Immediately write formula: F = RT as a heading on two charts, one labeled “alone” and the other labeled “together” 2. Make and label one line in both charts for everything “working” 3. Write “x” in the box for the most basic unknown 4. Fill out the remainder of that column based on information given in the problem 5. As you fill out other boxes in both charts based on information given always: Always put F = 1 in all boxes in the column in the alone chart Always put the same value for R in both the “alone” and “together” charts Use the formula to fill out the final box in a row when other boxes are known 6. Always write an equation based on the fact that when things are working together, the sum of their fractions, F, must equal 1 7. Solve the equation and answer the original question

Solve the Work Problem:
If A, working alone, takes 5 hours to complete a job, and B, working alone, takes 9 hours to complete the same job, how long should it take to do the job if they work together? Immediately write formula: F = RT as a heading on two charts, one labeled “alone” and the other labeled “together” Alone Together F= R T F= R T

Example Continued 2. Make and label one line in both charts for everything “working” Alone Together F= R T F= R T A B 3. Put “x” in box for most basic unknown. What is it? Time working together (same for both)

Example Continued 4. Fill out the remainder of that column based on information given in the problem Already done in this example since that together A and B worked the same time Alone Together F= R T F= R T A B 5. Fill out other boxes in both charts based on information given Always put F = 1 in all boxes in the column in the alone chart Use the formula to fill out the final box in a row when other boxes are known Always put the same value for R in both the “alone” and “together” charts

Example Continued 6. Always write an equation based on the fact that when things are working together, the sum of their fractions, F, must equal 1 Looking at the charts below, how would you write an equation that says “the sum of the fractions of their work is one”? Alone Together F= R T F= R T A B

Example Continued 7. Solve the equation and answer question:

Solve this problem: When A and B work together they can complete a job in 7 hours, but A is twice as fast as B. How long would it take B to do the job alone? Alone Together F= R T F= R T A B

Example Continued

Homework Problems Section: 1.2 Page: 103 Problems: 29 – 34

Solving Mixture Problems With Formula: IA = (IP) (SA )
This formula tells us that the amount of an ingredient, IA, is equal to the percent of the ingredient, IP, multiplied by the amount of the substance that includes the ingredient, SA Example: If a 20 gallon tank contains 15% gasoline, what is the amount of gasoline in the tank? IA = (IP)(SA) IA = (15%)(20) = (.15)(20) = 3 gallons Note: Like all other formulas, this formula can be solved for any of the variables as necessary For mixture problems it is important to realize that “mix” means “add”

Solving Mixture Problems with Formula and Chart
1. Immediately write formula: IA = (IP)(SA) as a heading on a chart 2. Make and label one line in the chart for everything “being mixed”, and another line for the “mixture” 3. Write “x” in the box for the most basic unknown 4. Fill out the remainder of that column based on information given in the problem 5. As you fill out other boxes in the chart based on information given always use the formula to fill out the final box in a row when other boxes are known 6. Always write an equation based on the fact that the sum of the individual ingredient amounts equals the amount of the ingredient in the mixture 7. Solve the equation and answer the question

Solve the mixture problem:
A chemist needs a mixture that is 30% alcohol, but has one bottle labeled 10% alcohol and another labeled 70%. How much 10% alcohol should be mixed with 5 liters of 70% to get a mixture that is 30% alcohol? IA = (IP) (SA) 10%A 70%A 30%A

Example Continued

Solving Simple Interest Problems With Formula: I = PRT
Formula means that the interest earned from an investment is equal to the amount of the investment, P, multiplied by the interest rate, R (percent), multiplied by the time, T, measured in years or parts of years Example: Calculate the interest earned on $2,000 invested at 5% interest for 3 years and 6 months: P = $2,000, R = 5%, T = 3.5 years I = ($2,000)(.05)(3.5) = $350

Solving Simple Interest Problems
1. Immediately write formula: I = PRT as a heading on a chart 2. Make and label one line in the chart for each investment 3. Write “x” in the box for the most basic unknown 4. Fill out the remainder of that column based on information given in the problem 5. As you fill out other boxes the chart based on information given always use the formula to fill out the final box in a row when other boxes are known 6. Always write an equation based on the fact that the “total interest” is the sum of the individual interests 7. Solve the equation and answer the question

Solve the Simple Interest Problem
A man invests some money at 6% interest and half that amount at 4% interest. If his annual income from the two investments is $400, how much did he invest at each rate? I = P R T 6%Inv 4%Inv

Example Continued Solve the equation:
The amount invested at 6% was $5,000 and the amount invested at 4% was half that amount, $2,500.

Imaginary Unit, i Introduction:
In the real number system an equation such as: x2 = - 1 has no solution. Why? The square of every real number is either 0 or positive. To solve this equation, a new kind of number, called an “imaginary unit”, i, has been defined as: Note: i is not a real number This definition is applied in the following way:

Complex Number A “complex number” is any number that can be written in the form: a + bi where “a” and “b” are real numbers and “i” is the imaginary unit (This is called “standard form” of a complex number) Based on this definition, why is every real number also a complex number? Every real number “a” can be written as: “a + 0i” Write in the standard form of a complex number: - 5 = i

Complex Number Continued
Are there some complex numbers that are not real? Yes, any number of the form “a + bi” where “b” is not zero. 7 + 3i is a complex number that is not real Every complex number that contains “i” is called “a non-real complex number” 2 - 5i is an “non-real” complex number 4 is a “real” complex number Every complex number that contains “i”, but is missing “a” is called “pure imaginary” 8i is a “pure imaginary” non- real complex number

Square Roots of Negative Radicands: Imaginary Numbers
Definition: Note: A square root of a negative radicand must immediately be changed to an imaginary number before doing any other operations Examples:

Homework Problems Section: 1.3 Page: 113
Problems: 1 – 16, Odd: 25 – 41 MyMathLab Homework Assignment 10

Addition and Subtraction of Complex Numbers
Pretend that “i” is a variable and that a complex number is a binomial Add and subtract as you would binomials Example: (2 + i) – (-5 + 7i) + (4 – 3i) 2 + i + 5 – 7i + 4 – 3i 11 – 9i

Multiplication of Complex Numbers
Pretend that “i” is a variable and that a complex number is a binomial Multiply as you would binomials Simplify by changing “i2” to “-1” and combining like terms Example: (-4 + 3i)(5 – i) i + 15i - 3i2 i + 15i + 3 i

Division of Complex Numbers
Write division problem in fraction form Multiply fraction by a special “1” where “1” is the conjugate of the denominator over itself Simplify and write answer in standard form: a + bi

Example:

Simplifying Integer Powers of “i”
Every integer power of “i” simplifies to one of four possible values: i, -1, -i, or 1 When “n” is an integer:

Simplifying “in” for Even Positive Integer “n”
Use the following procedure to simplify any even positive integer power of “i”, in If “n” is even write in = (i2)m for some integer “m” Change i2 to -1 and simplify Example:

Simplifying “in” for Odd Positive Integer “n”
Use the following procedure to simplify any odd positive integer power of “i”, in Write in = i(i)n-1 (Note: n – 1 will be even) Finish simplifying by using rules for simplifying even powers of “i” Example:

Simplifying “in” for Negative Integer “n”
First use definition of negative exponent Simplify according to rules for even and odd exponents as already explained If necessary perform any division (never leave an “i” in a denominator) Example:

Quadratic Equations Technical Definition: any equation in one variable that can be written in the form: ax2 + bx + c = 0 where “a”, “b”, and “c” are real and a ≠ 0 (This form is called the “standard form”) Practical Definition: A polynomial equation of degree 2 Examples: 5x2 + 7 = – 4x 9x2 = 4 2x(x – 3) = x – 1

Solving Quadratic Equations
There are four possible methods: Square root method Zero factor method Completing the square method Quadratic formula method The last two methods will solve any quadratic equation The first two work only in special situations

Square Root Method Can be used only when:
the first degree term is missing, or when the variable is found only within parentheses with an exponent of two on the parentheses Which of these can be solved by this method? .

Steps in Applying Square Root Method
Write equivalent equations to isolate either the variable squared, or the parenthesis squared Square root both sides, being sure to put a “plus and minus sign” on any real number that is square rooted (This step reduces the equation to two linear equations) Solve the linear equations

Example of Solving by the Square Root Method

Second Example of Solving by the Square Root Method

Zero Factor Method Put equation in standard form (one side zero other side in descending powers) Factor non-zero side (If it won’t factor this method won’t work!) Use zero factor property that says: ab = 0 if and only if a=0 or b=0 Set each factor equal to zero Solve resulting equations

Example Consider the following equation:
Is this equation linear or quadratic? Quadratic! (2nd degree) Could it be solved by square root method? No (first degree term is not missing and variable is not entirely inside parenthesis with a square) What other method might be used to solve it? Maybe zero factor method will work.

Solving by Zero Factor Method
Put in standard form: Factor non-zero side: Apply zero factor principle: Solve the equations:

Completing the Square Method
The third possible method of solving quadratic equations will solve every quadratic equation in practice this method is used only when directions dictate This method is essential in developing the fourth method: Quadratic Formula

Completing the Square Method
Isolate variables on one side of equal sign and number on the other side Divide both sides of equation by coefficient of second degree term (unless it is already one) Find “n” by: Add “n” to both sides of the equation (As a result of doing this, the trinomial on the left will always factor as the square of a binomial) Factor the side of the equation containing the trinomial Solve the resulting equation by means of the “square root method”

Example Consider the following equation:
Is this equation linear or quadratic? Quadratic! (2nd degree) Could it be solved by square root method? No (first degree term is not missing and variable is not entirely inside parenthesis with a square) Could it be solved by zero factor method? No (non-zero side won’t factor) What method will work?

Solve by Completing the Square Method
Isolate variables on one side: Divide both sides by coefficient of second degree term: Calculate “n” by taking ½ times coefficient of first degree term and squaring that: Add “n” on both sides of equation:

Example Continued Factor trinomial as a square of a binomial:
Solve by square root method:

Quadratic Formula Development
Solve standard form of quadratic equation by completing the square: .

Steps in Using the Quadratic Formula to Solve an Equation
Write the quadratic equation in standard form: Determine the values of “a”, “b”, and “c” Plug those values into the quadratic formula: Simplify

Solve by Using the Quadratic Formula
Write in standard form: Find “a”, “b”, and “c”: a = 3, b = -4, and c = -4 Plug these into quadratic formula: Simplify:

Solving Formulas Using the Quadratic Formula
If the formula to be solved is “quadratic” in the variable for which you wish to solve: Write the formula in standard form for that variable Identify “a”, “b”, and “c” Plug into quadratic formula Simplify

Example Solve for t: Is this equation linear or quadratic for t?
Put in standard form for t: Identify a, b and c: Plug into quadratic formula & simplify: .

“Discriminate” Determines the Number and Type of Solutions of a Quadratic Equation: ax2+bx +c = 0
The “discriminate” of a quadratic equation is the radicand of the quadratic formula: disc = b2 – 4ac If disc = 0, then whole quadratic formula becomes x = -b/2a, so the equation has one rational solution If disc is negative, then solution involves a square root of a negative radicand with a ± in front, so there will be two non-real complex solutions If disc is positive perfect square, then radical will disappear, but there is still a ±, so there will be two rational solutions If disc is positive but not a perfect square, then radical will remain with a ±, so there will be two irrational solutions

Examples of Using Discriminate to Determine Nature of Solutions
Disc = b2 – 4ac 5x2 – 3x + 2 = 0 Disc = (-3)2 – 4(5)(2) = 9 – 40 = - 31 Two non-real complex solutions 3x2 – 4x – 2 = 0 16 – 4(3)(-2) = = 40 (positive, but not perfect square) Two irrational solutions

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