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College Algebra

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 = a 2 + 2ab + b 2 (a – b) 2 = a 2 – 2ab + b 2 Both of these can be summarized as a formula: Square the first term

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College Algebra

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  1. College Algebra Exam 1 Material

  2. 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 firstterm times secondterm • Square the last term

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

  4. 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)2is NOT equal toa2 + b2 (a + b)2 = (a + b)mis NOT equal toam + bm (a – b)mis NOT equal to am – bm

  5. Patterns in Binomials Raised to Whole Number Powers

  6. Patterns in Binomials Raised to Whole Number Powers

  7. Binomial Theorem • Patterns observed on previous slide are the basis for the Binomial Theorem that gives a short cut method for raising any binomialto any whole number power:

  8. 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

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

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

  11. 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)

  12. 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

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

  14. Conditional Equation • An equation that is truefor onlycertain 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

  15. 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

  16. 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

  17. 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

  18. 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}

  19. Finding Solutions to Equations • One way to find the solutions to an equation is to write it as a simplerequivalent 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

  20. 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 .

  21. 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:

  22. 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

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

  24. 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:

  25. Homework Problems • Section: 1.1 • Page: 90 • Problems: Odd: 9 – 27 • MyMathLab Homework Assignment 2

  26. 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”

  27. 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”

  28. Homework Problems • Section: 1.1 • Page: 91 • Problems: Odd: 29 – 35 • MyMathLab Homework Assignment 3

  29. 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”

  30. 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

  31. Solving Formulasfor 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

  32. 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.)

  33. Solving a Formula for a Specified Variable 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?

  34. Solving a Formula for a Specified Variable

  35. Example Two Solve for y:

  36. Homework Problems • Section: 1.1 • Page: 91 • Problems: Odd: 39 – 57 • MyMathLab Homework Assignment 4 • MyMathLab Homework Quiz 1 will be due for a grade on the date of our next class meeting!!!

  37. 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 problemagain 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

  38. 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

  39. 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

  40. 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

  41. 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)

  42. 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

  43. 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 + 5 + 2x + 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

  44. 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

  45. 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

  46. 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)

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

  48. Homework Problems • Section: 1.2 • Page: 101 • Problems: Odd: 9 – 17 • MyMathLab Homework Assignment 5

  49. 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)

  50. 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

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