Quadratic Equations and Inequalities

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1 MA 134 Lecture Notes August 20, 2012

2 Introduction The purpose of this lecture is to...

3 Introduction The purpose of this lecture is to... Learn about different types of equations

4 Introduction The purpose of this lecture is to... Learn about different types of equations Discuss methods for solving quadratic equations

5 Introduction The purpose of this lecture is to... Learn about different types of equations Discuss methods for solving quadratic equations Introduce polynomial and rational inequalities

6 Introduction The purpose of this lecture is to... Learn about different types of equations Discuss methods for solving quadratic equations Introduce polynomial and rational inequalities Learn techniques for analyzing polynomial and rational inequalities

7 Different types of equations What is an equation?

8 Different types of equations What is an equation? - A mathematical statement that sets two expressions equal to each other. Example 1: x + 2 = 4

9 Different types of equations What is an equation? - A mathematical statement that sets two expressions equal to each other. Example 1: x + 2 = 4 What type of equation is Example 1?

10 Different types of equations What is an equation? - A mathematical statement that sets two expressions equal to each other. Example 1: x + 2 = 4 What type of equation is Example 1? - The equation in Example 1 is called a linear equation.

11 Different types of equations What is an equation? - A mathematical statement that sets two expressions equal to each other. Example 1: x + 2 = 4 What type of equation is Example 1? - The equation in Example 1 is called a linear equation. - The degree or highest power of x is 1.

12 Different types of equations What is an equation? - A mathematical statement that sets two expressions equal to each other. Example 1: x + 2 = 4 What type of equation is Example 1? - The equation in Example 1 is called a linear equation. - The degree or highest power of x is 1. What is an equation that has variable x with degree 2?

13 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0.

14 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0. Note: If a = 0 then the above equation would be a linear equation.

15 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0. Note: If a = 0 then the above equation would be a linear equation. Example 2: x 2 4x + 4 = 0

16 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0. Note: If a = 0 then the above equation would be a linear equation. Example 2: x 2 4x + 4 = 0 3x 2 + x 2 = 0

17 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0. Note: If a = 0 then the above equation would be a linear equation. Example 2: x 2 4x + 4 = 0 3x 2 + x 2 = x x = 0

18 Quadratic Equations Quadratic Equation A quadratic equation is a single variable equation with degree two that is generally written in the form ax 2 + bx + c = 0, where a, b, and c are constants such that a 0. Note: If a = 0 then the above equation would be a linear equation. Example 2: x 2 4x + 4 = 0 3x 2 + x 2 = x x = 0 How can we solve the equations in Example 2?

19 Methods for solving quadratic equations Factoring

20 Methods for solving quadratic equations Factoring - Greatest Common Factor

21 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0

22 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0

23 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error

24 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0

25 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6.

26 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0

27 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping)

28 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0

29 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = 144

30 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7

31 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7 Rewrite the middle term 6x x 9x 24 = 0

32 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7 Rewrite the middle term 6x x 9x 24 = 0 Factor out the GCF from each group 2x(3x + 8) 3(3x + 8) = 0 (2x 3)(3x + 8) = 0

33 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7 Rewrite the middle term 6x x 9x 24 = 0 Factor out the GCF from each group 2x(3x + 8) 3(3x + 8) = 0 (2x 3)(3x + 8) = 0 ALL methods depend upon Zero Property

34 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7 Rewrite the middle term 6x x 9x 24 = 0 Factor out the GCF from each group 2x(3x + 8) 3(3x + 8) = 0 (2x 3)(3x + 8) = 0 ALL methods depend upon Zero Property

35 Methods for solving quadratic equations Factoring - Greatest Common Factor - 2x 2 + 4x = 0 2x(x + 2) = 0 - Trial and Error x 2 5x + 6 = 0 Consider the possible factors for 6. (x 3) = 0 - AC Method (Factor By Grouping) 6x 2 + 7x 24 = 0 a c = 6 24 = = 144,16 + ( 9) = 7 Rewrite the middle term 6x x 9x 24 = 0 Factor out the GCF from each group 2x(3x + 8) 3(3x + 8) = 0 (2x 3)(3x + 8) = 0 ALL methods depend upon Zero Property a b = 0 either a = 0 or b = 0

36 Other Methods for Solving Quadratics

37 Other Methods for Solving Quadratics Square root property

38 Other Methods for Solving Quadratics Square root property x 2 = 9 x = ± 9 = ±3

39 Other Methods for Solving Quadratics Square root property x 2 = 9 x = ± 9 = ±3 Quadratic Formula

40 Other Methods for Solving Quadratics Square root property x 2 = 9 x = ± 9 = ±3 Quadratic Formula b ± b 2 4ac 2a Completing the square

41 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows:

42 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation.

43 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c.

44 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a.

45 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1.

46 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1. - This step should yield the form x 2 + b a x = c a.

47 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1. - This step should yield the form x 2 + b a x = c a. Step:3 Take 1 2 of b a, square, then add to both sides of the equation.

48 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1. - This step should yield the form x 2 + b a x = c a. Step:3 Take 1 2 of b a, square, then add to both sides of the equation. b - This step should yield, 2a,

49 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1. - This step should yield the form x 2 + b a x = c a. Step:3 Take 1 2 of b a, square, then add to both sides of the equation. b - This step should yield, 2a, ( b 2, 2a) and then

50 Completing the square Algorithm for completing the square Given the equation ax 2 + bx + c = 0 then the algorithm for completing the square is as follows: Step 1: Subtract c from both sides of the equation. - This step should yield the form ax 2 + bx = c. Step 2: Divide the equation by a. - Note that this step is not necessary if a = 1. - This step should yield the form x 2 + b a x = c a. Step:3 Take 1 2 of b a, square, then add to both sides of the equation. b - This step should yield, 2a, ( b 2, 2a) and then x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a).

51 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2a) 2.

52 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square.

53 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a).

54 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + b 2a) 2 = c a + ( b 2a) 2.

55 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2 2a). Step 5: Isolate our variable x.

56 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2a) 2. Step 5: Isolate our variable x. Take the square root of both sides and simplify.

57 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2 2a). Step 5: Isolate our variable x. Take the square root of both sides and simplify. x + b 2a = c a + ( b 2a ) 2

58 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2 2a). Step 5: Isolate our variable x. Take the square root of both sides and simplify. x + b 2a = c a + ( b 2a ) 2 Result: x = b 2a ± c a + ( ) b 2 2a

59 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2 2a). Step 5: Isolate our variable x. Take the square root of both sides and simplify. x + b 2a = c a + ( b 2a ) 2 Result: x = b 2a ± c a + ( ) b 2 2a What would we yield for x if we simplified?

60 Completing the square Algorithm for completing the square cont. From Step 3 we have x 2 + b a x + ( ) b 2 2a = c a + ( b 2 2a). Step 4: From the above equation, rewrite the left side as a perfect square. Then x 2 + b a x + ( ) b 2 ( 2a can be written as x + b 2 2a). Thus, ( x + 2a) b 2 = c a + ( b 2 2a). Step 5: Isolate our variable x. Take the square root of both sides and simplify. x + b 2a = c a + ( b 2a ) 2 Result: x = b 2a ± c a + ( ) b 2 2a What would we yield for x if we simplified? Quadratic Formula x = b ± b 2 4ac 2a

61 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x.

62 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over.

63 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2.

64 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3

65 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3

66 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of

67 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square 3 2

68 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square 3 2

69 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square

70 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square x 2 + 3x = Add 9 4 both sides.

71 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square x 2 + 3x = Add 9 4 both sides. ( ) x = 15 4 Write left side as perfect square.

72 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square x 2 + 3x = Add 9 4 both sides. ( ) x = 15 2 ( x Write left side as perfect square. ) = ± 15 4 Take the square root of both sides.

73 Completing the square Example 3: Solve 2x 2 + 6x 3 = 0 Solve for x. 2x 2 + 6x = 3 Move 3 over. 2 2 x x = 3 2 Divide by 2. x 2 + 3x = 3 2 Take 1 2 of 3 3 ( 2. 3 ) 2 2 Square x 2 + 3x = Add 9 4 both sides. ( ) x = 15 2 ( x Write left side as perfect square. ) = ± 15 4 Take the square root of both sides. x = 3 2 ± 15 2 Move 3 2 over. Solve 3x 2 5x 2 = 0 and 2x 2 + 3x + 5 = 0

74 Inequalities What is an inequality?

75 Inequalities What is an inequality? - It is the relation between two quantities when they are different, not equal.

76 Inequalities What is an inequality? - It is the relation between two quantities when they are different, not equal. Two types of inequalities:

77 Inequalities What is an inequality? - It is the relation between two quantities when they are different, not equal. Two types of inequalities: Polynomial Inequalities

78 Inequalities What is an inequality? - It is the relation between two quantities when they are different, not equal. Two types of inequalities: Polynomial Inequalities Rational Inequalities

79 Polynomials What is a polynomial?

80 Polynomials What is a polynomial? - Where have we encountered polynomials?

81 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers.

82 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers. - Note that a polynomial function is of the form f (x) = a n x n + a n 1 x n a 1 x + a 0.

83 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers. - Note that a polynomial function is of the form f (x) = a n x n + a n 1 x n a 1 x + a 0. We say that the degree of a polynomial function is the highest power n such that the coefficient a n is not 0.

84 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers. - Note that a polynomial function is of the form f (x) = a n x n + a n 1 x n a 1 x + a 0. We say that the degree of a polynomial function is the highest power n such that the coefficient a n is not 0. Example 4: x 2 + x 1 x + 1

85 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers. - Note that a polynomial function is of the form f (x) = a n x n + a n 1 x n a 1 x + a 0. We say that the degree of a polynomial function is the highest power n such that the coefficient a n is not 0. Example 4: x 2 + x 1 x Functions: f (x) = 3x 2 5x 2 f (x) = 2x 2 6x 3

86 Polynomials What is a polynomial? - Where have we encountered polynomials? A polynomial is an expression of the form a n x n + a n 1 x n a 1 x + a 0, where n is a non-negative integer and the coefficients a n, a n 1,...a 0 are real numbers. - Note that a polynomial function is of the form f (x) = a n x n + a n 1 x n a 1 x + a 0. We say that the degree of a polynomial function is the highest power n such that the coefficient a n is not 0. Example 4: x 2 + x 1 x Functions: f (x) = 3x 2 5x 2 f (x) = 2x 2 6x 3 - Inequalities: x 2 1 > 0 x 2 4x + 4 0

87 Geometric approach to analyzing polynomial inequalities Consider a polynomial inequality of the general form ax 2 + bx + c 0. Solving the inequality ax 2 + bx + c 0.

88 Geometric approach to analyzing polynomial inequalities Consider a polynomial inequality of the general form ax 2 + bx + c 0. Solving the inequality ax 2 + bx + c 0. Step 1: Use a method to factor ax 2 + bx + c 0.

89 Geometric approach to analyzing polynomial inequalities Consider a polynomial inequality of the general form ax 2 + bx + c 0. Solving the inequality ax 2 + bx + c 0. Step 1: Use a method to factor ax 2 + bx + c 0. Step 2: Label the roots, solutions for x, on a line graph.

90 Geometric approach to analyzing polynomial inequalities Consider a polynomial inequality of the general form ax 2 + bx + c 0. Solving the inequality ax 2 + bx + c 0. Step 1: Use a method to factor ax 2 + bx + c 0. Step 2: Label the roots, solutions for x, on a line graph. Step 3: Use test points and sign math to determine what values make the inequality true.

91 Geometric approach to analyzing polynomial inequalities Consider a polynomial inequality of the general form ax 2 + bx + c 0. Solving the inequality ax 2 + bx + c 0. Step 1: Use a method to factor ax 2 + bx + c 0. Step 2: Label the roots, solutions for x, on a line graph. Step 3: Use test points and sign math to determine what values make the inequality true. Step 4: Write the solution in the specified format (typically interval notation).

92 Analyzing Polynomial Inequalities Consider the polynomial inequality x 2 4 > 0.

93 Analyzing Polynomial Inequalities Consider the polynomial inequality x 2 4 > 0. Factor x 2 4 > 0.

94 Analyzing Polynomial Inequalities Consider the polynomial inequality x 2 4 > 0. Factor x 2 4 > 0. (x + 2) > 0

95 Analyzing Polynomial Inequalities Consider the polynomial inequality x 2 4 > 0. Factor x 2 4 > 0. (x + 2) > 0 Draw a number line

96 Analyzing Polynomial Inequalities Consider the polynomial inequality x 2 4 > 0. Factor x 2 4 > 0. (x + 2) > 0 Draw a number line

97 Polynomial Inequalities

98 Polynomial Inequalities (x + 2) = 0,

99 Polynomial Inequalities (x + 2) = 0, x = ±2.

100 Polynomial Inequalities (x + 2) = 0, x = ±2. Plot the roots, x = ±2, on the line.

101 Polynomial Inequalities (x + 2) = 0, x = ±2. Plot the roots, x = ±2, on the line. ) (

102 Polynomial Inequalities (x + 2) = 0, x = ±2. Plot the roots, x = ±2, on the line. ) ( Note that we solved (x + 2) = 0 NOT (x + 2) > 0 to find the roots!

103 Polynomial Inequalities (x + 2) = 0, x = ±2. Plot the roots, x = ±2, on the line. ) ( Note that we solved (x + 2) = 0 NOT (x + 2) > 0 to find the roots! We use open parentheses on the graph to represent strictly <, >.

104 Polynomial Inequalities Use test points and sign math

105 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign.

106 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) =

107 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) =

108 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) =

109 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = =

110 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = +

111 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =

112 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+

113 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+

114 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ =

115 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ =

116 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =

117 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+

118 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+

119 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+ + =

120 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+ + = +

121 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+ + = + Draw the signs above the corresponding test points.

122 Polynomial Inequalities Use test points and sign math Use 3, 0, and 3 then plug into x in (x + 2), noting the sign. (x + 2) = ( 3 + 2) ( 3 2) = = + (x + 2) = (0 + 2) (0 2) =+ = (x + 2) = (3 + 2) (3 2) =+ + = + Draw the signs above the corresponding test points. + ) - (

123 Polynomial Inequalities Recall our original inequality x 2 4 > 0.

124 Polynomial Inequalities Recall our original inequality x 2 4 > 0. Shade the region that makes the inequality true.

125 Polynomial Inequalities Recall our original inequality x 2 4 > 0. Shade the region that makes the inequality true ) (

126 Polynomial Inequalities Recall our original inequality x 2 4 > 0. Shade the region that makes the inequality true ) ( Our inequality is true when x < 2 and x > 2

127 Polynomial Inequalities Recall our original inequality x 2 4 > 0. Shade the region that makes the inequality true ) ( Our inequality is true when x < 2 and x > 2 Interval Notation: (, 2) (2, )

128 Polynomial Inequalities Recall our original inequality x 2 4 > 0. Shade the region that makes the inequality true ) ( Our inequality is true when x < 2 and x > 2 Interval Notation: (, 2) (2, ) Examples: x ( x 2 4 ) 0 x 2 3x 4 < 0

129 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials.

130 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials. Because the rational inequality has a denominator, we must be concerned with the domain of x.

131 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials. Because the rational inequality has a denominator, we must be concerned with the domain of x. - What do we mean by domain?

132 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials. Because the rational inequality has a denominator, we must be concerned with the domain of x. - What do we mean by domain? - So, the denominator q (x) CANNOT be equal to 0.

133 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials. Because the rational inequality has a denominator, we must be concerned with the domain of x. - What do we mean by domain? - So, the denominator q (x) CANNOT be equal to 0. How does solving a rational inequality differ from a polynomial inequality?

134 Rational Inequalities In general, a rational inequality is of the form p(x) q(x) 0, where p(x), q(x) are polynomials. Because the rational inequality has a denominator, we must be concerned with the domain of x. - What do we mean by domain? - So, the denominator q (x) CANNOT be equal to 0. How does solving a rational inequality differ from a polynomial inequality? - q (x) will have a value for x that is UNDEFINED

135 Analyzing Rational Inequalities Consider a rational inequality of the general form p(x) q(x) 0. Solving the inequality p(x) q(x) 0.

136 Analyzing Rational Inequalities Consider a rational inequality of the general form p(x) q(x) 0. Solving the inequality p(x) q(x) 0. Step 1: Use a method to factor p(x) 0, both top and bottom. q(x)

137 Analyzing Rational Inequalities Consider a rational inequality of the general form p(x) q(x) 0. Solving the inequality p(x) q(x) 0. Step 1: Use a method to factor p(x) 0, both top and bottom. q(x) Step 2: Label the roots and undefined places on a line graph.

138 Analyzing Rational Inequalities Consider a rational inequality of the general form p(x) q(x) 0. Solving the inequality p(x) q(x) 0. Step 1: Use a method to factor p(x) 0, both top and bottom. q(x) Step 2: Label the roots and undefined places on a line graph. Step 3: Use test points and sign math to determine what values make the inequality true.

139 Analyzing Rational Inequalities Consider a rational inequality of the general form p(x) q(x) 0. Solving the inequality p(x) q(x) 0. Step 1: Use a method to factor p(x) 0, both top and bottom. q(x) Step 2: Label the roots and undefined places on a line graph. Step 3: Use test points and sign math to determine what values make the inequality true. Step 4: Write the solution in the specified format (typically interval notation).

140 Solving Rational Inequalities

141 Solving Rational Inequalities x 2 Examples: x 2 2x 3 0, x 2 3x 10 1 x x 2 the inequality x 2 2x Consider

142 Solving Rational Inequalities x 2 Examples: x 2 2x 3 0, x 2 3x 10 1 x x 2 the inequality x 2 2x 3 0. x 2 Factor x 2 2x Consider

143 Solving Rational Inequalities x 2 Examples: x 2 2x 3 0, x 2 3x 10 1 x x 2 the inequality x 2 2x 3 0. x 2 Factor x 2 2x 3 0. (x 3) (x + 1) 0 0 Consider

144 Solving Rational Inequalities x 2 Examples: x 2 2x 3 0, x 2 3x 10 1 x x 2 the inequality x 2 2x 3 0. x 2 Factor x 2 2x 3 0. (x 3) (x + 1) 0 Draw a number line 0 Consider

145 Solving Rational Inequalities x 2 Examples: x 2 2x 3 0, x 2 3x 10 1 x x 2 the inequality x 2 2x 3 0. x 2 Factor x 2 2x 3 0. (x 3) (x + 1) 0 Draw a number line 0 Consider

146 Rational Inequalities

147 Rational Inequalities (x 3) (x + 1) 0,

148 Rational Inequalities (x 3) (x + 1) 0, x = 2, x 1, 3.

149 Rational Inequalities (x 3) (x + 1) 0, x = 2, x 1, 3. Plot the roots and values that are not defined, x = 2,x 1, 3, on the line.

150 Rational Inequalities (x 3) (x + 1) 0, x = 2, x 1, 3. Plot the roots and values that are not defined, x = 2,x 1, 3, on the line. nd nd

151 Solving Rational Inequalities Pick test points and use sign math in our factored inequality

152 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0

153 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only

154 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) =

155 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) =

156 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) =

157 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) =

158 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = =

159 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = =

160 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) =

161 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) =

162 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) =

163 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) =

164 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + =

165 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = +

166 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) =

167 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) =

168 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = +

169 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = +

170 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + =

171 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + =

172 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + =

173 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + = (x 3) (x + 1) = (4 2) (4 3) (4 + 1) =

174 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + = (x 3) (x + 1) = (4 2) (4 3) (4 + 1) = +

175 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + = (x 3) (x + 1) = (4 2) (4 3) (4 + 1) = + +

176 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + = (x 3) (x + 1) = (4 2) (4 3) (4 + 1) = =

177 Solving Rational Inequalities Pick test points and use sign math in our factored inequality (x 3) (x + 1) 0 Choose the test points 2, 0, 5 2, 4 and plug in for x noting the sign only (x 3) (x + 1) = ( 2 2) ( 2 3) ( 2 + 1) = = (x 3) (x + 1) = (0 2) (0 3) (0 + 1) = + = + (x 3) (x + 1) = ( 5 ( 2 2) 5 2 3) ( ) = + + = (x 3) (x + 1) = (4 2) (4 3) (4 + 1) = = +

178 Rational Inequalities

179 Rational Inequalities Label the signs on the line graph over their corresponding test points

180 Rational Inequalities Label the signs on the line graph over their corresponding test points nd nd

181 Rational Inequalities Label the signs on the line graph over their corresponding test points nd nd Shade the regions that make the inequality true and use interval notation

182 Rational Inequalities Label the signs on the line graph over their corresponding test points nd nd Shade the regions that make the inequality true and use interval notation nd nd ( ] (

183 Rational Inequalities Label the signs on the line graph over their corresponding test points nd nd Shade the regions that make the inequality true and use interval notation nd nd ( ] ( Write the solution in interval notation

184 Rational Inequalities Label the signs on the line graph over their corresponding test points nd nd Shade the regions that make the inequality true and use interval notation nd nd ( ] ( Write the solution in interval notation - ( 1, 2] (3, )

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