Math 1302 Notes 2. How many solutions? What type of solution in the real number system? What kind of equation is it?

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1 Math 1302 Notes 2 We know that x = 0 has How many solutions? What type of solution in the real number system? What kind of equation is it? What happens if we enlarge our current system? Remember the number system; natural numbers whole numbers integers rational numbers real numbers irrational numbers Def. Let i = 1 and let i 2 = - 1. Note: that we could interpret i 2 as follows: 1) i 2 = i i = 1 1 = -1 if we use the property of radicals: a a = a or 2) i 2 = 1 1 = 1 ( 1) = 1 = 1 -- which we choose not to use. Based on properties of exponents notice what happens with i n. i 1 =, i 2 =, i 3 =, i 4 =, because of this last exponent We get i 5 = i 4 i = i 6 = i 4 i 2 = i 7 = i 4 i 3 = i 14 = i 4 i 4 i 4 i 2 = i 25 = i 4 i 4 i 4 i 4 i 4 i 4 i 1 = Four Simple forms of i n There are only four types of results with i n, if n is a whole number. i = i, i 2 = -1, i 3 = -i, and i 4 = 1 all the other powers of n ( integer exponent) can be written in terms of these four results. 71

2 We know that is not a real number. Why? but since 9 = 1 9 = 1 9 = 3 1 = we can say that = We define this type of number as follows. Def. Let i be defined as above with the condition that i 2 = -1 and let a and b represent any real numbers. Then a + bi is called a complex number. The set of numbers of the form a +bi form the set of complex numbers. examples: 2 + 3i, i, ½ + ¼ i, 0 2i, i,... Also, 3 π i, i 7 We can work with complex numbers in a similar way to the way that we work with real numbers. First: Although 3 + 0i is a complex number which can be reduced to 3 + 0i = 3 we can not think of 3 + 0i as a real number. They behave similarly but they are not identical. Example: graph the real numbers: 3, - 7, 14 Now graph the complex numbers: 3 + 0i, i, and i If you encountered no problem the numbers above, then it should be fairly easy for you to graph the numbers 3 2i, i 72

3 Operations on Complex Numbers Now that you see one difference between the real numbers and the complex numbers let s look at some of their similar properties. Graph: 3 2i, i, 0 2i, and 3 + 0i Review: Find x + 2y + ( 3x 5y ) = ( x + 3 ) ( x 2 ) = We can add, subtract, multiply, and divide complex numbers ( also plot graph ) ex. ( 3 2i ) + ( 4-5i ) = ex. ( 4 i ) - ( 5 4i ) = ex. ( 3 + 2i ) ( 4 4i ) = ex. ( 4 i ) ( 3 + i ) = Graph 4 + 2i, -3i, 4 + 0i on the complex plane. 73

4 Other Definitions: Conjugate: Let a + bi be a given complex number. We define a - bi as the conjugate of a + bi and we say they are conjugates of each other. Find the conjugate of - 2 3i of -2i of 4-0i Real part of a + bi Let a + bi be a given complex number. Then the real part of a + bit is a, The imaginary part of a + bi is b Find the real part of - 4 7i, the imaginary part of - 4 7i i Real part: Imaginary part 2i Real part: Imaginary part: - 4 Real part: Imaginary part: Modulus of a + bi (absolute value) a + bi = 2 a + b 2-4 3i 4i - 3 A number of the form a + bi is called an number if b 0. ex i, 0 4i, ½ + ¼ i,. What kind of number is 4 + 0i? A number of the form a + bi, where a = 0 and b 0, is called a number ex. 2i, ½ i, 74

5 Def. Two complex numbers a + bi and c + di are equal provided a = c and b = d. ex. Find x and y ; if 2x + 3i = 4 - yi ( 2 x ) - ( x + y)i = 4 + i True or False. Every real number is also a complex number. 75

6 Problems set #11 76

7 Quadratic Equations (in one variable) An equation of the form ax 2 + bx + c = 0 is called a quadratic equation in x and it is said to be in standard form. Assume that a > 0. There are lots of quadratic equations that do not have a real solution. x = 0 does not have any real solution If we introduce the idea of complex numbers we can extend the list of equations that do have a solution in the complex number system Now x = 0 has a solution x = We can find a solution to any equation of the form ax 2 + bx + c = 0. The solution may be real, or imaginary but it has a solution two distinct solutions (usually sometimes they are identical but always 2 solutions) x 2 x 2 + 2x + 1 = 0 has two identical solutions. Can you guess what they are? - 4x - 5 = 0 has two distinct real solutions. Can you guess what they are? x 2 + 4x = 0 has how many solutions? What is(are) they? Write in standard form and identify the values of a, b, c ex. x 2 = 5x 2 a = b = c = ex. 1 x 2 = 2x a = b = c = To find the solution of equations of this form we may use one of the following methods: 1) factoring: either ax 2 + bx = 0, or ax 2 + bx + c = 0 2) square root method: ax 2 + c = 0 3) completing the square: any equation of the form ax 2 + bx + c = 0 4) Quadratic formula: any equation of the form ax 2 + bx + c = 0 77

8 Property: Given a b = 0 what can we say about a and b? The product of any two ( or more ) numbers equal to zero means that at least one of the numbers must be zero. Find solution of 2 ( x 2 ) = 0 x = 3 ( x + 5 ) = 2 x = x ( x + 2 ) = 0 x = 4x x 2 = 0 x = x 2 = 2x x = x( x + 3 ) = 4 x = x 2 12x + 32 = 0 2x 2 5x 63 = 0 These are all examples of the first method; factoring The following could be done by factoring but also by isolating the x 2 term. Find the solution of x 2-9 = 0 x = 4x 2-25 = 0 x = 4x 2-3 = 0 x = 9x = 0 x = 78

9 ( x ) 2 = 4 ( x + 1 ) 2 = 4 x = Finally, x 2-12x + 36 = 0 x = The previous method is of course called the square-root method. Very few polynomials are factorable or are in a form where the square root method can be used. We normally have to use one of the following two methods. But first recall how to recognize or how to construct a perfect square. x 2 + 4x + x 2 + 3x + 2x 2 + 4x + Completing the square: Solve : x 2 = 9 What about (x + 2 ) 2 = 9 x = Now, what about x 2 + 4x - 5 = 0 While we can factor the polynomial, let s try this approach x 2 + 4x

10 Repeat for the following two polynomials. a) x 2-8x + 12 = 0 b) 2x 2 - x - 4 = 0 80

11 Problem set #12 81

12 Quadratic Formula: When solving equations of the form ax 2 + bx + c we have seen that we first try to factor or use the square root method. When these two methods are not possible, then we have to complete the square. Use this method to construct a special formula. Construction: Make a perfect square on the left side: x 2 + b/a x + c/a = 0 x 2 + b/a x + = - c/a ( + ) 2 = -c/a + Now use the square root method (x + ) 2 = Find the square root of both sides x + = x = the quadratic formula: x = 82

13 Find the solution of each of the following equations by any method. 3x 2-12 = 0 x = 9x 2 = 6x x = ex. x 2 = 3 2x x = x 2-4x - 2 = 0 x = 4x 2 + 2x - 1 = 0 x = What about the solution of the equation x 4-8x 2-9 = 0? 83

14 Almost quadratic equations We have seen quadratic equations of the form ax 2 + bx + c = 0. We can find the solution of such equations by : Factoring, square root method, completing the square, and last by the quadratic formula Some equations may not look quadratic but can be solved as quadratic equations 1) x 4-3x 2-4 = 0 2) x + 5x x = 0 3) x = 0 4) ( x + 1) 2-2(x+1) 3 = 0 5) x 6 9x = 0 6) x - 3x 1/2 + 2 = 0 x + 1 x 2 x 2 x + 1 6) = 0 84

15 Problem set #13 ex. Find the solution of each of the following equations. 1) 4x = 0 2) 3x 2-4x + 6 = 0 3) x 2 + x - 4 = 0 Name Math Long Quiz March 26, State the quadratic formula used to solve equations of the form ax 2 + bx + c = 0 2. What is a) the real part of i ==> b) the imaginary part of 4-5i ==> c) the conjugate of i ==> of - 2i ==> c) the modulus, the absolute value, of i ==> 3. Plot 4 3i on the complex plane. Label the axes. 85

16 4. Simplify each of the following by performing the given property and writing in simplest form. a) ( 4 + i ) - ( 3 2 i ) = b) ( 2-3i ) 2 = c) ( i ) / ( 3 2i ) = 5. Reduce to one of the four simplest forms of i n. Answers should never have an i in the denominator. a) i 15 = b) i 20 = c) i -3 = 6. Find the solution of each of the following quadratic equations by any method. a) x 2 = - 4x b) x = 0 c) 4x 2-2x + 1 = 0 d) x + 1/x + 2 = 0 7. Solve by competing the square. No other method will do. a) x 2 + 4x + 4 = 9 b) 4x 2 + 8x + 3 = 0 86

17 Equation with radicals 1) x = 3 2) 1 x = 2 3) 1 5 = 3 2 x 4) x 3 = 1 x 5) + 41 = x 1 x 6) x + 2 x = 2 7) 3 1 x = x

18 Inequalities (in one variable) Inequalities -- we have seen how to solve simple inequalities; 1) 2 < 3 ( 2 4x ) 2) ( x - 1 ) / - 4 ( x + 1 ) / 2 3) -4 < 2 3x 12 Some examples(and/or): Consider all students in class: I want all students in the class that are left handed or male: I want all students in class that did their HW and filled up with gas this morning 88

19 More inequalities (Systems of inequalities); special attention to the words and and or ex. x < 4 and x > - 3, List all of the integers that satisfy these statements If we wanted to list every real number, we could not, unless we use a graph to represent them. ex. x < 4 or x > 8, list all of the whole numbers that satisfy these statements Find all real numbers: Notice that - 4 < x and x < 3 can be written differently while - 4 > x or x > 3 can not! Why!!! When we see the inequality < x - 1 < 4, we can solve it by either thinking of it as - 4 < x 1 and x 1 < 4 and solving it: Or just leave it as is and solve it - 4 < x - 1 < 4 89

20 Problem set #14 90

21 More Inequalities ex. Find the solution of ( x - 1 ) > 4 ex. Find the solution of 12 < 1 x < 20 ex. What about the solution of 1) x 2 < 4, think of all the integers that satisfy this statement 2) x 2-2x > 3, We have seen that the solution of x 2 < 4 is represented by If I told you the solution of x 2 < 25 was what would you guess the solution of x 2 < 49 to be? 91

22 Look at the inequality x 2-2x > 3. Let s solve as before Try the same approach with the inequality x( x 2 )( x + 2) ( x + 3 ) < 0 Try a different approach Begin with the inequality x 2 + 5x + 6 > 0 What about x ( x 2 ) ( x + 3 ) < 0? 92

23 Because the approach above has to do with signs and multiplication and division have the same rule of signs this approach works just as well with quotients. Find the solution of a) x < x 2 x + x 6 b) 0 x 2 x c) 1 x 1 Absolute Value Equations a) x = 4 b) x = 10, b) 2x = 12, d) x + 2 = 3, e) 1-2x = 3, f) 4 + 3x = - 2, 93

24 Problem set # 14 94

25 Absolute Value Inequalities a) x < 4 b) 2x < 4 c) 2 x < 2 More d) x > 4 e) x / 4 > 3 f) 1-2x > 4 What about 2 3x > - 2? or 4x - 3 < - 2? 95

26 Other Inequlities. 96

27 Word Problems

28 Problem set #15 98

29 Table of Contents Notes2 page 71 99

Solving Quadratic Equations Review

Math III Unit 2: Polynomials Notes 2-1 Quadratic Equations Solving Quadratic Equations Review Name: Date: Period: Some quadratic equations can be solved by. Others can be solved just by using. ANY quadratic