Chapter 7 Quadratic Equations

Transcription

1 Chapter 7 Quadratic Equations We have worked with trinomials of the form ax 2 + bx + c. Now we are going to work with equations of this form ax 2 + bx + c = 0 quadratic equations. When we write a quadratic equation in the form above we say it is in standard form. We do not require a >0, but we usually write it as such. Write in the standard form and identify a, b, and c. 1) 2x 4 = 4 2x 2 standard form :, a= b = c = 2) 4x 2 = 2x a= b= c= c) 1-4x 2 = 0 a = b =, c = Methods of Solutions. 0.1) Product of zero: A B = 0 means that either A = 0 or B = 0 or both = 0 This is the method that is used when we solve quadratic equations that can factor. 1) Square Root Method: if b=0, we are trying to solve an equation of the form ax 2 + c = 0. That s the type of equation that is suited for this method. 2) Factoring: Very few equations can be solved by factoring unless they are book problem but when they are, this method works well. 3) Completing the square: Awkward method but useful when the polynomial does not factor 4) Quadratic formula: Not as awkward as completing the square. 101

2 Product of zero: In the real number system if the product of any two numbers equals zero, then at least one of the numbers must be zero. 1) 2 3 = 0, false statement because? 2) 2x = 0, this is the product of the numbers 2 and x. The result is zero which means either 2 = 0 ( NO!) or x = 0 ( Yes ) The solution is zero. 3) x( x + 1 ) = 0, this is the product of two numbers x and x + 1. Since the result is zero, that means either x = 0 ( possible ) or x +1 = 0 ( possible ). The values of x which make these statements true are x = 0 or x = - 1 This last type of setup is when we find the method useful. Solve each of the following by using the method above ( Factoring) 1. 5x 2 + 2x = 0 2. x 2-4x = x 2 = 12 5x 4. 24x x x = x 2 = 9 x 7. x 2-7 = 0 8. x = 0 102

3 The last three examples above could have done with a different method since b = 0 solving equations of the form c ax 2 + c = 0 ax 2 = - c x 2 =, we can solve by taking the square root of both sides. a Your author calls these type of equations incomplete quadratics when either b= 0 or when c = 0 Square Root Method: Although it is very limited it is useful in constructing the two important methods that follow. 1. x 2 = 81 4x 2 25 = 0 3. ( x ) 2 = 49 x = ( x + 2 ) 2 = ( x 1 ) 2 = 3 x 2 2x + 1 = 3 Completing the Square. x 2 2x 2 = 0 It is not obvious to you but this is the last problem above We want to rewrite the polynomial so that it looks like number 6 and then

4 Completing the Square. Can be used on any type of quadratic equation with real coefficients of the form ax 2 + bx + c = 0. Before we go through the method let me remind you what a perfect square is 9 is a perfect square because we can find a number x so that x 2 = 9 x = 3 works. 121 is a perfect square because we can find a number x so that x 2 = 121 x = 121. The same idea works with polynomials x 4 is a perfect square if we can find a number so that when squared we get x 4. What is the number? Is x 9 a perfect square? Why or why not? Is x 2-8x + 16 a perfect square? Why would we want a polynomial to be a perfect square? It is easier to solve if it is! x 2 = 5 ( x + 1 ) 2 = 5 x 2 +2x + 1 = 5 Look at an example with a nice solution and work it out with this method. x 2 2x 3 = 0 Let s try to solve this equation by changing the polynomial so that it looks like a perfect square. 1) Isolate the x-tems: x 2 2x = 3 2) Make the left-side into a perfect square: 3) Use the square-root method ( take the square root of both sides ) : 4) Simplify to find the solution: 104

5 Try these next two examples. x 2-6x + 8 = 0 2x 2-3x + 1 = 0 ( What is different about this equation?) Try one last example to set us up for the last method we will look at. 2x 2 - x = 6 105

6 Use the same method for the general quadratic equation; find the solution(s) of ax 2 + bx + c = 0 by completing the square. --- this means that as long as the equation is quadratic and it has the form above we can solve it as follows or even better by using our conclusion. Solve ax 2 + bx + c = 0 1) Isolate x-tems: 2) Make the polynomial into a good form a =1. 3) Rework so that right side is a perfect square 4) Use the square-root method and simplify to solve This is called the quadratic formula 106

7 Quadratic Formula. A (the solution) to an equation of the form ax 2 + bx + c = 0 is given by x = b ± b 2 4ac 2a, there are two solutions which can be separated and in some cases must be separated in order to simplify your final solution x 1 b + 2 b 4ac 2a =, x 2 = b 2 b 4ac 2a x 1 and x 2 represent the two solutions of the equation ax 2 + bx + c = 0 a: the coefficient of x 2, b: coefficient of x, and c : is the constant it is possible for b or c to be zero ( not both ), we prefer that a > 0 ( positive ) but it does not have to be. Note: b 2-4ac, is useful so we give it a name: the discriminant Examples: 2x 2-4x - 3 = 0 (factoring would work but let s use quadratic formula) x 2 + 8x - 9 = 0 ( factoring would work but less use the quadratic formula) 107

8 More Examples x 2 + 6x = 0 ( It would be a lot easier to factor and solve and for checking purposes maybe we should but let s also use the quadratic formula to solve ) Could you solve by completing the square? What about the solution of the equation 5 1 x 2 x + = If x = 3 and x = -2 are the only solutions of the equation ax 2 + bx + c = 0, how would you find the equation? Is there anything wrong with the solution of the following quadratic equations. x 2 + 2x + 4 = 0? 108

9 Other Problems that may not look like quadratic equations but are similar. x 4-3x 2-4 = 0 5 x + 4 = 0 x 109

10 Nature of the roots of a quadratic equation - although we are usually interested in the actual roots ( solutions) of the equation ax 2 + bx + c = 0, it is possible to determine the nature of the roots without actually solving for the roots ( find what the roots look like ) We use the discriminant of the equation axp 2 + bx + c = 0 b 2-4ac Here are the possibilities: ( and we actually use b 2 4ac - to get an idea of the nature of the roots) 1) if b 2-4ac < 0 ( negative ), we do not have a real root we will discuss in the next section what the roots look like - this is the only case in which the roots are not real all the other cases, the roots will be real ex. 2x 2-4x + 5 = 0, find b 2 4ac. ( why would this result gives us no real root? ) 2) if b 2 4ac = 0, we have two roots they are real and identical (equal ) Why? This is the only case in which the roots happen to be equal all the other cases the roots are unequal. ex. x 2-12x + 36 = 0 3) if b 2 4ac > 0 and a perfect square ( think of b 2 4ac ), then we have two real roots that happen to be rational ( no radical in the solution ) and the original problem could have been factored 12x 2-13x + 3 = 0 4) If b 2-4ac > 0 and is not a perfect square ( again, think of b 2 4ac ), then we have two real roots that happen to have a radical two irrational roots. x 2 + 5x - 1 = 0 110

11 One more type of Equation Equations with radicals If I asked to find a number so that if you took the square root you would get an answer of 5, what would your number be? Can you write an equation that asks the same question? Find me a number that when you take the square root gives an answer of 13. Write an equation. Let s make this question a little bit more difficult Find a number so that if you subtract one and then take the square root of the result you end up with 8. What is the original number. Write an equation that asks this same question. One more question of this type. Find me a number so that when you take the square root, you will get a negative four. Write the equation that corresponds to this question. More Problems: 1. x 2 = 3 x = 2. x + 2 = 0, x = 3. 2x 5 = 2, x = 4. x x = 6, x = 2 5. x + 4 = x + 2, x= 111

12 More (difficult) examples: ( It may be a good idea to remember how to find ( a + b) 2 =, ( 2x + 3) 2 = ) - you should have used this idea on the last example - Find the solution of each throw out any extraneous solutions! 1. 3x 2 + 2x = 3x 2, x = 2. x + 7 = 1+ 2x, x = 3. 2x + 5 x + 2 = 3x 5, x = 112

13 Imaginary and Complex numbers So far we have avoided answers to problems that look like x = 0. So far our answer has been no real root. If I asked you the following questions, would they have an answer? 1. Find all natural numbers so that x ( x 2 ) = 0. You would say the answer (s) is ( are )? 2. Find all whole numbers that satisfy the following equation x 2 = 36, x = 3. Find all integers that make the statement 6x 2-13x + 6 = 0. x = 4. Find all rational numbers that give you a solution to the equation x 2 = 3, x = 5. Find all of the real numbers that work in the following equation ( 2x 5 ) 2 = - 4, x = 113

14 You saw the problem we had in each case in the previous page the number system had to be enlarged to find all of the solutions that we normally consider. Take our real number system and enlarge it into what we will call the complex numbers(set of complex numbers) Def. Let i = 1 and we will choose to define i 2 = -1. Now we have an answer for 9 =, we find the answer by 9 = ( 1)(9 ) = 1 9 Using the definition above (i = 1 and the fact that 9 = 3, 9 = 3i Find 64 = 5 = 400 = 4 = if x is a positive real number, 2 x = ( - 25 ) ½ = Def. A complex number is a number of the form a + bi where a and b are real numbers i, i, 1 + i 2 which we usually write 1 i 2 + (that way you can see i is not inside the radical) also, 3 7 3i, 2 + i 3,... We can add, subtract, multiply and divide such numbers, graph them as well. 114

15 Addition and subtraction of complex numbers is nothing unusual as long as we remember how to add and subtract polynomials. Find the sum of ( 2 + 3x ) and ( 4 7x ). Find ( 2-3y ) - ( 4 + 6y ) = Same idea with complex numbers. Find ( 2 + 3i ) + ( 5 4i ). ( 3 i ) - ( 7 + 5i ). 3( 2 + i) - 4( 3-2i). What about the product of two complex numbers? If we remember how to multiply polynomial, it is the same idea with one added condition. Remember that i 2 = - 1 Find ( 2 + 3i) ( 4 i ) = 3i ( 1 2i ) = ( 2 3i ) 2 = -2 ( 1 + 3i ) = That leaves only division and graphing of complex numbers. 115

16 Division of complex numbers. Remember two ideas from real numbers; opposite and absolute value. We have the corresponding ideas with complex numbers not exactly the same, but similar. We do have the idea of opposite and it s what you would think Find the opposite ( the additive inverse ) of 2 3i. - ( 2-3i) = i Find the opposite of i., of 6i = We want to look at the conjugate of a complex number and it s not the same thing as the opposite but there is some similarity. Def. Let a + bi be a given complex number. The conjugate of a + bi is defined by a bi. Find the conjugate of each of the complex numbers. 2 3i 0 4i i we use this idea when we divide two complex numbers Def. The absolute value of a complex number a + bi is defined as 2 2 a + b. Find the absolute value ( modulus) of 3 4i i = - 3 4i = 2 + 3i 2i 4 + 0i Now let s see how we divide two complex numbers. ( 3 + i ) 2 = ( 3 + 4i ) (2i) = 3 + 4i 2i = 2 i (2 i) (1 2i) = = 1 2i 116

17 Some additional problems with solutions in the complex number system which means you can not answer a question in the form no real root. 1. x 4-3x 2-4 = 0 x = ( Notice: there are up to 4 different possible solutions) 2. x 4-7x = 0 x = 3. x 2-4x + 5 = 0, x = 4. x 1/2-4x 1/4-5 = 0, x = 117

18 Word Problems Odd numbers 1-15 page 329. ( plus a couple of the even problems) 1. The length of a rectangle is two more than twice its width. Find the dimensions of the rectangle if its area is A triangle of area 35 has altitude which is three less than its base. Find the base and altitude. 4. Find two consecutive positive odd integers whose product is Find three consecutive positive integers if the sum of their squares is

19 6. The diagonal of a rectangle is ten centimeters and the width is two centimeters less than the length. Find the dimensions of the rectangle. Use Pythagorean property The sum of the reciprocals of two consecutive even integers is. Find the integers One number is four greater than the other. The sum of their squares is 106. Find the numbers. 9. Find the number that is six greater than its positive square root. 119

20 10. One side of a right triangle is four centimeters. The other side is seven centimeters less than twice the length of the hypotenuse. Find the length of the hypotenuse. 11. The area of a square is 36 square meters. If the area is to be increased by 28 square meters, how many meters should each side be increased by? 12. The length of a rectangle is twice the width. IF each dimension is increased by three, the new area would be 104 square meters. Find the original dimensions. 13. A polygon of n sides has n( n 3) 2 diagonals. How many sides does a polygon have if it has 54 diagonals? 120

21 14. A small motorboat can travel ten kilometers per hour in still water. The boat travels eight kilometers upstream and returns in one hour and forty five minutes. What is the speed of the current? 15. Two pumps can fill a swimming pool in three hours. It would take pump A 8 hours longer than pump B to fill the pool alone. How many hours would it take each pump to fill the pool alone? 121

22 Review and chapter 7 page 333 and chapter 7 test page

23 Chapter 8 System of Equations and Inequalities Simple definition and examples of relations and functions Your author explains the idea of a relation in a very nice way. Def. A relation is a set of ordered pairs of numbers between two sets. A rule that assigns members of one set (first set) to member(or members) of a second set. Here is his explanation: Let A = { 1, 3, 5 } and B = { 2, 4, 6, 8, 10 }. We will call the first set the domain of the relation and the elements of the second set that are actually used will be called the range of the relation or the image set of the relation. 1) Let the first rule ( relation ) be defined as follows: a member of set A is related to to B by doubling the value. 1 is related to 2 we write ( 1, 2), 3 is related to 6 and we write (3, 6), and 5 is related to 10 written ( 5, 10) The complete relation can now be written as a set of pairs of numbers; { ( 1, 2 ), ( 3, 6), ( 5, 10 ) } Domain: { 1, 3, 5 } and Range: { 2, 6, 10 } 2) a second example: relate a member of set A to member(s) of the second set by selecting the numbers that are larger than the given number. 1 is related to 2, 4, 6, 8, 10 we write ( 1, 2), (1, 4), (1, 6), ( 1, 8), ( 1, 10 ) 3 is related to 4, 6, 8, 10 we write ( 3, 4), ( 3, 6 ), (3, 8), (3, 10) 5 is related to 6, 8, 10 we write (5, 6 ), ( 5, 8 ), ( 5, 10 ) Complete relation: { ( 1, 2 ), ( 1, 4 ),... ( 5, 8), ( 5, 10 ) } Domain: { 1, 3, 5 } Range: { 2, 4, 6, 8, 10 } Look at relations by writing in a different form 3) 4) { } { } Domain and Range? Domain and Range? 123

24 Def. If every member of the domain of a relation is related to only one number ( produces one image), then we say the relation is a function. (or ) Def. A function is a relation such that each element of the domain has exactly one image. A function is a special kind of relation. All functions can be classified as relations but not all relations are functions Go back over the four examples we did and identify as functions or just relations. 1) 2) 3) 4) Most of the time that we see relations we see them in equation form ( sometimes in graph form) Determine which of the following equation(relations) also represent functions. Find the domain of each function. 1) y = -3x + 2 2) y = x 2 Function or Relation Domain: Function or Relation Domain: 3x 3) y = x 5 Function or Relation Domain: 4) y 2 = x + 2 Function or Relation Domain: 5) y = 1 x 3 Function or Relation 6) y = 1 x 2 4 Function or Relation Domain: Domain: 7) x 2 + y 2 = 4 Relation or Function Domain: 124

25 The previous seven examples contained the variables x and y. In the form that they are written we consider x to be the independent value within certain limitations it can have any value. The variable y is considered to be the dependent variable once x is given, then y has no choice but to depend on that value of x. Cartesian Coordinate System ( Rene / Descartes: ) Allows to create a connection between equations ( functions and relations and their graphs) Given a flat surface ( normally infinite in length) we can label each point on the surface in terms of an ordered pair of the form (x,y) Construction: 1) begin with two number lines drawn perpendicular to each other. 2) Label each of the axes accordingly; horizontal axis x axis, vertical axis y axis. 3) Label each quadrant properly; Quadrant I, II, III, and IV every point (x,y) can be plotted and every point P on the plane can now be labeled. Example: 1) Draw a coordinate system and plot each of the following points; A(3, -2), B(-4, 0 ), C( -2, -5 ) 2) Which quadrant is each of the following points in P(-4, 1), Q( 1, 2 ), T( 2, 0 ) 3) Label the points on the plane 125

26 We were able to solve equations before. 1) Find the value of x that solves the equation x + 3 = -1 x = 2) What is the solution set of x 2-3 = 1 3) Find all values of x that make the following statement true. x 2 + 3x + 1 Continue with this type of questions. Find the solution of x + y = 3 x 2 + 2y = 3 x 2 + y 2 = As you can see, the last three problems have too many solutions so how do we show the solution set? 126

27 Graphs: Sketch the graph of the following relations ( the solution set of each equation ) y = x + 3 y = 4 2x x = y 2-4y + 2 y = x y = - x 2 + 2x - 3 y = x 3 Some graphs are easier to find than others. Let s look at one of the simplest; lines. 127

28 X and Y intercepts Any graph that we look at may have a place where it crosses the x and the y axes. If a curve crosses the x-axis at (a,0 ), we say it has an x-intercept at x = a a curve crosses the y-axis at (0, b ), we say it has a y-intercept at y = b. to find an x-intercept: set y =0 and solve for x. What is( are) the x intercepts of 2x + y = 9 x 2 + y 2 = 9 y = x 2 4x 5 y = x + 2 y = 4 x to find the y-intercept: set x = 0 and solve for y. y = 4 x y = x 5 2x y = - 4 x y = 2x 4 Def. Let P(x 1, y 2 ) and Q(x 2, y 2 ) be given points. We can find the slope of a line segment connecting them ( or the slope of a line passing through them) by y2 y1 slope = m =, the difference of the y s over the difference of the x s x x

29 Straight Line Lines have the standard form ax + by = c, where a and b can not both be zero at the same time. Lines can be of three types; vertical, horizontal, and slant vertical lines Equation: x = a, a is the x-intercept do not have slopes: convince yourself by selecting two points on any vertical line and try to find the slope horizontal lines Equations: y = b, b is the y-intercept have slope m = 0 slant lines Equations: either ax + by = c or y =mx + b do have slopes and can be found by writing the equation in the form y =mx + b or by finding two points and using the formula m = y x 2 2 y x 1 1 Find the slope of a line that 1) passes through the points ( -2, 4) and ( -4, 8) 2) Find the slope of the line that is represented by the equation 2x 3y = 5 3) What is the slope of a line that is parallel to the line y = - 3? 4) If a line is parallel to the y-axis and passes through the point ( 3, 1), then its slope is given by m = Conclusion: a) if m > 0, the line leans to the b) if m < 0 the line leans to the c) if m = 0, then the line is d) if the line does not have a slope, then the line is 129

30 Different forms of a line 1. slope-intercept form: y =mx +b provides a good way to find the slope and the y-intercept can be used whenever you are given the slope and a point (preferably the y-intercept) What is the slope and the y-intercept of the line represented by 2x 3y = 7? Find the equation of the line that has slope 3/2 and passes through the point ( 0, - 3 ) 2. point slope form: y y 1 = m (x x 1 ) can be used whenever you are given a point and a slope or if you are given two points, you can find the slope and then use one the two points Find the equation of the line passes through the points A( -3, 4) and B( 7, 4 ) Find the equation of the line that passes through the point P(-4, 5 ) and has y-intercept=3 x y 3. Intercept form + = 1, a = x-intercept, b = y-intercept a b Can be used to quickly write an equation when the x and y-intercepts are given Find the equation of the line that has no y-intercept and passes through the point ( 3, -2) Find the x and y-intercepts of the equation x y = Find the equation of the line that has x-intercept = 3 and y-intercept = 3/2. Write in standard form ax + by = c 130

31 Note: a). Lines that are parallel have slopes that are b). Lines that are perpendicular have slopes that are 4. Standard form: ax + by = c gives the line a nice, normal appearance Find the equation of the line a)that has an undefined slope and passes through the point ( 5, -4) b) is parallel to the line 2x y = 3 and passes through the point ( 2, -2 ) c) is perpendicular to the line x + 4y = - 2 and passes through the origin. d) Find the value of k so that the resulting line has slope = 4 2x ky = 4 e) Find the value of t so that the resulting equation has no slope( undefined slope) tx + 2y = 4 f) Find the value of p so that the following equation represents a line with x-intercept = 1/5. 4x 5y = p 131

32 Graphing linear equation(lines) and linear inequalities(half-planes) Linear equations(functions): Now that we know about x and y-intercepts we can make use of these to sketch the graphs. The slope is also useful. 1) Sketch the graph of 2 y = x ) 2x + y = 8 3) x = - 4 Linear Inequalities: 2 1) y < x + 4 2) 2x + y 8 3 3) x <

33 Once we learn how to graph individual linear equations and linear inequalities, we can look at systems of each. Find the solution of each system of linear equation by graphing and circling the point of intersection ( solution of the system) 1) x + 2y = 2 2x y = 4 2) y = - 2 x + 3y = 3 System of Inequalities: To save on work we will use inequalities similar to the equations above. 1) x + 2y 2 2x y 4 2) y > - 2 x + 3y < 3 133

34 Methods of solution ( System of Equations) ( two equations: two lines) There is more than one way to find the solution of a system of linear equations. 1. Graphing: 2. substitution 3. Elimination: ( your author calls the addition method) Graphing we just did Find all the points where the lines x = -3 2x + y = 12 intersect Substitution Method 1. given two equations solve for one variable in a chosen equation 2. substitute for that variable in the other equation which has not been used 3. solve for remaining variable 4. substitute back to get original variable. ex. a) 2x + y = -3 b) 3x 2y = 3 3x + 2y = 4 6x 4y = 1 c) 2x + 3y = 1 d) 3x 6y = 21 3x + 4y = 2 2x 4y =

35 Elimination Method ( addition method) Sometimes it is easier to use this method Process: 1) select a variable to eliminate 2) change the coefficient on each equation so that they are equal except for the same sign ( Least common multiple) 3) Add the equations and eliminate the chosen variable 4) Solve for the remaining variable 5) Substitute into either of the original equations to find the value of the missing variable. a) 3x + 2y = 8 b) x 2y = 13 2x 3y = 14 2x - 4y = 1 c) 4x + 3y = 1 d) 9x 15y = 9 2x 3y = 1 6x 10y = 6 dependent: identity independent: unique ( conditional) inconsistent: contradiction 135

36 More examples: (including word problems from page 385 and 386, plus problems from page ) 12/385 The length of a rectangular lot is eight feet less than three times the width. Find the length and width if the perimeter of the lot is 600 feet. 14/385 An airliner took 6 hours to travel 3864 kilometers from New York to Las Vegas. The return trip took 4 hours. Find the speed of the plane and the wind 16/386 A total of 5000 tickets were sold for a football game. The price was $5.00 per adult ticket and $2.50 per child. If the total receipts were $21, , find the number of each type of ticket sold. 19/386 Mike has two more dimes than nickels. The total value of the coins is $2.60 How many of each type of coin does he have? 136

37 Solution of a system of three equations and three unknowns. We knew how to solve a system of one equation with one unknown; 5-3( x + 2 ) = 2 We just saw how to solve a system of 2 equations and 2 unknowns; reduce to a single equation with one variable by using the substitution or the elimination method What do we do if the system is larger, say a system of three equations and three unknowns? Substitution method would work but we will solve by using an elimination method. 1) 2x 3y + z = 6 3x + 2y 2z = 6 4x - 3y + 2z = 9 2) x + 2y = 7 3x 2z = 7 y + 3z = 5 3) x y = 3 x + 2y z = 2 2x 2y + z = 4 4) The sum of three numbers is 58. Twice the first number added to the sum of the second and third numbers is 71. If the first number is added to four times the second number and the sum is decreased by three times the third number, the result is 18. Find the numbers. 137

38 Review and chapter test: page 396, 403, and end-of-book test 138