SECTION 6-3 Systems Involving Second-Degree Equations

Transcription

1 446 6 Systems of Equations and Inequalities SECTION 6-3 Systems Involving Second-Degree Equations Solution by Substitution Other Solution Methods If a system of equations contains any equations that are not linear, then the system is called a nonlinear system. In this section we investigate nonlinear systems involving second-degree terms such as y 3 y 1 y y 3y y 0 y y 0 It can be shown that such systems have at most four solutions, some of which may be imaginary. Since we are interested in finding both real and imaginary solutions to the systems we consider, we now assume that the replacement set for each variable is the set of comple numbers, rather than the set of real numbers. Solution by Substitution The substitution method used to solve linear systems of two equations in two variables is also an effective method for solving nonlinear systems. This process is best illustrated by eamples. EXAMPLE 1 Solving a Nonlinear System by Substitution Solve the system: y 3 y 1 Solution Solve the second equation for y in terms of ; then substitute for y in the first equation to obtain an equation that involves alone. 3 y 1 y 1 3 aeebeec y (1 3) ( 1)( ) 0 Substitute this epression for y in the first equation. Simplify and write in standard quadratic form. Divide through by to simplify further. 1, If we substitute these values back into the equation y 1 3, we obtain two solutions to the system:

2 6-3 Systems Involving Second-Degree Equations 447 y 1 3 y 1 y 1 3(1) y 1 3( 11 ) y A check, which you should provide, verifies that (1, ) and (, ) are both solutions to the system. These solutions are illustrated in Figure 1. However, if we substitute the values of back into the equation y, we obtain 11 FIGURE y y 4 y ( ) y y 11 y 11 It appears that we have found two additional solutions, (1, ) and (, ). But neither of these solutions satisfies the equation 3 y 1, which you should verify. So, neither is a solution of the original system. We have produced two etraneous roots, apparent solutions that do not actually satisfy both equations in the system. This is a common occurrence when solving nonlinear systems. It is always very important to check the solutions of any nonlinear system to ensure that etraneous roots have not been introduced. 11 Matched Problem 1 Solve the system: y 10 y 1 EXPLORE-DISCUSS 1 In Eample 1, we saw that the line 3 y 1 intersected the circle y in two points. (A) Consider the system y 3 y 10 Graph both equations in the same coordinate system. Are there any real solutions to this system? Are there any comple solutions? Find any real or comple solutions. (B) Consider the family of lines given by 3 y b b any real number What do all these lines have in common? Illustrate graphically the lines in this family that intersect the circle y in eactly one point. How many such lines are there? What are the corresponding value(s) of b? What are the intersection points? How are these lines related to the circle?

3 448 6 Systems of Equations and Inequalities EXAMPLE Solving a Nonlinear System by Substitution Solve: y y Solution Solve the second equation for y, substitute in the first equation, and proceed as before. y y u u 8 0 (u 4)(u ) 0 u 4, Multiply both sides by and simplify. Substitute u to transform to quadratic form and solve. FIGURE y y y Thus, For, y 1. For 4 or or, y 1. i For i, y i. i For i, y i. i Thus, the four solutions to this system are (, 1), (, 1), (i, i ), and ( i, i ). Notice that two of the solutions involve imaginary numbers. These imaginary solutions cannot be illustrated graphically (see Fig. ); however, they do satisfy both equations in the system (verify this). Matched Problem Solve: 3 y 6 y 3 EXPLORE-DISCUSS (A) Refer to the system in Eample. Could a graphing utility be used to find the real solutions of this system? The imaginary solutions?

4 6-3 Systems Involving Second-Degree Equations 449 (B) In general, eplain why graphic approimation techniques can be used to approimate the real solutions of a system, but not the comple solutions. EXAMPLE 3 Design 19 inches FIGURE 3 y Solution An engineer is to design a rectangular computer screen with a 19-inch diagonal and a 17-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch. Sketch a rectangle letting be the width and y the height (Fig. 3). We obtain the following system using the Pythagorean theorem and the formula for the area of a rectangle: This system is solved using the procedures outlined in Eample. However, in this case, we are only interested in real solutions. We start by solving the second equation for y in terms of and substituting the result into the first equation. y 17 y 19 y , ,6 0 Multiply both sides by and simplify Quadratic in Solve the last equation for using the quadratic formula, then solve for : (1)(30,6) 1.0 inches or 11.7 inches Substitute each choice of into y 17/ to find the corresponding y values: For 1.0 inches, 17 y 11.7 inches 1 For 11.7 inches, y inches 11.7 Assuming the screen is wider than it is high, the dimensions are 1.0 by 11.7 inches.

5 40 6 Systems of Equations and Inequalities Matched Problem 3 An engineer is to design a rectangular television screen with a 1-inch diagonal and a 09-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch. Since Eample 3 is only concerned with real solutions, graphic techniques can also be used to approimate the solutions (see Fig. 4). As we saw in Section 3-1, graphing a circle on a graphing utility requires two functions, one for the upper half of the circle and another for the lower half. [Note: since and y must be nonnegative real numbers, we ignore the intersection points in the third quadrant see Fig 4(a).] FIGURE 4 Graphic solution of y 19, y (a) y (b) Intersection point: (c) Intersection point: y 361 (11.7, 1.0) (1.0, 11.7) y 3 17 Other Solution Methods We now look at some other techniques for solving nonlinear systems of equations. EXAMPLE 4 Solving a Nonlinear System by Elimination Solve: y y 17 Solution This type of system can be solved using elimination by addition. Multiply the second equation by 1 and add: y y 17 3y 1 y 4 y Now substitute y and y back into either original equation to find. For y, For y, () 3 ( ) 3

6 6-3 Systems Involving Second-Degree Equations 41 Thus, (3, ), (3, ), ( 3, ), and ( 3, ), are the four solutions to the system. The check of the solutions is left to you. Matched Problem 4 Solve: 3y 3 4y 16 EXAMPLE Solving a Nonlinear System Using Factoring and Substitution Solve: 3y y 0 y y 0 Solution Factor the left side of the equation that has a 0 constant term: y y 0 y( y) 0 y 0 or y Thus, the original system is equivalent to the two systems: y 0 3y y 0 or or y 3y y 0 These systems are solved by substitution. First System y 0 3y y 0 Substitute y 0 in the second equation, and solve for. 3(0) (0) Second System y 3y y 0 Substitute y in the second equation and solve for Substitute these values back into y to find y.

7 4 6 Systems of Equations and Inequalities For, y. For, y. The solutions for the original system are (, 0), (, 0), (, ), and (, ). The check of the solutions is left to you. Matched Problem Solve: y y 9 y 0 Eample is somewhat specialized. However, it suggests a procedure that is effective for some problems. EXAMPLE 6 Graphic Approimations of Real Solutions Use a graphing utility to approimate real solutions to two decimal places: 4y y 1 y y 6 Solution Before we can enter these equations in our graphing utility, we must solve for y: 4y y 1 y 4y ( 1) 0 y y 6 y y ( 6) 0 Applying the quadratic formula to each equation, we have y ( 1) y 4 4( 6) Since each equation has two solutions, we must enter four functions in the graphing utility, as shown in Figure (a). Eamining the graph in Figure (b), we see that there are four intersection points. Using the built-in intersection routine repeatedly (details omitted), we find that the solutions to two decimal places are (.10, 0.83), ( 0.37,.79), (0.37,.79), and (.10, 0.83). FIGURE (a) (b)

8 6-3 Systems Involving Second-Degree Equations 43 Matched Problem 6 Use a graphing utility to approimate real solutions to two decimal places: 8y y 70 y y 0 Answers to Matched Problems 9 1. ( 1, 3), (, 13 ). ( 3, 3), ( 3, 3), (i, 3i), ( i, 3i) by 1. in 4. (, 1), (, 1), (, 1), (, 1). (0, 3), (0, 3), ( 3, 3), ( 3, 3) 6. ( 3.89, 1.68), ( 0.96,.3), (0.96,.3), (3.89, 1.68) EXERCISE 6-3 A Solve each system in Problems y 169. y 1 y y y. 3 y 6. y 0 7. y 8. y y 3 y 9. y y 3 y 1 y 11. y y 1 16 y 4y B Solve each system in Problems y y 6 0 y y 4 1. y y y y 1 4y 3 y 0 y 18 y y y 10 4 y 8 4y y 0. y 0 y y y 1. y 9. y 16 y 9 y y 4 3. y 3 4. y 1 y y An important type of calculus problem is to find the area between the graphs of two functions. To solve some of these problems it is necessary to find the coordinates of the points of intersections of the two graphs. In Problems 3, find the coordinates of the points of intersections of the two given equations.. y, y 6. y, y 3 7. y, y 8. y, y 3 9. y 6 9, y 30. y 3, y y 8 4, y 3. y 4 10, y Consider the circle with equation y and the family of lines given by y b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in eactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle. (C) How is the line with equation y 0 related to this family of lines? How could this line be used to find the intersection points in part B? 34. Consider the circle with equation y and the family of lines given by 3 4y b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in eactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle.

9 44 6 Systems of Equations and Inequalities (C) How is the line with equation 4 3y 0 related to this family of lines? How could this line be used to find the intersection points and the values of b in part B? of the circle is 6. inches, and the area of the rectangle is 1 square inches. Find the dimensions of the rectangle. C Solve each system in Problems y 7y y y 16 y 3 0 y y y y 39. y y y 0 y y y 3 y y 36 y y 3y 3 4. y y 16 4y 3y 0 y y 0 In Problems 43 48, use a graphing utility to approimate the real solutions of each system to two decimal places. 43. y y y y 3 4y y 8 y y y y 46. 4y y 4 y y 9 4 y y y y 9 4 4y y y y 1 4 4y y y 9. Construction. A rectangular swimming pool with a deck feet wide is enclosed by a fence as shown in the figure. The surface area of the pool is 7 square feet, and the total area enclosed by the fence (including the pool and the deck) is 1,1 square feet. Find the dimensions of the pool. ft Pool ft 6. inches Fence APPLICATIONS ft ft 49. Numbers. Find two numbers such that their sum is 3 and their product is Numbers. Find two numbers such that their difference is 1 and their product is 1. (Let be the larger number and y the smaller number.) 1. Geometry. Find the lengths of the legs of a right triangle with an area of 30 square inches if its hypotenuse is 13 inches long.. Geometry. Find the dimensions of a rectangle with an area of 3 square meters if its perimeter is 36 meters long. 3. Design. An engineer is designing a small portable television set. According to the design specifications, the set must have a rectangular screen with a 7.-inch diagonal and an area of 7 square inches. Find the dimensions of the screen. 4. Design. An artist is designing a logo for a business in the shape of a circle with an inscribed rectangle. The diameter 6. Construction. An open-topped rectangular bo is formed by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides. The area of the cardboard before the corners are removed is 768 square inches, and the volume of the bo is 1,440 cubic inches. Find the dimensions of the original piece of cardboard. 6 in. 6 in. 6 in. 6 in. 6 in. 6 in. 6 in. 6 in. 7. Transportation. Two boats leave Bournemouth, England,

10 6-4 Systems of Linear Inequalities in Two Variables 4 at the same time and follow the same route on the 7-mile trip across the English Channel to Cherbourg, France. The average speed of boat A is miles per hour greater than the average speed of boat B. Consequently, boat A arrives at Cherbourg 30 minutes before boat B. Find the average speed of each boat. 8. Transportation. Bus A leaves Milwaukee at noon and travels west on Interstate 94. Bus B leaves Milwaukee 30 minutes later, travels the same route, and overtakes bus A at a point 10 miles west of Milwaukee. If the average speed of bus B is 10 miles per hour greater than the average speed of bus A, at what time did bus B overtake bus A? SECTION 6-4 Systems of Linear Inequalities in Two Variables Graphing Linear Inequalities in Two Variables Solving Systems of Linear Inequalities Graphically Application Many applications of mathematics involve systems of inequalities rather than systems of equations. A graph is often the most convenient way to represent the solutions of a system of inequalities in two variables. In this section, we discuss techniques for graphing both a single linear inequality in two variables and a system of linear inequalities in two variables. Graphing Linear Inequalities in Two Variables We know how to graph first-degree equations such as y 3 and 3y but how do we graph first-degree inequalities such as y 3 and 3y Actually, graphing these inequalities is almost as easy as graphing the equations. But before we begin, we must discuss some important subsets of a plane in a rectangular coordinate system. A line divides a plane into two halves called half-planes. A vertical line divides a plane into left and right half-planes [Fig. 1(a)]; a nonvertical line divides a plane into upper and lower half-planes [Fig. 1(b)]. FIGURE 1 Half-planes. y y Left half-plane Right half-plane Upper half-plane Lower half-plane (a) (b)