Solving Quadratic Systems - ClassZone

1 Page 1 of 2632 Chapter 10 Quadratic Relations and Conic SectionsSolving Quadratic SystemsSOLVING ASYSTEM OFEQUATIONSIn Lesson you studied two algebraic techniques for Solving a system of linear equations . You can use the same techniques (substitution and linearcombination) to solve Quadratic Points of IntersectionFind the points of intersection of the graphs of x2+ y2= 13 and y = x+ find the points of intersection, substitute x+ 1for yin the equation of the + y2= 13 Equation of circlex2+ (x+ 1)2= 13 Substitute x+1 for + x2+ 2x + 1 = 13 Expand the + 2x 12 = 0 Combine like (x 2)(x+ 3) = 2 or x= 3 Zero product propertyYou now know the x-coordinates of the points of intersection. To find the y-coordinates, substitute x= 2 and x = 3 into the linear equation and solve for y.

2 The points of intersection are (2, 3) and ( 3, 2). CHECK You can check your answer algebraicallyby substituting the coordinates of the points intoeach equation. Another way to check your answeris to graph the two equations . You can see from thegraph shown that the line and the circle intersect intwo points, at (2, 3) and at ( 3, 2).EXAMPLE 1 GOAL1 Solve Systems ofquadratic quadraticsystems to solve real-lifeproblems, such as determiningwhen one car will catch up toanother in Ex. 58. To model real-lifesituations with quadraticsystems, such as finding theepicenter of an earthquake in Example you should learn itGOAL2 GOAL1 What you should Points of IntersectionThe circle and line in Example 1 intersect in two points. A circle and a line can also intersect in one point or no points.

3 Sketch examples to illustrate thedifferent numbers of points of intersection that the following graphs can and and and and lineDevelopingConceptsACTIVITY12yx( 3, 2)(2, 3)Look Back For help with solvingsystems, see p. 148 .STUDENTHELPPage 1 of Quadratic Systems633 Solving a system by SubstitutionFind the points of intersection of the graphs in the + 4y2 4 = 0 Equation 1 2y2+ x + 2 = 0 Equation 2 SOLUTIONB ecause Equation 2 has no x2-term, solve that equation for x. 2y2+ x+ 2 = 0x= 2y2 2 Next, substitute 2y2 2for xin Equation 1 and solve for + 4y2 4 = 0 Equation 1(2y2 2)2+ 4y2 4 = 0 Substitute for 8y2+ 4 + 4y2 4 = 0 Expand the 4y2= 0 Combine like (y2 1) = 0 Factor common (y 1)(y+ 1) = 0 Difference of 0, y= 1, or y= 1 Zero product propertyThe corresponding x-values are x= 2, x= 0, and x= 0.

4 The graphs intersect at ( 2, 0), (0, 1), and (0, 1), as a system by Linear CombinationFind the points of intersection of the graphs in the +y2 16x+39=0 Equation 1x2 y2 9=0 Equation 2 SOLUTIONYou can eliminate the y2-term by adding the two equations . The resulting equationcan be solved forx because it contains no other +y2 16x+39=0x2 y2 9=02x2 16x + 30 = (x 3)(x 5) = 3 or x=5 Zero productpropertyThe corresponding y-values are y= 0 andy= 4. The graphs intersect at (3, 0), (5, 4), and (5, 4), as 3 EXAMPLE 21yx( 2, 0)(0, 1)(0, 1)262yx(5, 4)(3, 0)(5, 4)Look Back For help with factoring,see p. 1 of 2634 Chapter 10 Quadratic Relations and Conic SectionsSEISMOLOGISTA seismologistdetermines the location andintensity of an earthquakeusing an instrument whichmeasures energy wavesresulting from movements in the Earth s ONCAREERSSOLVINGQUADRATICSYSTEMS INREALLIFES olving a system of Quadratic ModelsSEISMOLOGYA seismograph measures the intensity of an earthquake.

5 Although a seismograph can determine the distance to the earthquake s epicenter, it cannotdetermine in what direction the epicenter is located. Use the following informationfrom three seismographs to find an earthquake s 1:500 miles from the epicenterLocation 2:100 miles west and 400 miles south of Location 1400 miles from the epicenterLocation 3:300 miles east and 600 miles south of Location 1200 miles from the epicenterSOLUTIONLet each unit represent 100 miles. If Location 1 is at (0, 0), then Location 2 is at ( 1, 4) and Location 3 is at (3, 6). Write the equation of each 1:x2+y2=25 Location 2:(x+1)2+(y+4)2= 16, orx2+2x+1+y2+8y+16=16 Location 3:(x 3)2+(y+6)2= 4, orx2 6x+9+y2+12y+36=4 Subtract the equation for Location 1 from the equation for Location +2x+1+y2+8y+ 16 = 16 (x2+y2= 25)2x+8y+ 17 = 92x+8y= 26, or x+4y= 13 Then subtract the equation for Location 1 from the equation for Location 6x+9 +y2+12y+ 36 = 4 (x2+y2= 25) 6x+12y+ 45 = 21 6x+12y= 66, or x+2y= 11 You are left with two linear equations .

6 Solve this linear system to find the +4y= 13 x+2y= 116y= 24y= 4x= 3 The epicenter of the earthquake is 300 miles east and 400 miles south of Location 4 GOAL222yx123 Page 1 of Quadratic this statement: The equations x2+3y2 2y= 4 and x2+y2= 5 arean example of a(n) ? an example of a circle and a line intersecting in a single what method you would use to find the points of intersection of thegraphs in the following system . Do not solve the + y2 16x= 0 Equation 1x2 y2+ 7 = 0 Equation 2 Find the points of intersection, if any, of the graphs in the + y2= + y2+8x 20y+7=0y= x+3x2+ 9y2+8x+4y+7= + y2 3x= 2x+2y+ 2 = 02x2 y2= 10 x2+2x y+ 3 = back at Example 4. Why are three (not just two)seismographs needed to determine the location of the epicenter?

7 CHECKINGPOINTS OFINTERSECTIOND etermine whether the given point is apoint of intersection of the graphs in the + y2= + y2= + 4x 4y 16 = 0y= 3y= x 1 2x+ y+ 1 = 0 Point: ( 3, 4)Point: (4, 5)Point: (6, 11) 5y2+2y= 4y = 5x+8y2+y=23y= 2x+10y= 2x+3y= x 1 Point: ( 3, 4)Point: ( 5, 7)Point: (2, 1)SOLVINGSYSTEMSFind the points of intersection, if any, of the graphs in the y= + y2= 1817. 3x2+ y2= 9 3x+ y= 7x y= 0 2x+ y= + 4y2= + y2= + 2y2= 6 x+ y= 4y= 2xx+ 8y= + 3y2= 5y2= + 2y2= 15 x+ y= 13x+ y= 6x+ 2y= +y2= + y2= + y2= 5x+ y= 1y= x 4y = 3x + +y2= + y2= 7y= xx 3y = 3y = x 2x2= +3y2= y2= 6y= 2xy= x+2y = 2x+1 PRACTICEANDAPPLICATIONSGUIDEDPRACTICEV ocabulary Check Concept Check Skill Check STUDENTHELPHOMEWORK HELPE xample 1:Exs.

8 9 32 Examples 2, 3:Exs. 33 51 Example 4:Exs. 52 55,58 63 Extra Practiceto help you masterskills is on p. 1 of 2636 Chapter 10 Quadratic Relations and Conic SectionsSOLVINGSYSTEMSFind the points of intersection, if any, of the graphs in the + y2= 1634. 3x2+ y2 3x= 035. x2+ y2+ 10 = 0x2 5y= 5x2 y2+ 27 = 0 3y2+ x+ 1 = + 2y2 10 = + x + 4 = 04x y = 24x2 6 = +x= 16y2= +y2= 813x+ y= 8x2+y2= 9x+y= y2+ 16y 128 = y2 8x+ 8y 24 = 0y2 48x 16y 32 = 0x2+ y2 8x 8y + 24 = + 4y2 4x 8y + 4 = 56x+ 9y2+ 160 = 0x2+ 4y 4 = 04x2+ y2 64 = + y2 16x + 39 = 4y2 20x 64y 172 = 0x2 y2 9 = 04x2+ y2 80x+ 16y+ 400 = 2x+4+y2 10= y2 8x+6y 9=02y2 x+3=02x2 3y2+4x+18y 43 = 25y2 100x= y 4=0y2 2x+ 16 = 0x2+ 3y2 4y 10 = 0 Systems OFTHREEEQUATIONSFind the points, if any.

9 That the graphs of allthree equations have in + y2+8x+7= + y2 8=0x2+y2+4x+4y 5=0x2+ y2 3x+y=0x2+ y2= 12x2+ 2y2 5x 10= + 3y2= + y2 4x 4y= 263x2+ y2= 16x2+ y2 4x= 54y= xy=3x a line intersects a circle whose center is at theorigin, and the line passes through the origin. If you know one of the points ofintersection, how do you know what the other point of intersection is withoutsolving the system algebraically? examples to illustrate the different numbers ofpoints of intersection that a circle and an ellipse can have if both are centered atthe a car is traveling down the highway at aconstant rate of 60 miles per hour. It passes a police car parked at the side of theroad. To catch up to the car, the police officer accelerates at a constant rate.

10 Thedistance d (in miles) the police car has traveled as a function of time t(in hours)since the other car has passed it is given by d= 3600t2. Write and solve a system ofequations to calculate how long it takes the police car to catch up to the other range of a radio station is bounded by a circlegiven by the following equation:x2+ y2 1620 = 0A straight highway can be modeled by the following equation:y = 13 x+30 Find the length of the highway that lies within the range of the radio OFFICERThe duties of apolice officer vary. An offi-cer in a large city is oftenassigned to a specific typeof duty, while an officer in a small community usuallyperforms a variety of ONCAREERSPage 1 of Quadratic be eligible to ride the school bus to East HighSchool, a student must live at least 1 mile from the school.