Mathematical Problem Solving: Quadratic Equations and Geometry Examples

1. Chabot Mathematics §8.2 QuadraticEquation Apps Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. MTH 55 8.2 Review § • Any QUESTIONS About • §8.2 → Complete-the-Square • Any QUESTIONS About HomeWork • §8.2 → HW-37

3. §8.2 Quadratic Formula • The Quadratic Formula • Problem Solving with the Quadratic Formula

4. The Quadratic Formula • The solutions of ax2 + bx + c = 0 are given by This is one of theMOST FAMOUSFormulas in allof Mathematics

5. Example Circular WalkWay • A Circular Pond 10 feet in diameter is to be surrounded by a “Paver” walkway that will be 2 feet wide. Find the AREA of the WalkWay • Familiarize: Recall the Formula for the area, A ,of a Circle based on it’s radius, r • Also the diameter, d, is half of r. Thus A in terms of d

6. Example Circular WalkWay • Familiarize: Makea DIAGRAM • Translate: Use Diagram of Subtractive Geometry = −

7. Example Circular WalkWay • Translate: Diagram to Equation = −

8. Example Circular WalkWay • CarryOUT: Solve Eqn for Awalk • Using π ≈ 3.14 find

9. Example Circular WalkWay • Check: Use Acircle = πr2  • State: The Area of the Paver Walkway is about 75.4 ft2 • Note that UNITS must be included in the Answer Statement

10. Example  Partition Bldg • A rectangular building whose depth (from the front of the building) is three times its frontage is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall. If the rear portion of the building contains 2100 square feet, find the dimensions of the building. • Familiarize: REALLY needs a Diagram

11. Example  Partition Bldg • Familiarize: by Diagram • Now LET x≡ frontage of building, in feet. • Translate: The other statements into Equations involving x

12. Example  Partition Bldg • The Bldg depth is three times its frontage, x→ 3x = depth of building, in feet • The Bldg Depth is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall → 3x – 45 = depth of rear portion, in ft

13. Example  Partition Bldg • Now use the 2100 ft2Area Constraint • Area of rear = 2100 • Area = x(3x−45), so

14. Example  Partition Bldg • So x is either 35ft or −20ft • But again Distances canNOT be negative • Thus x = 35 ft • Check: Use 2100 ft2 Area 

15. Example  Partition Bldg • State: • The Bldg Frontage is 35ft • The Bldg Depth is 3(35ft) = 105ft 60’ 105’ 35’

16. Example  Bike Tire BlowOut • Devon set out on a 16 mile Bike Ride. Unfortunately after 10 miles of Biking BOTH tires Blew Out. Devon Had to complete the trip on Foot. • Devon biked four miles per hour (4 mph) faster than she walked • The Entire journey took 2hrs and 40min Find Devon’s Walking Speed (or Rate)

17. Example  Bike Tire BlowOut • Familiarize: Make diagram • LET w≡ Devon’s Walking Speed • Recall the RATE Equation for Speed Distance = (Speed)·(Time)

18. Example  Bike Tire BlowOut • Translate: The BikingSpeed, b, is 4 mphfaster than theWalking Speed → • From the Diagram note Distances by Rate Equation: • Biking Distance = 10 miles = b·tbike • Walking Distance = 6 miles = w·twalk

19. Example  Bike Tire BlowOut • Translate: Now the Total Distance of16mi is the sum of theBiking & Walking Distances → • From the Spd Eqn: Time = Dist/Spd

20. Example  Bike Tire BlowOut • Translate: Thus bySpeed Eqn: • Next, the sum of the Biking and Walking times is 2hrs & 40min = 2-2/3hr →

21. Example  Bike Tire BlowOut • CarryOut: Clear Fractions in the last Eqn by multiplying by the LCD of 3·(w+4) ·w:

22. Example  Bike Tire BlowOut • CarryOut: Divide the last Eqn by 8 to yield a Quadratic Equation • This Quadratic eqn does NOT factor so use Quadratic Formula with a = 1, b = −2, and c = −9

23. Example  Bike Tire BlowOut • CarryOut: find w by Quadratic Formula • So Devon’s Walking Speed is

24. Example  Bike Tire BlowOut • CarryOut: Since SPEED can NOT be Negative find: • Check: Test to see that the time adds up to 2.67 hrs 

25. Example  Bike Tire BlowOut • State: Devon’s Walking Speed was about 4.16 mph (quite a brisk pace)

26. Example  Golden Rectangle • Let’s Revisit the Derivation of the GOLDEN RATIO • A rectangle of length p and width q, with p > q, is called a golden rectangle if you can divide the rectangle into a square with side of length q and a smaller rectangle that is geometrically-similar to the original one. • The GOLDEN RATION then = p/q

27. Example  Golden Rectangle • Familiarize: Make a Diagram

28. Example  Golden Rectangle • Translate: Use Diagram

29. Example  Golden Rectangle • CarryOut: LETΦ ≡ GoldenRatio= p/q

30. Example  Golden Rectangle • Carry Out: Since both p & q are distances they are then both POSITIVE • Thus Φ = p/q must be POSITIVE • State: GOLDEN RATIO as defined by the Golden Rectangle

31. Example  Pythagorus • The hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs Familarize.First make a drawing and label it. Lets = length, in yards, of one leg. Then s + 28 = the length, in yards, of the other leg. 52 s s + 28

32. 52 s s + 28 Pythagorean Triangle • Translate. We use the Pythagorean theorem:s2 + (s + 28)2 = 522 Carry Out. Identify the Quadratic Formula values a, b, & c s2 + (s + 28)2 = 522 s2 + s2 + 56s + 784 = 2704 2s2 + 56s− 1920 = 0 s2 + 28s − 960 = 0

33. Pythagorean Triangle • Carry Out: With s2 + 28s− 960 = 0 Find: a = 1, b = 28, c = −960 Evaluate the Quadratic Formula

34. 52 s s + 28 Pythagorean Triangle • Carry Out: Continue Quadratic Eval

35. Pythagorean Triangle • Check. Length cannot be negative, so −48 does not check. • State. One leg is 20 yards and the other leg is 48 yards. 52 yds s = 20 yds s + 28yd = 48 yds

36. Vertical Ballistics • In Physics 4A, you will learn the general formula for describing the height of an object after it has been thrown upwards: • Where • g≡ the acceleration due to gravity • A CONSTANT = 32.2 ft/s2 = 9.81 m/s2 • t≡ the time in flight, in s • v0≡ the initial velocity in ft/s or m/s • h0≡ the initial height in ft or m

37. Example  X-Games Jump • In an extreme games competition, a motorcyclist jumps with an initial velocity of 80 feet per second from a ramp height of 25 feet, landing on a ramp with a height of 15 feet. • Find the time the motorcyclist is in the air.

38. Example  X-Games Jump • Familiarize: We are given the initial velocity, initial height, and final height and we are to find the time the bike is in the air. Can use the Ballistics Formula • Translate: Use the formula with h = 15, v0 = 80, and h0 = 25. • Use h = 15, v0 = 80, and h0 = 25

39. Example  X-Games Jump • CarryOut: • Use the Quadratic Formula to find t • a = −16.1, b = 80 and c = 10

40. Example  X-Games Jump • CarryOut: • Since Times can NOT be Negative t ≈ 5.09 seconds • Check: Check by substuting 5.09 for t in the ballistics Eqn. • The Details are left for later • State: The MotorCycle Flight-Time is very nearly 5.09 seconds

41. Example  Biking Speed • Kang Woo (KW to his friends) traveled by Bicycle 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less. Find KW’s average speed

42. Example  Biking Speed • Familiarize: In this can tabulating the information can help. Let r represent the rate, in miles per hour, and t the time, in hours for Kang Woo’s trip • Uses the Rate/Spd Eqn → Rate = Qty/Time

43. Example  Biking Speed • Translate: From the Table Obtain two Equations in r & t and • Carry out: A system of equations has been formed. We substitute for r from the first equation into the second and solve the resulting equation

44. Example  Driving Speed • Carry out: Next Clear Fractions Multiplying by the LCD

45. Example  Driving Speed • Carry out: The Last Eqn is Quadratic in t: • Solve by Quadratic Forumula with: • a = 1, b = −1, and c = −12

46. Example  Driving Speed • Carry out: By Quadratic Formula • Since TIMES can NOT be NEGATIVE, then t = 4 hours • Return to one of the table Eqns to find r 12 mph.

47. Example  Driving Speed • Check: To see if 12 mph checks, we increase the speed 4 mph to16 mph and see how long the trip would have taken at that speed: • The Answer checks a 3hrs is indeed 1hr less than 4 hrs that the trip actually took • State: KW rode his bike at an average speed of 12 mph

48. ReCall The WORK Principle • Suppose that A requires a units of time to complete a task and B requires b units of time to complete the same task. • Then A works at a rate of 1/a tasks per unit of time. • B works at a rate of 1/b tasks per unit of time, • Then A and B together work at a rate of [1/a + 1/b] per unit of time.

49. The WORK Principle • If A and B, working together, require t units of time to complete a task, then their rate is 1/t and the following equations hold:

50. Example  Empty Tower Tank • A water tower has two drainpipes attached to it. Working alone, the smaller pipe would take 20 minutes longer than the larger pipe to empty the tower. If both drainpipes work together, the tower can be drained in 40 minutes. • How long would it take the small pipe, working alone, to drain the tower?