Presentation on theme: "Chapter 2: Equations and Inequalities 2.3: Applications of Equations"— Presentation transcript:
1Chapter 2: Equations and Inequalities 2.3: Applications of Equations Essential Question: What are the different methods to solve a quadratic equation?
22.3 Applications of Equations Guidelines for solving applied problemsRead the problem carefully, and determine what is asked for.Label the unknown quantities with variablesDraw a picture of the situation, if appropriateTranslate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language.Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed.Solve for at least one of the unknown quantities.Find all remaining unknown quantities by using the relationships given in the problem.Check and interpret all quantities found in the original problem.
32.3 Applications of Equations Example 1: Number RelationsThe average of two real numbers is , and their product is Find the two numbers.SolutionTwo equations:Solve one equation for one variable, and then substitute.
42.3 Applications of Equations Example 1 (Continued)a a+1683=0Option 1 → GraphAnswers are where the graph crosses the x-axisOption 2 → Ye old Quadratic Equation
52.3 Applications of Equations Example 1 (Finishing)Check the answersIf a = 44Plug into the ab=1683 equation to find that b=38.25If a = 38.25Plug into the same function to find that b=44CheckThe average of 44 & isThe two numbers are 44 and 38.25
62.3 Applications of Equations Example 2: Dimensions of a RectangleA rectangle is twice as wide as it is high. If it has an area of 24.5 square inches, what are its dimensions?Two equationsSubstitute and solve
72.3 Applications of Equations Example 2 (Continued)It’s not possible to have a negative height, so the only value we have to check is h=3.5 inchesIf the width is twice the height, then the width = 2(3.5) = 7 inchesCheck your answer: (3.5 in)(7 in) = 24.5 in2The width is 7 inches and the height is 3.5 inches
82.3 Applications of Equations Example 4: Interest ApplicationsI = PrtI = InterestP = Principal (initial invested amount)r = rate (percentage written as a decimal)t = time (in years)A high-risk stock pays dividends at a rate of 12% per year, and a savings account pays 6% interest per year. How much of a $9000 investment should be put in the stock and how much should be put in savings to obtain a return of 8% per year on the total investment?
92.3 Application of Equations Example 4: Interest Applications (continued)Let s be the amount invested in stock. The rest of the money ($9000 – s) is the amount invested in savingsTranslate English into math:Individual interest added together = total interest earned(stock at 12%) + (savings at 6%) = 8% of $90000.12s (9000 – s) = 0.08(9000) [Distribute]0.12s – 0.06s = 720 [Combine like terms]0.06s =0.06s ÷ 0.06 = 180 ÷ 0.06s = 3000
10Application of Equations Example 5: Distance Applicationsd = rtd = distancer = ratet = timeYou can convert the equation if necessaryr = d/tt = d/rA pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30mph going to Peoria, and it is estimated that there will be a 40mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown?
112.3 Application of Equations Example 5: Distance Applications (continued)Let r be the engine speed of the planeHeadwind slows the velocity by 30 mphTailwind increases the velocity by 40 mphCleveland to PeoriaDistance = 420Actual velocity = r – 30Time = 420/(r – 30)Peoria to ClevelandActual velocity = r + 40Time = 420/(r + 40)
12Application of Equations Example 5: Distance Applications (continued)The total time is going to be 5 hours, so
132.3 Application of Equations Example 5: Distance Applications (concluded)r r = 0This can be factored (you can use the quadratic equation as well of course)(r – 170)(r + 12) = 0r – 170 = 0 or r + 12 = 0r = 170 or r = -12Because you can’t fly at a negative rate, the plane must fly at a constant rate of 170 mph
142.3 Application of Equations Example 8: Mixture ProblemA car radiator contains 12 quarts of fluid, 20% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the resulting mixture is 50% antifreeze?Let x be the number of quarts of fluid to be replaced by pure antifreeze.When x quarts are drained, there are 12 – x quarts of fluid left in the radiator, 20% of which is antifreeze.
152.3 Application of Equations Example 8: Mixture Problem (continued)Translate English into math20% of (12 – x) + x = 50% of 120.2(12 – x) + x = 0.5(12) [Distribute]2.4 – 0.2x + x = 6 [Combine like terms]0.8x – 2.4 = 6 – 2.40.8x ÷ 0.8 = 3.6 ÷ 0.8x = 4.54.5 quarts should be drained and replaced with pure antifreeze
162.3 Applications of Equations AssignmentPagesWednesday: 9, 15, 17, 25Thursday: 11, 13, 19, 21, 23For #25, you’re going to want to solve it by graphing and finding the x-intercept(s). Make sure you alter your window so that you can see a solution.