Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 1 Equations and Inequalities Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2 1.5 Applications and Modeling with Quadratic Equations Geometry Problems Using the Pythagorean Theorem Height of a Projected Object Modeling with Quadratic Equations

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Example 1 SOLVING A PROBLEM INVOLVING VOLUME A piece of machinery produces rectangular sheets of metal such that the length is three times the width. Equal-sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in. 3, find the dimensions of the original piece of metal.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Step 1 Read the problem. We must find the dimensions of the original piece of metal.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Step 2 Assign a variable. We know that the length is three times the width. Let x = the width (in inches) and thus, 3x = the length. The box is formed by cutting = 10 in. from both the length and the width. Solution x – 10 3x – x – 10 x 3x3x

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Step 3 Write an equation. The formula for the volume of a box is V = lwh. Volume = length  width  height (Note that the dimensions of the box must be positive numbers, so 3x – 10 and x – 10 must be greater than 0, which implies and These are both satisfied when x > 10.)

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Multiply. Step 4 Solve the equation. Subtract 1435 from each side. Divide each side by 5. Factor.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Step 4 Solve the equation. or Zero-factor property or Solve each equation. The width cannot be negative.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Step 5 State the answer. Only 17 satisfies the restriction x > 10. Thus, the dimensions of the original piece should be 17 in. by 3(17) = 51 in.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Example 1 SOLVING A PROBLEM INVOLVING VOLUME Solution Step 6 Check. The length of the bottom of the box is 51 – 2(5) = 41 in. The width is 17 – 2(5) = 7 in. The height is 5 in. (the amount cut on each corner), so the volume of the box is

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Using the Pythagorean Theorem Example 2 requires the use of the Pythagorean theorem for right triangles. Recall that the legs of a right triangle form the right angle, and the hypotenuse is the side opposite the right angle. Leg a Leg b Hypotenuse c

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. Leg a Leg b Hypotenuse c

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 2 APPLYING THE PYTHAGOREAN THEOREM A piece of property has the shape of a right triangle. The longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 2 APPLYING THE PYTHAGOREAN THEOREM Solution Step 1 Read the problem. We must find the lengths of the three sides. Step 2 Assign a variable. Let x = the length of the shorter leg (in meters). Then 2x + 20 = the length of the longer leg, and (2x + 20) + 10 or 2x + 30 = the length of the hypotenuse. x 2x x + 30

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 2 APPLYING THE PYTHAGOREAN THEOREM Solution Step 3 Write an equation. The hypotenuse is c. Substitute into the Pythagorean theorem.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 2 APPLYING THE PYTHAGOREAN THEOREM Solution Remember the middle term here. Square the binomials. Step 4 Solve the equation. Remember the middle term here.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Example 2 APPLYING THE PYTHAGOREAN THEOREM Solution Standard form or Factor. Zero-factor property or Solve each equation. Step 4 Solve the equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 2 APPLYING THE PYTHAGOREAN THEOREM Solution Step 5 State the answer. Since x represents a length, – 10 is not reasonable. The lengths of the sides of the triangular lot are 50 m, 2(50) + 20 = 120 m, and 2(50) + 30 = 130 m. Step 6 Check. The lengths 50, 120, and 130 satisfy the words of the problem, and also satisfy the Pythagorean theorem.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Height of a Projected Object If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height of s 0 feet, with initial velocity v 0 feet per second is given by the equation Here t represents the number of seconds after the object is projected. The coefficient of t 2, – 16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT If a projectile is launched vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT Solution (a) After how many seconds will it be 50 ft above the ground? We must find value(s) of t so that height s is 50 ft. Standard form Divide by −2. Let s = 50 in the given equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT Solution (a) After how many seconds will it be 50 ft above the ground? Quadratic formula Simplify. or Use a calculator.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT Solution (a) After how many seconds will it be 50 ft above the ground? or Both solutions are acceptable, since the projectile reaches 50 ft twice—once on its way up (after 0.55 sec) and once on its way down (after 5.70 sec).

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 (b) How long will it take for the projectile to return to the ground? Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT Solution When the projectile returns to the ground, the height s will be 0 ft. Factor. or Zero-factor property or Solve each equation. Let s = 0.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 (b) How long will it take for the projectile to return to the ground? Example 3 SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT Solution The first solution, 0, represents the time at which the projectile was on the ground prior to being launched, so it does not answer the question. The projectile will return to the ground 6.25 sec after it is launched.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 4 The I-Ride Trolley service carries passengers along the International Drive Resort Area of Orlando, Florida. The bar graph shows I-Ride Trolley ridership in millions. The quadratic equation ANALYZING TROLLEY RIDERSHIP models ridership from 2000 to 2010, where y represents ridership in millions and x = 0 represents 2000, x = 1 represents 2001, and so on.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 (a) Use the model to determine ridership in Compare the result to the actual ridership figure of 2.1 million. Example 4 ANALYZING TROLLEY RIDERSHIP Solution Since x = 0 represents the year 2000, use x = 8 to represent Given model Let x = 8. Use a calculator. The prediction is about 0.1 million (that is, 100,000) less than the actual figure of 2.1 million.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 4 ANALYZING TROLLEY RIDERSHIP Solution Let y = 1.9 in the model. Standard form (b) According to the model, in what year did ridership reach 1.9 million? Quadratic formula or Use a calculator.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 The year 2003 corresponds to x = 3.3. Thus, according to the model, ridership reached 1.9 million in the year This outcome closely matches the bar graph and seems reasonable. The year 2011 corresponds to x = The model predicts that ridership will be 1.9 million again in the year Example 4 ANALYZING TROLLEY RIDERSHIP Solution (b) According to the model, in what year did ridership reach 1.9 million?