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COLLEGE ALGEBRA

Explore real-world problems involving quadratic equations and geometric concepts, with step-by-step solutions and practical examples. From volume calculations to Pythagorean theorem applications, sharpen your problem-solving skills with this comprehensive guide.

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COLLEGE ALGEBRA

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  1. COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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

  3. Geometry Problems Problem Solving When solving problems that lead to quadratic equations, we may get a solution that does not satisfy the physical constraints of the problem. For example if x represents a width and the two solutions of the quadratic equation are –9 and 1, the value –9 must be rejected since a width must be a positive number.

  4. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 A piece of machinery is capable of producing 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 their 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, what should the dimensions of the original piece of metal be?

  5. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 Solution Step 1 Read the problem. We must find the dimensions of the original piece.

  6. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 Solution 3x Step 2 Assign a variable. We know the length is three times the width, so let x = width (in inches) and thus 3x = the length. The box is formed by cutting 5 + 5 = 10 in. from both the length and the width. 5 3x –10 5 5 5 5 x x –10 x – 10 5 5 5 5 3x – 10

  7. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 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

  8. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 Solution Step 3 Write an equation. The formula for volume of a box is V = lwh. Volume length  width  height = These are both satisfied when x > 10.

  9. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 Solution Step 4 Solve the equation. Multiply. Subtract. Divide by 5. Factor.

  10. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 Solution Step 4 Solve the equation. Factor. or Zero-factor property or Solve. A length cannot be negative.

  11. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 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.

  12. SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX Example 1 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

  13. Using Pythagorean Theorem Pythagorean theorem from geometry is used in Example 2. The two sides that meet at the right angle are the legs and the side opposite the right angle is the hypotenuse. Hypotenuse c Leg a Leg b

  14. 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. Hypotenuse c Leg a Leg b

  15. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Erik Van Erden finds a piece of property in the shape of a right triangle. He finds that 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.

  16. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Solution Step 1 Read the problem. We must find the lengths of the three sides. Step 2 Assign a variable. Let s = the length of the shorter leg (in meters). Then 2s + 20 = length of the longer leg, and (2s + 20) + 10 or 2s + 30 = the length of the hypotenuse. 2s + 30 s 2s + 20

  17. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Solution Step 3 Write an equation. Be sure to substitute correctly here. Pythagorean theorem

  18. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Solution Step 4 Solve the equation. Remember the middle term here. Remember the middle term here. Square the binomials.

  19. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Solution Step 4 Solve the equation. Standard form Factor. or Zero-factor property or Solve.

  20. SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM Example 2 Solution Step 5 State the answer. Since s 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.

  21. 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 s0 feet, with initial velocity v0 feet per second is where t is the number of seconds after the object is projected. The coefficient of t2, –16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets.

  22. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 If a projectile is shot 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

  23. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 a. After how many seconds will it be 50 ft above the ground? Solution We must find the values of t so that height s is 50 ft. Let s = 50 in the given equation. Standard form Divide by 2.

  24. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 a. After how many seconds will it be 50 ft above the ground? Solution Divide by 2. Quadratic formula

  25. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 a. After how many seconds will it be 50 ft above the ground? Solution or Use a calculator.

  26. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 a. After how many seconds will it be 50 ft above the ground? Solution or Both solutions are acceptable, since the projectile reaches 50 ft twice: once on the way up (after .55 sec) and once on the way down (after 5.70 sec).

  27. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 b. How long will it take for the projectile to return to the ground? SolutionWhen the projectile returns to the ground, the height s will be 0 ft, so let s = 0 in the given equation. Factor. or Zero-factor property or Solve.

  28. SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 3 b. When the projectile returns to the ground, the height s will be 0 ft, so let s = 0 in the given equation. 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.

  29. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 The bar graph shows sales of SUVs in the United States, in millions. The quadratic equation models sales of SUVs from 1992 to 2003, where S represents sales in millions, and x = 0 represents 1992, x = 1 represents 1993 and so on.

  30. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 a. Use the model to determine sales in 2002 and 2003. Compare the results to the actual figures of 4.2 million and 4.4 million. Solution Original model For 2002, x = 10.

  31. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 a. Use the model to determine sales in 2002 and 2003. Compare the results to the actual figures of 4.2 million and 4.4 million. Solution Original model For 2003, x = 11.

  32. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 a. Use the model to determine sales in 2002 and 2003. Compare the results to the actual figures of 4.2 million and 4.4 million. Solution The prediction is .1 million less than the actual figure of 4.2 in 2002 and .1 million more than the actual prediction of 4.4 million in 2003.

  33. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 b. According to the model, in what year do sales reach 3.5 million? (Round down to the nearest year.) Is the result accurate? Solution Let S = 3.5 in the original model. Standard form

  34. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 b. According to the model, in what year do sales reach 3.5 million? (Round down to the nearest year.) Is the result accurate? Solution Standard form Quadratic formula

  35. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 b. According to the model, in what year do sales reach 3.5 million? (Round down to the nearest year.) Is the result accurate? Solution Quadratic formula or

  36. ANALYZING SPORT UTILITY (SUV) VEHICLES Example 4 b. According to the model, in what year do sales reach 3.5 million? (Round down to the nearest year.) Is the result accurate? Solution Reject the negative solution and round 8.3 down to 8. The year 2000 corresponds to x = 8. Thus, according to the model, the number of SUVs reached 3.5 million in the year 2000. The model closely matches the graph, so it is accurate.

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