Homework 13 Quadratic Equation Word Problems

Decoding the Mysteries: Mastering Quadratic Equation Word Problems

Quadratic equations, with their elegant curves and multiple solutions, often appear in the form of word problems. These problems can initially seem daunting, but with a systematic approach and a solid understanding of quadratic equations, you can unlock their secrets and confidently arrive at the correct answers. This guide will walk you through the essential steps, provide practical examples, and equip you with the tools to conquer even the most challenging quadratic equation word problems.

Understanding the Fundamentals

Before diving into the problems themselves, it's crucial to revisit the core concepts of quadratic equations. A quadratic equation is a polynomial equation of the second degree, generally represented in the standard form as:

ax² + bx + c = 0

where a, b, and c are constants and a ≠ 0.

Key Concepts:

• Roots/Solutions: The values of x that satisfy the equation. A quadratic equation can have two distinct real roots, one real root (a repeated root), or two complex roots.

• Methods for Solving: There are several methods to solve quadratic equations:

  • Factoring: Breaking down the quadratic expression into two linear factors. This method is efficient when the factors are easily identifiable.

  • Completing the Square: Transforming the equation into a perfect square trinomial, allowing you to isolate x.

  • Quadratic Formula: A universal formula that provides the solutions for any quadratic equation:

    x = (-b ± √(b² - 4ac)) / 2a

  • Graphing: Finding the x-intercepts of the parabola represented by the quadratic equation.

• Discriminant: The part of the quadratic formula under the square root, b² - 4ac, determines the nature of the roots:

  • If b² - 4ac > 0: Two distinct real roots
  • If b² - 4ac = 0: One real root (repeated)
  • If b² - 4ac < 0: Two complex roots

The Art of Translating Words into Equations

The biggest hurdle in solving word problems is translating the given information into mathematical equations. Here's a breakdown of the process:

1. Read Carefully and Understand: Read the problem multiple times, paying close attention to the details and what the problem is asking you to find. Identify the knowns and unknowns.
2. Assign Variables: Represent the unknown quantities with variables. Often, x is used, but you can choose any letter that helps you remember what it represents. Define your variables clearly. For example: "Let x be the width of the rectangle."
3. Formulate the Equation: Use the information provided in the problem to create a quadratic equation that relates the variables. Look for keywords and phrases that suggest mathematical operations, such as:
   • "Sum": Addition (+)
   • "Difference": Subtraction (-)
   • "Product": Multiplication (*)
   • "Quotient": Division (/)
   • "Is/Equals": Equal (=)
   • "Square": Raising to the power of 2
4. Solve the Equation: Use one of the methods mentioned earlier (factoring, completing the square, quadratic formula) to find the values of the variable(s).
5. Check Your Answer: Substitute the solutions back into the original word problem to ensure they make sense in the context of the problem. Discard any solutions that are not realistic (e.g., negative lengths).
6. State Your Answer Clearly: Answer the question asked in the problem, including the appropriate units.

Example Problems and Solutions

Let's illustrate these steps with several example problems:

Problem 1: The Area of a Rectangle

The length of a rectangle is 5 meters more than its width. The area of the rectangle is 84 square meters. Find the length and width of the rectangle.

Solution:

1. Understand: We need to find the length and width of a rectangle given the relationship between them and the area.
2. Variables:
   • Let w be the width of the rectangle (in meters).
   • Then, the length l is w + 5 (in meters).
3. Equation: The area of a rectangle is given by A = l * w. We know A = 84, so: (w + 5)w = 84 w² + 5w = 84 w² + 5w - 84 = 0
4. Solve: We can factor this quadratic equation: (w + 12)(w - 7) = 0 This gives us two possible solutions for w:
   • w = -12
   • w = 7
5. Check: Since the width cannot be negative, we discard w = -12. Therefore, w = 7 meters. The length is l = w + 5 = 7 + 5 = 12 meters.
6. Answer: The width of the rectangle is 7 meters, and the length is 12 meters.

Problem 2: The Consecutive Integers

The product of two consecutive positive integers is 132. Find the two integers.

Solution:

1. Understand: We need to find two consecutive positive integers whose product is 132.
2. Variables:
   • Let n be the first integer.
   • Then, the next consecutive integer is n + 1.
3. Equation: The product of the two integers is 132: n(n + 1) = 132 n² + n = 132 n² + n - 132 = 0
4. Solve: We can factor this quadratic equation: (n + 12)(n - 11) = 0 This gives us two possible solutions for n:
   • n = -12
   • n = 11
5. Check: Since the problem specifies positive integers, we discard n = -12. Therefore, n = 11. The next consecutive integer is n + 1 = 11 + 1 = 12.
6. Answer: The two consecutive positive integers are 11 and 12.

Problem 3: Projectile Motion

A ball is thrown vertically upwards from a height of 2 meters with an initial velocity of 20 m/s. The height h (in meters) of the ball after t seconds is given by the equation h = -5t² + 20t + 2. At what time(s) will the ball be 17 meters above the ground?

Solution:

1. Understand: We need to find the time(s) when the ball's height is 17 meters.
2. Variables:
   • h = height (meters)
   • t = time (seconds)
3. Equation: We are given the equation h = -5t² + 20t + 2. We want to find t when h = 17: 17 = -5t² + 20t + 2 0 = -5t² + 20t - 15 0 = t² - 4t + 3 (Dividing by -5)
4. Solve: We can factor this quadratic equation: (t - 1)(t - 3) = 0 This gives us two possible solutions for t:
   • t = 1
   • t = 3
5. Check: Both solutions are positive and make sense in the context of the problem.
6. Answer: The ball will be 17 meters above the ground at 1 second and 3 seconds.

Problem 4: The Pythagorean Theorem

The hypotenuse of a right triangle is 13 cm. One leg is 7 cm longer than the other. Find the lengths of the legs.

Solution:

1. Understand: We need to find the lengths of the legs of a right triangle given the hypotenuse and the relationship between the legs.
2. Variables:
   • Let x be the length of the shorter leg (in cm).
   • Then, the length of the longer leg is x + 7 (in cm).
3. Equation: Using the Pythagorean Theorem (a² + b² = c²), where c is the hypotenuse: x² + (x + 7)² = 13² x² + (x² + 14x + 49) = 169 2x² + 14x + 49 = 169 2x² + 14x - 120 = 0 x² + 7x - 60 = 0 (Dividing by 2)
4. Solve: We can factor this quadratic equation: (x + 12)(x - 5) = 0 This gives us two possible solutions for x:
   • x = -12
   • x = 5
5. Check: Since the length cannot be negative, we discard x = -12. Therefore, x = 5 cm. The longer leg is x + 7 = 5 + 7 = 12 cm.
6. Answer: The lengths of the legs are 5 cm and 12 cm.

Problem 5: The Border Around a Garden

A rectangular garden is 8 meters long and 6 meters wide. A path of uniform width is built around the garden. The area of the path is equal to the area of the garden. Find the width of the path.

Solution:

1. Understand: We need to find the width of a path around a rectangular garden, given the dimensions of the garden and the relationship between the areas.

2. Variables:

   • Let w be the width of the path (in meters).

3. Equation:

   • Area of the garden: 8 * 6 = 48 square meters.
   • Dimensions of the garden plus path: length = 8 + 2w, width = 6 + 2w
   • Area of the garden plus path: (8 + 2w)(6 + 2w)
   • Area of the path: (8 + 2w)(6 + 2w) - 48

   The problem states the area of the path is equal to the area of the garden, so:

   (8 + 2w)(6 + 2w) - 48 = 48 (8 + 2w)(6 + 2w) = 96 48 + 16w + 12w + 4w² = 96 4w² + 28w + 48 = 96 4w² + 28w - 48 = 0 w² + 7w - 12 = 0 (Dividing by 4)

4. Solve: This quadratic equation doesn't factor easily, so we'll use the quadratic formula:

   w = (-b ± √(b² - 4ac)) / 2a w = (-7 ± √(7² - 4 * 1 * -12)) / (2 * 1) w = (-7 ± √(49 + 48)) / 2 w = (-7 ± √97) / 2

   This gives us two possible solutions for w:

   • w = (-7 + √97) / 2 ≈ 1.42
   • w = (-7 - √97) / 2 ≈ -8.42

5. Check: Since the width cannot be negative, we discard w ≈ -8.42. Therefore, w ≈ 1.42 meters.

6. Answer: The width of the path is approximately 1.42 meters.

Strategies for Tackling Complex Problems

While the examples above provide a foundation, some quadratic equation word problems can be more intricate. Here are some strategies for tackling those:

• Draw a Diagram: Visualizing the problem with a diagram can be extremely helpful, especially for geometry-related problems. Label the diagram with the given information.
• Break Down the Problem: Divide the problem into smaller, more manageable parts. Identify the key relationships and information needed to form the equation.
• Use a Table: For problems involving rates, time, and distance, creating a table can help organize the information and identify the equation.
• Look for Hidden Information: Sometimes, the problem may not explicitly state all the necessary information. Look for clues or assumptions that can be used to derive additional equations or relationships.
• Practice, Practice, Practice: The more you practice solving different types of quadratic equation word problems, the better you will become at recognizing patterns and applying the appropriate techniques.

Common Mistakes to Avoid

• Incorrectly Translating Words: Pay close attention to the wording of the problem and ensure you are accurately translating the information into mathematical expressions.
• Forgetting Units: Always include the appropriate units in your answer.
• Not Checking Your Answer: Always substitute your solutions back into the original word problem to ensure they make sense and are realistic.
• Ignoring Negative Solutions: While negative solutions are often discarded in problems involving physical quantities like length or time, be careful to consider the context of the problem. Sometimes, a negative solution might have a valid interpretation.
• Choosing the Wrong Method: While the quadratic formula always works, factoring or completing the square can be more efficient for certain types of equations. Choose the method that best suits the problem.

Real-World Applications

Quadratic equations are not just abstract mathematical concepts; they have numerous real-world applications in various fields, including:

• Physics: Projectile motion, calculating the trajectory of objects thrown into the air.
• Engineering: Designing bridges, optimizing structures, and analyzing circuits.
• Economics: Modeling supply and demand curves, calculating profit maximization.
• Computer Science: Creating algorithms, developing game physics.
• Finance: Calculating compound interest, analyzing investments.

By mastering quadratic equation word problems, you are not only developing your mathematical skills but also gaining valuable problem-solving abilities that can be applied to a wide range of real-world scenarios.

Conclusion

Quadratic equation word problems can be challenging, but with a systematic approach, a solid understanding of the underlying concepts, and plenty of practice, you can conquer them with confidence. Remember to read carefully, translate accurately, solve methodically, and check thoroughly. By following the steps outlined in this guide and practicing regularly, you will be well-equipped to decode the mysteries and unlock the solutions to even the most complex quadratic equation word problems. So, embrace the challenge, persevere through the difficulties, and enjoy the satisfaction of mastering this important mathematical skill. Good luck!