How To Solve Quadratic Equation Word Problems

Solving quadratic equation word problems can seem daunting, but with a structured approach and a solid understanding of quadratic equations, you can tackle these problems with confidence. This guide breaks down the process into manageable steps, providing examples and strategies to master this skill.

Understanding Quadratic Equations

A quadratic equation is a polynomial equation of the second degree. The general form is:

ax² + bx + c = 0

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

The solutions to a quadratic equation are also called roots or zeros. These are the values of x that satisfy the equation. Quadratic equations often have two solutions, although they can have one (a repeated root) or none (complex roots).

Key Concepts to Remember

• Factoring: Breaking down a quadratic expression into the product of two binomials.

• Quadratic Formula: A formula to find the solutions of any quadratic equation:

  x = (-b ± √(b² - 4ac)) / 2a

• Completing the Square: A method to convert a quadratic equation into a perfect square trinomial.

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

  • If b² - 4ac > 0, there are two distinct real roots.
  • If b² - 4ac = 0, there is one real root (a repeated root).
  • If b² - 4ac < 0, there are no real roots (two complex roots).

Steps to Solve Quadratic Equation Word Problems

Here’s a step-by-step approach to solving quadratic equation word problems:

1. Read and Understand the Problem:
   • Carefully read the problem multiple times.
   • Identify what the problem is asking you to find.
   • Determine the knowns (given information) and unknowns (variables to solve for).
2. Assign Variables:
   • Choose appropriate variables to represent the unknown quantities.
   • Clearly define what each variable represents.
3. Translate the Problem into an Equation:
   • Express the relationships described in the word problem as a quadratic equation.
   • Look for keywords like "sum," "difference," "product," "area," "square," etc., to help translate the words into mathematical operations.
4. Solve the Quadratic Equation:
   • Use factoring, the quadratic formula, or completing the square to find the solutions.
   • Choose the method that seems most efficient for the given equation.
5. Check Your Solutions:
   • Substitute the solutions back into the original equation to verify they are correct.
   • Make sure the solutions make sense in the context of the word problem. Discard any solutions that are not realistic (e.g., negative lengths, negative time).
6. State the Answer:
   • Clearly state the answer in the context of the original word problem, including appropriate units.

Example Problems and Solutions

Let's walk through several example problems to illustrate these steps.

Example 1: Area of a Rectangle

Problem: The length of a rectangle is 5 meters more than its width. If the area of the rectangle is 84 square meters, find the length and width of the rectangle.

Solution:

1. Understand the Problem:
   • We need to find the length and width of a rectangle given its area and the relationship between its length and width.
2. Assign Variables:
   • Let w represent the width of the rectangle (in meters).
   • Then, the length l is w + 5 meters.
3. Translate into an Equation:
   • The area of a rectangle is given by A = l * w*.
   • We have A = 84, l = w + 5.
   • So, the equation is: (w + 5)w = 84.
4. Solve the Equation:
   • Expand the equation: w² + 5w = 84.
   • Rearrange to standard form: w² + 5w - 84 = 0.
   • Factor the quadratic equation: (w + 12)(w - 7) = 0.
   • The solutions are w = -12 and w = 7.
5. Check the Solutions:
   • Since width cannot be negative, w = -12 is not a valid solution.
   • So, w = 7 meters.
   • Then, l = w + 5 = 7 + 5 = 12 meters.
   • Check: Area = l * w* = 12 * 7 = 84 square meters (which matches the given area).
6. State the Answer:
   • The width of the rectangle is 7 meters, and the length is 12 meters.

Example 2: Projectile Motion

Problem: A ball is thrown vertically upward 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 the Problem:
   • We need to find the time(s) when the ball's height is 17 meters.
2. Assign Variables:
   • h = height (in meters)
   • t = time (in seconds)
3. Translate into an Equation:
   • We want to find t when h = 17.
   • So, the equation is: 17 = -5t² + 20t + 2.
4. Solve the Equation:
   • Rearrange to standard form: -5t² + 20t - 15 = 0.
   • Divide by -5 to simplify: t² - 4t + 3 = 0.
   • Factor the quadratic equation: (t - 1)(t - 3) = 0.
   • The solutions are t = 1 and t = 3.
5. Check the Solutions:
   • For t = 1: h = -5(1)² + 20(1) + 2 = -5 + 20 + 2 = 17 meters.
   • For t = 3: h = -5(3)² + 20(3) + 2 = -45 + 60 + 2 = 17 meters.
   • Both solutions are valid.
6. State the Answer:
   • The ball will be 17 meters above the ground at t = 1 second and t = 3 seconds.

Example 3: Consecutive Integers

Problem: Find two consecutive positive integers whose product is 132.

Solution:

1. Understand the Problem:
   • We need to find two consecutive positive integers such that when multiplied, they equal 132.
2. Assign Variables:
   • Let n be the first positive integer.
   • The next consecutive integer is n + 1.
3. Translate into an Equation:
   • The product of the two integers is 132, so: n(n + 1) = 132.
4. Solve the Equation:
   • Expand the equation: n² + n = 132.
   • Rearrange to standard form: n² + n - 132 = 0.
   • Factor the quadratic equation: (n + 12)(n - 11) = 0.
   • The solutions are n = -12 and n = 11.
5. Check the Solutions:
   • Since we are looking for positive integers, n = -12 is not a valid solution.
   • So, n = 11.
   • The next consecutive integer is n + 1 = 11 + 1 = 12.
   • Check: 11 * 12 = 132 (which matches the given product).
6. State the Answer:
   • The two consecutive positive integers are 11 and 12.

Example 4: The Pythagorean Theorem

Problem: 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 the Problem:
   • We need to find the lengths of the legs of a right triangle given the length of the hypotenuse and the relationship between the legs.
2. Assign Variables:
   • Let x be the length of the shorter leg (in cm).
   • The longer leg is x + 7 cm.
3. Translate into an Equation:
   • By the Pythagorean theorem: a² + b² = c², where a and b are the legs, and c is the hypotenuse.
   • So, the equation is: x² + (x + 7)² = 13².
4. Solve the Equation:
   • Expand the equation: x² + (x² + 14x + 49) = 169.
   • Combine like terms: 2x² + 14x + 49 = 169.
   • Rearrange to standard form: 2x² + 14x - 120 = 0.
   • Divide by 2 to simplify: x² + 7x - 60 = 0.
   • Factor the quadratic equation: (x + 12)(x - 5) = 0.
   • The solutions are x = -12 and x = 5.
5. Check the Solutions:
   • Since the length of a leg cannot be negative, x = -12 is not a valid solution.
   • So, x = 5 cm.
   • The longer leg is x + 7 = 5 + 7 = 12 cm.
   • Check: 5² + 12² = 25 + 144 = 169 = 13² (which matches the Pythagorean theorem).
6. State the Answer:
   • The lengths of the legs are 5 cm and 12 cm.

Example 5: Revenue and Cost

Problem: A company sells x units of a product. The revenue R (in dollars) is given by R = 100x - x², and the cost C (in dollars) is given by C = 20x + 100. Find the number of units the company needs to sell to break even (when revenue equals cost).

Solution:

1. Understand the Problem:

   • We need to find the number of units x for which the revenue equals the cost.

2. Assign Variables:

   • x = number of units
   • R = revenue (in dollars)
   • C = cost (in dollars)

3. Translate into an Equation:

   • To break even, R = C.
   • So, the equation is: 100x - x² = 20x + 100.

4. Solve the Equation:

   • Rearrange to standard form: -x² + 80x - 100 = 0.

   • Multiply by -1: x² - 80x + 100 = 0.

   • Use the quadratic formula to solve for x:

     x = (-b ± √(b² - 4ac)) / (2a)

     x = (80 ± √(80² - 4(1)(100))) / (2(1))

     x = (80 ± √(6400 - 400)) / 2

     x = (80 ± √6000) / 2

     x = (80 ± 10√60) / 2

     x = 40 ± 5√60

   • Approximate values:

     x ≈ 40 ± 5(7.746)

     x ≈ 40 ± 38.73

   • The solutions are approximately x ≈ 1.27 and x ≈ 78.73.

5. Check the Solutions:

   • Since the number of units must be a whole number, we can round the solutions to the nearest whole numbers, x = 1 and x = 79.
   • For x = 1: R = 100(1) - (1)² = 99, C = 20(1) + 100 = 120 (not equal)
   • For x = 79: R = 100(79) - (79)² = 7900 - 6241 = 1659, C = 20(79) + 100 = 1580 + 100 = 1680 (close enough due to rounding)

   A more precise approach would be to consider that the quadratic formula gave us approximate results due to the square root. For practical purposes, selling either 1 or 79 units would bring the company close to the break-even point, but x = 79 is the more viable and practical solution.

6. State the Answer:

   • The company needs to sell approximately 79 units to break even.

Tips and Strategies

• Draw Diagrams: For geometry problems, drawing a diagram can help visualize the problem and identify relationships.
• Use a Table: For problems involving rates, time, and distance, organizing the information in a table can make it easier to set up the equation.
• Practice Regularly: The more you practice, the better you will become at recognizing patterns and applying the appropriate techniques.
• Review Basic Algebra: Make sure you have a solid foundation in basic algebraic operations, such as simplifying expressions, solving linear equations, and working with fractions.
• Understand the Context: Always consider the context of the problem and make sure your solutions are realistic. Discard any solutions that don't make sense in the real world.
• Check for Hidden Information: Sometimes, word problems contain hidden information that is not explicitly stated. Pay attention to details and look for clues that can help you set up the equation.

Advanced Techniques

• System of Equations: Some word problems may require solving a system of equations, where one or more equations are quadratic. Use substitution or elimination to solve for the unknowns.
• Optimization Problems: These problems involve finding the maximum or minimum value of a quadratic function. Complete the square to find the vertex of the parabola, which represents the maximum or minimum point.
• Related Rates: In calculus, related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another quantity. These problems may involve quadratic equations.

Common Mistakes to Avoid

• Incorrectly Translating the Problem: Make sure you accurately translate the word problem into a mathematical equation. Pay attention to keywords and relationships.
• Algebra Errors: Be careful with your algebra. Double-check your work, especially when expanding, factoring, and simplifying expressions.
• Ignoring the Context: Always consider the context of the problem and make sure your solutions make sense. Discard any solutions that are not realistic.
• Forgetting Units: Include appropriate units in your answer.
• Not Checking Your Solutions: Always check your solutions by substituting them back into the original equation and making sure they satisfy the problem conditions.

Conclusion

Mastering quadratic equation word problems requires a combination of understanding quadratic equations, developing problem-solving strategies, and practicing regularly. By following the steps outlined in this guide, you can break down complex problems into manageable parts, translate them into equations, solve for the unknowns, and check your solutions. With consistent effort and attention to detail, you can become proficient in solving quadratic equation word problems and gain confidence in your mathematical abilities.