9.4 Writing and Modeling Real-World Problems with Quadratic Equations
Key Takeaways
A.4.b tests writing one-variable quadratic equations from context, most often using projectile motion, profit/revenue, or area scenarios.
The GED's projectile motion model is h(t) = −16t² + v₀t + h₀, where h(t) is height in feet, t is time in seconds, v₀ is initial velocity, and h₀ is initial height.
A parabola's vertex, found at t = −b/(2a), represents the maximum value when the leading coefficient is negative and the minimum value when it is positive.
Quadratic word problems produce two mathematical solutions, but only one is usually valid in context — always reject negative time, negative length, or negative width.
Area problems where length and width are both expressed in the same variable (e.g., length = width + 3) create a quadratic equation when set equal to a given area.
Why Writing and Modeling Quadratic Context Problems Matters
Indicator A.4.b ("Write one-variable quadratic equations to represent context") is the applied half of quadratic equations on the GED. Rather than handing you an equation to solve, these items describe a real situation — a ball thrown in the air, a company's profit over time, or a rectangular area — and ask you to build the correct quadratic equation, then interpret the solution in terms of the scenario. This is where quadratics connect to two of the GED's favorite real-world models: projectile motion and area/profit modeling.
Model 1: Projectile Motion
The most frequently tested quadratic context on the GED is an object thrown, dropped, or launched into the air. The standard model is:
h(t) = −16t² + v₀t + h₀
where:
- h(t) = height of the object (in feet) at time t (in seconds)
- −16 is a constant related to gravity's effect in feet per second squared (this coefficient is always −16 on the GED, since it uses feet and seconds)
- v₀ = initial velocity (how fast the object is launched upward)
- h₀ = initial height (how high off the ground the object starts)
Worked Example: A ball is thrown upward from ground level with the height modeled by h(t) = −16t² + 64t. At what time does the ball hit the ground?
- The ball "hits the ground" when height = 0, so set h(t) = 0: −16t² + 64t = 0
- Factor out the greatest common factor, −16t: −16t(t − 4) = 0
- Zero Product Property: t = 0 or t = 4
- Interpretation: t = 0 is the moment the ball is thrown (height is 0 at the start too), which is not the answer being asked for. The ball hits the ground at t = 4 seconds.
This example illustrates a key GED skill beyond the algebra: rejecting the solution that doesn't fit the real-world context. Quadratic equations mathematically produce two roots, but a word problem often only has one root that makes physical sense.
Model 2: Profit, Revenue, and Business Context
Quadratic models also appear in business scenarios where profit rises, peaks, and falls over time — the parabola's vertex represents the maximum (or minimum) value.
Worked Example: A company's profit (in thousands of dollars) is modeled by P(t) = −t² + 6t − 5, where t is the number of years since 2020. In which year did the company reach maximum profit?
- Since the leading coefficient (−1) is negative, the parabola opens downward, so its vertex is a maximum point.
- The t-coordinate of the vertex of ax² + bx + c is found with t = −b / (2a).
- Here a = −1, b = 6: t = −6 / (2 × −1) = −6 / −2 = 3
- Since t = 3 means 3 years after 2020, maximum profit occurred in 2023.
Model 3: Area Problems
A rectangle's area (length × width) creates a quadratic relationship whenever length and width are both expressed in terms of the same variable.
Worked Example: A rectangular garden's length is 3 feet more than its width, and its area is 54 square feet. Write an equation and find the width.
- Let w = width, so length = w + 3.
- Area equation: w(w + 3) = 54, which expands to w² + 3w − 54 = 0
- Factor: two numbers that multiply to −54 and add to 3 are 9 and −6: (w + 9)(w − 6) = 0
- Solutions: w = −9 or w = 6
- Interpretation: A width cannot be negative, so reject w = −9. The width is 6 feet (and the length is 9 feet).
Recognizing Which Model Applies
| Scenario Clue | Quadratic Model | What to Solve For |
|---|---|---|
| "thrown," "launched," "dropped," "height," "seconds" | Projectile: h(t) = −16t² + v₀t + h₀ | Time when height = 0 (hits ground), or vertex for max height |
| "profit," "revenue," "cost over time" | Profit/business model | Vertex (t = −b/2a) for maximum or minimum |
| "length is __ more/less than width," "area equals" | Area model: dimension × dimension = area | The dimension (reject negative solutions) |
Common Traps to Avoid
- Reporting both mathematical roots without checking which one fits the real-world situation (negative time, negative length, or negative width are never valid answers).
- Confusing the vertex formula −b/(2a) with a root of the equation — the vertex tells you when the maximum/minimum occurs, not when the value is zero.
- Mixing up which variable is time vs. height (or width vs. length) when translating the word problem into the equation.
- Forgetting to add the starting year, starting height, or other offset back into the final answer after solving for t (e.g., t = 3 years after 2020 is the year 2023, not "year 3").
Key Takeaways
Writing quadratic equations from context is really two skills combined: correctly building the equation from the scenario (projectile, profit, or area), and correctly interpreting which of the two mathematical solutions actually answers the question — since negative time, negative length, and other impossible values must always be rejected. Practice identifying the scenario type from its keywords first, then apply the matching formula (h(t) = −16t² + v₀t + h₀ for projectiles, −b/2a for a vertex/maximum, or dimension × dimension for area) before solving.
A ball's height is modeled by h(t) = −16t² + 48t + 4, where h(t) is in feet and t is in seconds. What does the value 4 represent in this equation?
The ball's initial height above the ground when thrown
The ball's initial velocity
The time it takes the ball to reach its maximum height
The maximum height the ball reaches
A rectangular patio's length is 2 feet more than its width, and its area is 80 square feet. What is the width of the patio?
6 feet
8 feet
9 feet
10 feet
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