Homework 13: Quadratic Equation Word Problems — 5 Worked Examples
Homework 13 quadratic equation word problems are where many algebra students first discover that solving x² + 5x + 6 = 0 is only half the job — the harder half is building the equation from a paragraph of English in the first place. Word problems require a translation step that turns a real-world scenario into a quadratic model, and that translation step gets far less explicit practice than the algebra itself. This guide covers five fully worked examples drawn from the most common homework 13 quadratic equation word problem types — area, projectile motion, number relationships, revenue, and distance-rate-time — with every calculation shown so you can follow along and repeat the method on your own problems.
Contents
- 01What Are Quadratic Word Problems and Why Do They Appear on Homework 13?
- 02The 4-Step Framework for Any Quadratic Word Problem
- 03Area Problems: The Most Common Quadratic Word Problem Type
- 04Projectile Motion Word Problems: Height and Time
- 05Number Relationship Problems Using Quadratic Equations
- 06Revenue and Pricing: Business Quadratic Word Problems
- 07Distance, Rate, and Time Problems That Lead to Quadratic Equations
- 08Common Mistakes Students Make on Homework 13 Quadratic Problems
- 09Five Practice Quadratic Word Problems with Full Solutions
- 10Strategies and Shortcuts for Solving Quadratic Word Problems Faster
- 11Frequently Asked Questions About Homework 13 Quadratic Word Problems
What Are Quadratic Word Problems and Why Do They Appear on Homework 13?
A quadratic word problem is any application problem whose mathematical model includes a term with a variable squared (x²). Unlike linear word problems, where the relationship between quantities is proportional and the graph is a straight line, quadratic word problems model situations where two quantities multiply together — the length and width of a rectangle, the time and initial velocity of a thrown object, the number of items sold and the price per item. Homework 13 quadratic equation word problems typically arrive after students have mastered solving quadratic equations algebraically, so the assignment is designed to test whether you can recognize a quadratic relationship inside a story. The five categories that show up most often are: area and geometry problems, projectile motion problems, consecutive number problems, revenue and optimization problems, and distance-rate-time problems where speed changes. Each category has a standard setup pattern, and once you know those patterns, the translation step becomes much more systematic.
A quadratic word problem always contains a quantity multiplied by itself or two related quantities multiplied together — look for area, products of unknowns, or squared terms in any formula given.
The 4-Step Framework for Any Quadratic Word Problem
Whether the problem asks about a flying ball or a rectangular garden, every homework 13 quadratic equation word problem follows the same four-step translation and solution process. Skipping Step 1 — defining the variable clearly — is the single biggest source of errors, because students either forget what x represents or choose x as a quantity that makes the algebra unnecessarily complicated. Work through these four steps in order every time.
1. Step 1 — Define your variable precisely
Choose one unknown to call x, and write it down explicitly: 'Let x = the width of the garden in meters.' If a second quantity appears, express it in terms of x — for example, 'length = x + 3'. Never use two separate variables when you can express one in terms of the other; that keeps the problem as a single equation in one unknown.
2. Step 2 — Build the equation from the word problem
Identify the relationship the problem states (area = l × w, or distance = rate × time, or product of two numbers = given value), substitute your expressions from Step 1, and set up the equation. Most quadratic word problems give you a numerical value that the product equals — that's your equation. Expand any brackets so you can see the x² term.
3. Step 3 — Solve the quadratic equation
Rearrange to standard form ax² + bx + c = 0, then choose your method: factoring if the numbers are friendly, completing the square if the leading coefficient is 1, or the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) for any equation. You will often get two solutions — that's normal.
4. Step 4 — Interpret the answer and reject impossible values
Ask: does this solution make sense in context? A negative length, a negative number of seconds before the ball is thrown, or a negative number of people are all mathematically valid solutions to a quadratic but physically impossible answers. Reject the negative (or otherwise nonsensical) root and state your final answer in the units the problem asked for. Then verify by substituting back into the original word problem description — not just the equation you wrote.
Always write 'Let x = ____' before writing any equation. Students who skip this step almost always end up confused about which root to keep.
Area Problems: The Most Common Quadratic Word Problem Type
Area problems are the most frequently assigned quadratic word problems because they arise naturally from the formula Area = length × width. When the length and width are expressed in terms of the same variable, multiplying them produces an x² term. The standard setup is: one dimension is defined as x, the other as x plus (or minus) some constant, the area is given as a number, and you must find both dimensions. Here is a complete worked example of this problem type.
1. Problem
A rectangular garden has a length that is 3 meters longer than its width. The area of the garden is 40 m². Find the width and length.
2. Step 1 — Define the variable
Let x = the width of the garden in meters. Then the length = x + 3 meters.
3. Step 2 — Build the equation
Area = length × width, so (x + 3)(x) = 40. Expanding: x² + 3x = 40.
4. Step 3 — Solve
Rearrange to standard form: x² + 3x − 40 = 0. Factor: look for two numbers that multiply to −40 and add to +3. Those numbers are +8 and −5. So: (x + 8)(x − 5) = 0. Set each factor to zero: x + 8 = 0 → x = −8, or x − 5 = 0 → x = 5.
5. Step 4 — Interpret
Width cannot be negative, so reject x = −8. Width = 5 m, Length = 5 + 3 = 8 m. Check: 5 × 8 = 40 m² ✓. The garden is 5 meters wide and 8 meters long.
For area problems: always set up Area = length × width using your variable expressions, expand, move everything to one side, and factor.
Projectile Motion Word Problems: Height and Time
Projectile motion problems are the second major category on homework 13 quadratic equation problem sets. They rely on the physics formula h = −(g/2)t² + v₀t + h₀, where h is height, t is time, v₀ is initial upward velocity, h₀ is initial height, and g is the gravitational acceleration (approximately 10 m/s² in metric or 32 ft/s² in imperial units). Most homework versions are given pre-simplified, so you just use the formula as given and solve for t when h = 0 (ground level) or h = some target height. Here is a clean example with round numbers that let you factor rather than use the formula.
1. Problem
A ball is thrown upward from ground level with an initial velocity of 20 m/s. Its height after t seconds is h = −5t² + 20t. At what times is the ball at ground level?
2. Step 1 — Define the variable
t = time in seconds after the ball is thrown. Ground level means h = 0.
3. Step 2 — Build the equation
Set h = 0: −5t² + 20t = 0.
4. Step 3 — Solve
Factor out −5t: −5t(t − 4) = 0. Set each factor to zero: −5t = 0 → t = 0, or t − 4 = 0 → t = 4.
5. Step 4 — Interpret
t = 0 is the moment the ball is thrown (it starts at ground level). t = 4 is when it returns to the ground. The ball is at ground level at t = 0 seconds (launch) and t = 4 seconds (landing). Check: h(4) = −5(16) + 20(4) = −80 + 80 = 0 ✓.
6. Extension: When does the ball reach maximum height?
The maximum height occurs at the midpoint between the two roots: t = (0 + 4)/2 = 2 seconds. Maximum height = −5(2²) + 20(2) = −20 + 40 = 20 m. This is a useful fact many homework 13 projectile problems ask for as a follow-up.
For projectile problems: set h = 0 to find when the object hits the ground. The two roots are launch time and landing time. Maximum height occurs at the vertex, t = −b/(2a).
Number Relationship Problems Using Quadratic Equations
Number relationship problems ask you to find two unknown numbers based on their sum, difference, or product. When the problem gives you the product of the two numbers, you almost always end up with a quadratic equation. The most common versions involve consecutive integers (like 8 and 9, or 7 and −8), consecutive odd integers (like 5 and 7), or two numbers with a stated difference. These problems look simple but they require careful setup — the second number must be expressed in terms of x before you can write the equation.
1. Problem
The product of two consecutive positive integers is 72. Find the integers.
2. Step 1 — Define the variable
Let x = the smaller integer. Then the next consecutive integer = x + 1.
3. Step 2 — Build the equation
Product of the two integers = 72: x(x + 1) = 72. Expanding: x² + x = 72.
4. Step 3 — Solve
Rearrange: x² + x − 72 = 0. Factor: find two numbers that multiply to −72 and add to +1. Those are +9 and −8. So: (x + 9)(x − 8) = 0. Solutions: x = −9 or x = 8.
5. Step 4 — Interpret
The problem says positive integers, so reject x = −9. x = 8, and x + 1 = 9. The integers are 8 and 9. Check: 8 × 9 = 72 ✓.
6. Variation: Consecutive odd integers
If the problem said 'two consecutive odd integers whose product is 63', let x = first odd integer and x + 2 = second odd integer (odd integers differ by 2). Then x(x + 2) = 63 → x² + 2x − 63 = 0 → (x + 9)(x − 7) = 0 → x = 7. The integers are 7 and 9. Check: 7 × 9 = 63 ✓.
Consecutive integers differ by 1: use x and x + 1. Consecutive odd or even integers differ by 2: use x and x + 2. Write this at the top of every number problem before doing anything else.
Revenue and Pricing: Business Quadratic Word Problems
Revenue problems appear frequently in Homework 13 quadratic equation problem sets because revenue = price × quantity sold, and when price and quantity are linearly related to each other (raising the price reduces quantity sold), their product is a quadratic. These problems often ask for the price that maximizes revenue, which means finding the vertex of the parabola. The vertex of y = ax² + bx + c occurs at x = −b/(2a). Here is a complete example.
1. Problem
A theater charges $8 per ticket and sells 200 tickets per show. For each $1 increase in ticket price, 10 fewer tickets are sold. What ticket price produces the maximum revenue? What is the maximum revenue?
2. Step 1 — Define the variable
Let x = the number of $1 price increases. Then ticket price = (8 + x) dollars and tickets sold = (200 − 10x).
3. Step 2 — Build the revenue equation
Revenue R = price × tickets sold = (8 + x)(200 − 10x). Expand: R = 1600 − 80x + 200x − 10x² = −10x² + 120x + 1600.
4. Step 3 — Find the vertex
R = −10x² + 120x + 1600 is a downward parabola (a = −10 < 0), so the vertex is the maximum. x = −b/(2a) = −120 / (2 × −10) = −120 / −20 = 6. So the optimal number of price increases is 6.
5. Step 4 — Interpret
Optimal price = 8 + 6 = $14. Tickets sold = 200 − 10(6) = 140. Maximum revenue = 14 × 140 = $1,960. Check using the formula: R = −10(36) + 120(6) + 1600 = −360 + 720 + 1600 = $1,960 ✓.
For revenue maximization: write R = (price)(quantity), expand to get ax² + bx + c, then find the vertex at x = −b/(2a). The vertex gives the input that produces maximum (or minimum) revenue.
Distance, Rate, and Time Problems That Lead to Quadratic Equations
Distance-rate-time problems typically produce linear equations (d = rt), but they become quadratic when the problem involves two legs of a trip at different speeds that are related to each other, or when you add two time expressions with different denominators and the denominators contain x. The key formula is time = distance ÷ rate. When you have two fractions with x in the denominator and you clear the denominators by multiplying through, you produce a quadratic equation. This type of problem frequently appears in homework 13 quadratic equation problem sets because it combines two skills: rational expressions and quadratics.
1. Problem
A motorboat travels 24 km upstream and then 24 km back downstream. The river current flows at 3 km/h. If the total trip takes 6 hours, find the boat's speed in still water.
2. Step 1 — Define the variable
Let v = the boat's speed in still water (km/h). Upstream speed = v − 3 km/h (fighting the current). Downstream speed = v + 3 km/h (helped by the current).
3. Step 2 — Build the equation
Time = distance ÷ rate. Upstream time = 24 / (v − 3). Downstream time = 24 / (v + 3). Total time = 6 hours: 24/(v − 3) + 24/(v + 3) = 6.
4. Step 3 — Clear the denominators
Multiply every term by (v − 3)(v + 3): 24(v + 3) + 24(v − 3) = 6(v − 3)(v + 3). Expand the left side: 24v + 72 + 24v − 72 = 48v. Expand the right side: 6(v² − 9) = 6v² − 54. Equation: 48v = 6v² − 54.
5. Step 4 — Solve
Rearrange: 6v² − 48v − 54 = 0. Divide by 6: v² − 8v − 9 = 0. Factor: (v − 9)(v + 1) = 0. Solutions: v = 9 or v = −1.
6. Step 5 — Interpret
Speed cannot be negative, so reject v = −1. The boat's speed in still water is 9 km/h. Check: upstream time = 24/6 = 4 h, downstream time = 24/12 = 2 h, total = 6 h ✓.
Distance-rate-time problems become quadratic when you add two fractions (time = d/r) with x in both denominators and clear them by cross-multiplying. Always check that the denominator does not equal zero for your answer.
Common Mistakes Students Make on Homework 13 Quadratic Problems
Homework 13 quadratic equation word problems have predictable failure points. Most errors happen before any algebra is written — in the setup stage. Here are the six mistakes that account for the majority of wrong answers, along with concrete ways to avoid each one.
1. Mistake 1: Not defining the variable before writing the equation
Jumping straight to writing an equation without stating 'Let x = ___' leads to confusion when two solutions appear. You won't know which quantity x represents or why one answer should be rejected. Fix: always write 'Let x = [specific quantity and units]' as the first line of your solution.
2. Mistake 2: Keeping both roots without checking context
Quadratic equations produce two solutions. Students sometimes report both without checking which makes sense in the problem. A rectangle cannot have a negative width. A ball cannot land before it is thrown. Fix: after solving, ask 'does each root make physical sense?' and reject the one that does not.
3. Mistake 3: Forgetting to move everything to one side
After expanding, students try to factor something like x² + 3x = 40 instead of x² + 3x − 40 = 0. Factoring only works reliably when one side is zero. Fix: always rearrange to ax² + bx + c = 0 before factoring or applying the quadratic formula.
4. Mistake 4: Sign errors when expanding (a + b)(a − b) vs (a − b)²
In revenue problems, expanding (8 + x)(200 − 10x) produces a mix of positive and negative terms. Students commonly drop a minus sign. Fix: write out every multiplication step explicitly and circle the sign of each term before combining.
5. Mistake 5: Using the wrong formula for projectile problems
Some textbooks use h = −16t² + v₀t + h₀ (feet, g = 32 ft/s²) and others use h = −5t² + v₀t + h₀ (meters, approximate). Using the wrong constant produces a completely wrong answer. Fix: read the problem to see whether it gives the formula explicitly, or note the units — feet usually means −16, meters usually means −5 or −4.9.
6. Mistake 6: Not checking the answer in the original word problem
Students check their answer in the equation they wrote, but if they set up the equation wrong, a correct algebraic check still gives a wrong answer to the word problem. Fix: after finding x, substitute back into the original problem description (the English sentences) and verify that the stated condition is satisfied.
The setup step takes less than two minutes but eliminates the majority of errors. Writing 'Let x = ___' and rearranging to standard form before anything else is worth more than speed.
Five Practice Quadratic Word Problems with Full Solutions
Use these five problems to test the framework before submitting your homework. They are arranged from straightforward to more involved. Cover the solution, attempt the problem yourself, and then compare your work step by step.
1. Practice Problem 1 — Area
A rectangle's length is twice its width. Its area is 98 cm². Find the dimensions. Solution: Let x = width. Length = 2x. Equation: x(2x) = 98 → 2x² = 98 → x² = 49 → x = 7 (reject −7). Width = 7 cm, Length = 14 cm. Check: 7 × 14 = 98 ✓.
2. Practice Problem 2 — Number relationship
Two positive numbers differ by 5. Their product is 84. Find the numbers. Solution: Let x = smaller number. Larger = x + 5. Equation: x(x + 5) = 84 → x² + 5x − 84 = 0. Factor: (x + 12)(x − 7) = 0 → x = 7 (reject −12). Numbers are 7 and 12. Check: 7 × 12 = 84, 12 − 7 = 5 ✓.
3. Practice Problem 3 — Projectile
A rocket is fired upward. Its height in feet after t seconds is h = −16t² + 96t. When does it reach a height of 128 feet? Solution: Set h = 128: −16t² + 96t = 128 → −16t² + 96t − 128 = 0. Divide by −16: t² − 6t + 8 = 0. Factor: (t − 2)(t − 4) = 0 → t = 2 or t = 4. The rocket reaches 128 feet at 2 seconds (on the way up) and again at 4 seconds (on the way down). Both answers are valid and both should be stated.
4. Practice Problem 4 — Revenue
A store sells 300 units per week at $5 each. For every $0.50 price increase, it sells 20 fewer units. What price maximizes revenue? Solution: Let x = number of $0.50 increases. Price = 5 + 0.5x, Units = 300 − 20x. Revenue R = (5 + 0.5x)(300 − 20x) = 1500 − 100x + 150x − 10x² = −10x² + 50x + 1500. Vertex: x = −50/(2 × −10) = 2.5 increases. Price = 5 + 0.5(2.5) = $6.25. Units = 300 − 20(2.5) = 250. Revenue = 6.25 × 250 = $1,562.50.
5. Practice Problem 5 — Distance-rate-time
A cyclist rides 30 km to a town. On the return trip she rides 5 km/h faster and takes 1 hour less. Find her speed on the outward trip. Solution: Let v = speed going out (km/h). Return speed = v + 5. Time out = 30/v, Time back = 30/(v + 5). Difference = 1: 30/v − 30/(v + 5) = 1. Multiply by v(v + 5): 30(v + 5) − 30v = v(v + 5) → 30v + 150 − 30v = v² + 5v → 150 = v² + 5v → v² + 5v − 150 = 0. Factor: (v + 15)(v − 10) = 0 → v = 10 (reject −15). Speed going out = 10 km/h. Check: Time out = 3 h, time back = 30/15 = 2 h, difference = 1 h ✓.
Strategies and Shortcuts for Solving Quadratic Word Problems Faster
Once you recognize the category of a quadratic word problem, the setup becomes almost automatic. These strategies help you move through quadratic word problems on any assignment efficiently without sacrificing accuracy.
1. Identify the category first
Before writing anything, classify the problem: area (look for 'rectangular', 'dimensions', 'area = '), projectile (look for 'thrown', 'height', 'falls', 'seconds'), number relationship (look for 'product', 'consecutive', 'two numbers'), revenue (look for 'price', 'sold', 'revenue', 'profit'), or distance-rate-time (look for 'upstream', 'downstream', 'faster', 'slower', 'trip'). Each category has a known equation structure, so classification saves time.
2. Try factoring before the quadratic formula
Factoring is faster when the discriminant b² − 4ac is a perfect square. Quickly compute b² − 4ac: if it is 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc., the equation factors cleanly. If not, go directly to the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) and save the factoring attempt.
3. Keep units throughout every step
Write the units on every quantity: x meters, v km/h, t seconds. If the units in your equation don't make sense (e.g., adding meters to meters² without noticing), that's an early signal that your setup has an error. Catching this at Step 2 is much better than catching it after a full solution.
4. Use the discriminant to predict solution type
For ax² + bx + c = 0, compute Δ = b² − 4ac. If Δ > 0: two real solutions (most word problems). If Δ = 0: exactly one solution (the ball just touches the ground, the dimensions are equal, etc.). If Δ < 0: no real solution, which means either the problem has no physical answer or you set up the equation incorrectly — go back and recheck.
5. For optimization problems, skip the quadratic formula
Revenue and area maximization problems ask for the vertex, not the roots. Use x = −b/(2a) directly — no need to set the equation to zero and solve. Compute x, substitute back to get the maximum or minimum value, and interpret in context.
Δ = b² − 4ac tells you everything before you solve: positive means two roots, zero means one, negative means recheck your setup.
Frequently Asked Questions About Homework 13 Quadratic Word Problems
These questions come up repeatedly when students work through homework 13 quadratic equation word problems for the first time. The answers address the most common points of confusion.
1. When should I use the quadratic formula vs. factoring?
Use factoring when the discriminant b² − 4ac is a perfect square, because the roots will be rational numbers and factoring is faster. Use the quadratic formula when the discriminant is not a perfect square, when the leading coefficient is large, or when you're unsure whether it factors. The formula always works; factoring only sometimes works quickly.
2. What if both roots are positive — which one do I use?
When both roots are positive, both may be valid mathematical answers, but usually the problem context rules one out. For example, if the problem says 'the smaller integer', take the smaller root. If the problem asks for 'dimensions' and both give valid positive dimensions, check which one satisfies any additional constraint (like 'the width is less than 10'). If neither constraint rules one out, both are valid and you should state both.
3. How do I know what x should represent?
Define x as the quantity the problem asks you to find. If the problem asks 'find the width', let x = the width. If the problem asks 'find both numbers', let x = the smaller number. Choosing x as the quantity you want makes interpreting the final answer trivial — you just read off x = [answer].
4. My equation doesn't factor — did I set it up wrong?
Not necessarily. Many real quadratic equations do not factor over integers, especially distance-rate-time problems and some projectile problems. Compute the discriminant: if Δ > 0, use the quadratic formula and leave the answer in simplified radical form or as a decimal. If Δ < 0, recheck your setup — that usually means an error in the equation.
5. How should I check my final answer?
Plug your value of x back into the original word problem sentence, not just the equation. For the garden problem: 'Does a garden of width 5 m and length 8 m have an area of 40 m²? Yes, 5 × 8 = 40.' For the boat problem: 'Does a boat going 9 km/h upstream (speed 6 km/h) cover 24 km in 4 hours and then 24 km downstream (speed 12 km/h) in 2 hours, totaling 6 hours? Yes.' This two-sentence check catches setup errors that algebraic substitution misses.
6. What is the hardest type of quadratic word problem?
Most students find distance-rate-time problems the most difficult because they require building two fractions (time = d/r), adding them, and then clearing denominators before any quadratic algebra begins. The two extra steps — fraction setup and denominator clearing — make errors more likely. Practice these specifically: write time = d/r for each leg, add the expressions, set equal to total time, and multiply both sides by the LCD.
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