Quadratic Word Problems

Quadratic equations are not just abstract algebra — they model real situations where something goes up and comes back down, where a quantity has a peak or a valley, or where two changing amounts multiply together. Projectile motion, fencing problems, revenue optimization, and falling objects all produce quadratic equations. Learning to translate a word problem into a quadratic equation, solve it, and interpret the result is one of the most practical skills in Algebra 1.

This page covers the four most common categories of quadratic word problems, with a systematic approach for each one.

Strategy for Quadratic Word Problems

Every quadratic word problem follows the same general process:

1. Read the problem carefully and identify what you are solving for
2. Define a variable for the unknown quantity
3. Write a quadratic equation that models the situation
4. Solve the equation using factoring, completing the square, or the quadratic formula
5. Interpret the answer in context — check units, discard nonsensical solutions, and answer the original question

The hardest part is usually Step 3: setting up the equation. The rest is computation you already know.

Projectile Motion Problems

When an object is launched upward (thrown, kicked, shot from a cannon), its height over time follows a quadratic model. Near Earth’s surface, the standard formula is:

$h(t) = -16t^2 + v_0 t + h_0$

where:

• $h(t)$ is the height in feet at time $t$ seconds
• $v_0$ is the initial upward velocity in feet per second
• $h_0$ is the initial height in feet
• The $-16$ comes from half of Earth’s gravitational acceleration ($-32$ ft/s$^2$ divided by 2)

If you work in meters, the formula uses $-4.9$ instead of $-16$:

$h(t) = -4.9t^2 + v_0 t + h_0$

Example 1: Ball thrown upward

A ball is thrown straight up from the ground with an initial velocity of 48 feet per second. (a) When does it hit the ground? (b) What is the maximum height? (c) When does it reach maximum height?

Set up the equation. Here $v_0 = 48$ and $h_0 = 0$ (ground level):

$h(t) = -16t^2 + 48t$

(a) When does it hit the ground? Set $h(t) = 0$:

$-16t^2 + 48t = 0$

Factor: $-16t(t - 3) = 0$

$t = 0 \quad \text{or} \quad t = 3$

$t = 0$ is the launch time. The ball hits the ground at $t = 3$ seconds.

(b) and (c) Maximum height. The vertex of the parabola gives the maximum. Since $a = -16$ and $b = 48$:

$t = -\frac{b}{2a} = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5$

$h(1.5) = -16(1.5)^2 + 48(1.5) = -16(2.25) + 72 = -36 + 72 = 36$

Answer: The ball reaches a maximum height of 36 feet at 1.5 seconds.

Example 2: Object launched from a platform

A model rocket is launched from a platform 20 feet above the ground with an initial velocity of 64 feet per second. When does the rocket hit the ground?

$h(t) = -16t^2 + 64t + 20$

Set $h(t) = 0$:

$-16t^2 + 64t + 20 = 0$

Divide everything by $-4$:

$4t^2 - 16t - 5 = 0$

Apply the quadratic formula with $a = 4$, $b = -16$, $c = -5$:

$\Delta = (-16)^2 - 4(4)(-5) = 256 + 80 = 336$

$t = \frac{16 \pm \sqrt{336}}{8} = \frac{16 \pm 4\sqrt{21}}{8} = \frac{4 \pm \sqrt{21}}{2}$

$t \approx \frac{4 + 4.583}{2} \approx 4.29 \quad \text{or} \quad t \approx \frac{4 - 4.583}{2} \approx -0.29$

The negative time does not make physical sense.

Answer: The rocket hits the ground after approximately 4.29 seconds.

Maximum Area Problems

These problems ask you to find the dimensions that maximize (or occasionally minimize) an area. They typically involve a fixed amount of fencing, rope, or material.

The key pattern: if you have a fixed perimeter and want to maximize the enclosed area, the problem produces a quadratic function whose vertex gives the answer.

Example 3: Classic fencing problem

A farmer has 120 feet of fencing and wants to enclose a rectangular garden along a barn wall (so only three sides need fencing). What dimensions maximize the garden area?

Step 1 — Define variables. Let $x$ be the length of each side perpendicular to the barn. The side parallel to the barn is $120 - 2x$ (the remaining fencing after two sides of length $x$).

Step 2 — Write the area function:

$A(x) = x(120 - 2x) = 120x - 2x^2$

This is a quadratic with $a = -2$, which opens downward — confirming a maximum exists.

Step 3 — Find the vertex:

$x = -\frac{120}{2(-2)} = -\frac{120}{-4} = 30$

The parallel side is $120 - 2(30) = 60$ feet.

$A(30) = 30 \times 60 = 1800$

Answer: The maximum area is 1,800 square feet with dimensions 30 ft by 60 ft.

Example 4: Two adjacent pens

A rancher has 200 feet of fencing to build two adjacent rectangular pens (they share a middle fence). What dimensions maximize the total enclosed area?

Step 1 — Define variables. Let $x$ be the width (there are three parallel widths: two outer walls and the shared divider). Let $y$ be the length (two sides). Then:

$3x + 2y = 200 \implies y = \frac{200 - 3x}{2}$

Step 2 — Total area:

$A = x \cdot y = x \cdot \frac{200 - 3x}{2} = \frac{200x - 3x^2}{2} = 100x - 1.5x^2$

Step 3 — Vertex:

$x = -\frac{100}{2(-1.5)} = -\frac{100}{-3} = \frac{100}{3} \approx 33.33 \text{ ft}$

$y = \frac{200 - 3(100/3)}{2} = \frac{200 - 100}{2} = 50 \text{ ft}$

$A = \frac{100}{3} \times 50 = \frac{5000}{3} \approx 1666.67 \text{ sq ft}$

Answer: Each pen is approximately 33.33 ft wide by 50 ft long, for a total area of about 1,666.67 square feet.

Revenue and Profit Optimization

Businesses use quadratic models when a price increase reduces the number of units sold. Revenue = price times quantity, and if both depend on the same variable, the result is a quadratic.

Example 5: Ticket pricing

A theater sells 200 tickets at $25 each. Market research shows that for every $1 increase in price, 4 fewer tickets are sold. What price maximizes revenue?

Step 1 — Define the variable. Let $x$ be the number of $1 price increases.

• Price per ticket: $25 + x$
• Tickets sold: $200 - 4x$

Step 2 — Revenue function:

$R(x) = (25 + x)(200 - 4x)$

Expand:

$R(x) = 5000 - 100x + 200x - 4x^2 = -4x^2 + 100x + 5000$

Step 3 — Find the vertex:

$x = -\frac{100}{2(-4)} = -\frac{100}{-8} = 12.5$

Since $x$ must be a whole number (you cannot raise the price by half a dollar in this scenario), check $x = 12$ and $x = 13$:

$R(12) = (37)(152) = 5624$

$R(13) = (38)(148) = 5624$

Both give the same revenue. The optimal price is either $37 or $38, both producing a maximum revenue of $5,624.

Example 6: Retail pricing optimization

A retail store sells phone cases for $15 each and averages 80 sales per day. For every $2 increase in price, daily sales drop by 5 cases. What price maximizes daily revenue?

Step 1 — Let $x$ be the number of $2 price increases.

• Price: $15 + 2x$
• Units sold: $80 - 5x$

Step 2 — Revenue:

$R(x) = (15 + 2x)(80 - 5x)$

$R(x) = 1200 - 75x + 160x - 10x^2 = -10x^2 + 85x + 1200$

Step 3 — Vertex:

$x = -\frac{85}{2(-10)} = \frac{85}{20} = 4.25$

Check $x = 4$ and $x = 5$:

• $x = 4$: Price $= 23$, units $= 60$, $R = 1380$
• $x = 5$: Price $= 25$, units $= 55$, $R = 1375$

Answer: A price of $23 per case maximizes daily revenue at $1,380.

Falling Objects and Gravity Problems

These are similar to projectile problems but involve objects dropped (no initial upward velocity) or falling from a height.

Example 7: Dropped object

A tool is dropped from a scaffold 144 feet above the ground. When does it hit the ground?

Since the tool is dropped (not thrown), $v_0 = 0$:

$h(t) = -16t^2 + 144$

Set $h(t) = 0$:

$-16t^2 + 144 = 0$

$16t^2 = 144$

$t^2 = 9$

$t = 3 \quad (\text{discard } t = -3)$

Answer: The tool hits the ground after 3 seconds.

Example 8: When does a falling object pass a certain height?

Using the same scenario, when is the tool 80 feet above the ground?

$-16t^2 + 144 = 80$

$-16t^2 = -64$

$t^2 = 4$

$t = 2 \quad (\text{discard } t = -2)$

Answer: The tool passes through 80 feet at 2 seconds after being dropped.

Setting Up Equations from Word Descriptions

Sometimes the problem does not fit neatly into the categories above. The key skill is translating English into algebra.

Example 9: Consecutive integers

The product of two consecutive positive integers is 182. Find the integers.

Let $n$ be the first integer. The next consecutive integer is $n + 1$.

$n(n + 1) = 182$

$n^2 + n - 182 = 0$

Using the quadratic formula:

$n = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm \sqrt{729}}{2} = \frac{-1 \pm 27}{2}$

$n = \frac{26}{2} = 13 \quad \text{or} \quad n = \frac{-28}{2} = -14$

Since the problem specifies positive integers, $n = 13$ and $n + 1 = 14$.

Check: $13 \times 14 = 182$. Confirmed.

Answer: The two consecutive positive integers are 13 and 14.

Example 10: Geometry setup

A rectangular garden has a length that is 3 feet more than twice its width. The area is 90 square feet. Find the dimensions.

Let $w$ be the width. Then the length is $2w + 3$.

$w(2w + 3) = 90$

$2w^2 + 3w - 90 = 0$

Using the quadratic formula with $a = 2$, $b = 3$, $c = -90$:

$\Delta = 9 + 720 = 729$

$w = \frac{-3 \pm 27}{4}$

$w = \frac{24}{4} = 6 \quad \text{or} \quad w = \frac{-30}{4} = -7.5$

Width cannot be negative, so $w = 6$ feet and length $= 2(6) + 3 = 15$ feet.

Check: $6 \times 15 = 90$. Confirmed.

Answer: The garden is 6 feet wide and 15 feet long.

Real-World Application: Carpentry — Maximum Deck Area

A carpenter has 40 feet of railing to enclose three sides of a rectangular deck attached to a house (the house forms the fourth side). What dimensions maximize the deck area?

Let $x$ be the depth of the deck (the two sides perpendicular to the house). The side parallel to the house is $40 - 2x$.

$A(x) = x(40 - 2x) = 40x - 2x^2$

Find the vertex:

$x = -\frac{40}{2(-2)} = 10$

$A(10) = 10(40 - 20) = 10 \times 20 = 200$

Answer: The deck should be 10 feet deep and 20 feet wide for a maximum area of 200 square feet. This is a common calculation when a carpenter needs to get the most usable space from a fixed amount of railing material.

Real-World Application: HVAC — Sizing a Duct Opening

An HVAC technician needs a rectangular duct opening with a perimeter of 24 inches. The airflow capacity is proportional to the cross-sectional area. What dimensions maximize airflow?

Let $x$ be the width. The perimeter gives $2x + 2y = 24$, so $y = 12 - x$.

$A(x) = x(12 - x) = 12x - x^2$

Vertex:

$x = -\frac{12}{2(-1)} = 6$

$A(6) = 6(12 - 6) = 36$

Answer: A 6-inch by 6-inch square opening maximizes the area at 36 square inches. This is a well-known result: for a fixed perimeter, a square always maximizes the area of a rectangle. The technician should specify a square duct when maximizing airflow is the priority.

Real-World Application: Retail — Pricing Strategy

A bookstore sells novels for $12 each and sells 150 per week. Experience shows that each $0.50 price increase reduces weekly sales by 5 books. What price maximizes weekly revenue?

Let $x$ be the number of $0.50 increases.

• Price: $12 + 0.5x$
• Books sold: $150 - 5x$

$R(x) = (12 + 0.5x)(150 - 5x)$

$R(x) = 1800 - 60x + 75x - 2.5x^2 = -2.5x^2 + 15x + 1800$

$x = -\frac{15}{2(-2.5)} = -\frac{15}{-5} = 3$

Optimal price: $12 + 0.5(3) = 13.50$

$R(3) = (13.50)(135) = 1822.50$

Answer: Setting the price at $13.50 maximizes weekly revenue at $1,822.50.

Common Mistakes to Avoid

1. Forgetting to define the variable clearly. Always write “Let $x$ = …” before setting up the equation. Vague variables lead to setup errors.
2. Not discarding nonsensical solutions. Quadratic equations often produce two answers, but negative time, negative length, or a number of items exceeding the total supply should be thrown out.
3. Confusing “when does it hit the ground” with “maximum height.” Hitting the ground means $h = 0$ (solve the equation). Maximum height means finding the vertex.
4. Setting up the wrong equation for area problems. Draw a picture and label every side. Write the constraint (total fencing, total perimeter) and the objective (area to maximize) as separate expressions.
5. Using the wrong gravity constant. Use $-16$ for feet and $-4.9$ for meters. Mixing units produces wildly wrong answers.
6. Rounding too early. Keep exact values through the calculation and round only at the final answer. Early rounding compounds errors.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Quadratic word problems fall into common categories: projectile motion, maximum area, revenue/profit optimization, falling objects, and general setup problems
• For projectile motion, use $h(t) = -16t^2 + v_0t + h_0$ (feet) or $-4.9t^2 + v_0t + h_0$ (meters)
• For maximum/minimum problems, find the vertex: $x = -b/(2a)$ gives the optimal value
• For area problems with a fixed perimeter, write area as a function of one variable using the perimeter constraint, then find the vertex
• For revenue problems, Revenue = (price)(quantity) — express both in terms of one variable
• Always define your variable, discard nonsensical solutions, and check your answer in the original problem context

Return to Algebra 1 for more topics in this section.