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Writing Quadratic Equations from Word Problems

Turn area, number and projectile situations into quadratic equations, then solve and interpret the answers.

Practice this skill 30 min Write quadratic equations from context Leer en español
Key idea

What you'll learn

In the last lesson you solved quadratic equations. In this lesson you will write them from a word problem — and then solve them and make sense of the answer. You will practice the three kinds of situations the GED uses most:

  1. Area problems — rectangles whose sides are related, and borders, frames or walkways around a shape.
  2. Number problems — consecutive integers, or "a number and its square."
  3. Motion (projectile) problems — the height of a thrown or dropped object over time.

Why it matters on the GED: many questions ask "Which equation could be used to find…?" You don't even have to solve — you have to translate correctly. Other questions ask for the final number, so you must write the equation, solve it, and throw away answers that make no sense (like a negative width).

Real life: planning a garden or patio with a fixed area, sizing a picture frame, or figuring out when a ball thrown in the air will land.

Words to know

  • Variable — a letter for the unknown number. Example: let w = the width in feet.
  • Area of a rectangle — length × width: A = lw. Example: a 7 ft by 12 ft room has area 84\text{ ft}^2.
  • "More than" / "less than" — add or subtract. Example: "3 more than the width" is w + 3; "2 less than twice the width" is 2w - 2.
  • Consecutive integers — whole numbers in a row, 1 apart: n, n + 1, n + 2. Example: 7, 8, 9.
  • Consecutive even (or odd) integers — 2 apart: n, n + 2. Example: 12 and 14, or 11 and 13.
  • Product — the result of multiplying. Sum — the result of adding.
  • Uniform border — a strip of the same width x on every side of a shape. It adds 2x to each dimension.
  • Projectile — an object thrown or dropped. Its height (in feet) is often h = -16t^2 + v_0 t + h_0.
  • Initial velocity v_0 — the starting upward speed (ft/s). Initial height h_0 — the starting height (ft).
  • Unreasonable (rejected) solution — a solution of the equation that doesn't fit the situation. Example: a width of -12 ft.

All Math words in the glossary

1. Area problems

Recipe: (1) Let a variable be one side. (2) Write the other side in terms of it. (3) Write length × width = area. (4) Put it in standard form and solve. (5) Keep only positive lengths.

A rectangle with width labeled w and length labeled w + 3. The area is 40 square feet.Area = 40 ft²w + 3w

Example: A rectangle's length is 3 ft more than its width, and its area is 40\text{ ft}^2.

  • Width = w; length = w + 3.
  • Equation: w(w + 3) = 40.
  • Standard form: w^2 + 3w - 40 = 0 → (w + 8)(w - 5) = 0 → w = 5 (reject -8).
  • The rectangle is 5 ft by 8 ft. Check: 5 \times 8 = 40 ✓

Translating the side:

WordsExpression
length is 4 more than the widthl = w + 4
length is twice the widthl = 2w
length is 1 less than twice the widthl = 2w - 1
a square with side sarea = s^2

Area vs. perimeter. Area is multiply (l \times w) and gives a quadratic. Perimeter is add (2l + 2w) and gives a linear equation. A common wrong answer choice uses perimeter.

Borders, frames and walkways. A border of width x on every side adds x on the left and x on the right, so each dimension grows by 2x.

Example: A 10 in by 8 in photo has a frame of width x. The outside of the frame measures (10 + 2x) by (8 + 2x). If the total area is 120\text{ in}^2:

(10 + 2x)(8 + 2x) = 120
A 10 inch by 8 inch photo inside a frame of the same width x on all four sides. Each outside dimension grows by 2x: the outside measures 10 + 2x by 8 + 2x, so the total area equation is (10 + 2x)(8 + 2x) = 120.photo 10 in × 8 in108xxxx10 + 2x8 + 2xFrame of width x on every side(10 + 2x)(8 + 2x) = 120

If the problem gives only the area of the border itself, subtract the inside:

(\text{outer area}) - (\text{inner area}) = \text{border area}

2. Number problems

Consecutive integers. Let the first be n. Then:

TypeNext numbersExample
consecutive integersn + 1, n + 27, 8, 9
consecutive even integersn + 2, n + 412, 14, 16
consecutive odd integersn + 2, n + 411, 13, 15

Yes — odd numbers are also 2 apart, so the expression is the same as for even numbers. What makes them odd is the starting value.

Example: The product of two consecutive integers is 56.

n(n + 1) = 56 \Rightarrow n^2 + n - 56 = 0 \Rightarrow (n + 8)(n - 7) = 0

So n = 7 (the numbers are 7 and 8) or n = -8 (the numbers are -8 and -7). Both pairs work unless the problem says "positive."

Number line from -10 to 10 with two pairs of consecutive integers marked: 7 and 8 (7 times 8 = 56) and -8 and -7 (-8 times -7 = 56). Both pairs solve n(n + 1) = 56.−10−8−7−6−4−202467810(−8)(−7) = 567 × 8 = 56n(n + 1) = 56: two pairs of neighbors, 1 apart

"Product" means multiply; "sum" means add. "The product of two consecutive integers is 56" is n(n + 1) = 56, while "the sum is 15" is n + (n + 1) = 15 — a linear equation.

A number and its square. "The sum of a number and its square is 90": x + x^2 = 90, or x^2 + x - 90 = 0. "A number squared, plus 3 times the number, is 40": x^2 + 3x = 40.

Watch the squaring words: "the square of a number plus 1" is x^2 + 1, but "the square of one more than a number" is (x + 1)^2.

3. Projectile (height) problems

For an object thrown straight up (or dropped) near Earth, with height in feet and time t in seconds:

h = -16t^2 + v_0 t + h_0
  • -16 comes from gravity (in feet). It is negative because gravity pulls the object down.
  • v_0 = initial upward velocity (ft/s). Use 0 if the object is simply dropped.
  • h_0 = starting height (ft). Use 0 if it starts on the ground.

The GED usually gives you this equation (or the full equation for the object) in the question. Your job is to set h equal to the right number:

Question asksSet h equal to
When does it hit the ground?0
When is it 22 ft high?22
When does it return to its starting height?h_0

Example: A ball is thrown up from 6 ft with v_0 = 40 ft/s. When is it 22 ft high?

-16t^2 + 40t + 6 = 22

Subtract 22: -16t^2 + 40t - 16 = 0. Divide by -8: 2t^2 - 5t + 2 = 0 → (2t - 1)(t - 2) = 0 → t = 0.5 s (going up) or t = 2 s (coming down). Both make sense here.

Graph of a ball's height h = -16t squared + 40t + 6, starting at 6 feet. The dashed line h = 22 crosses the curve twice: at t = 0.5 s on the way up and t = 2 s on the way down.t (s)h (ft)0.511.522.5351015202530350h = 22t = 0.5 (up)t = 2 (down)start 6 fth = −16t² + 40t + 6−16t² + 40t + 6 = 22 → t = 0.5 s or t = 2 s

Example: A rocket launched from the ground at 96 ft/s: h = -16t^2 + 96t. It lands when -16t^2 + 96t = 0 \Rightarrow -16t(t - 6) = 0 \Rightarrow t = 0 (launch) or t = 6 s (landing).

Graph of a rocket's height h = -16t squared + 96t. It is launched from the ground at t = 0, reaches 144 feet at t = 3, and lands at t = 6 seconds.t (s)h (ft)123456740801201600launch t = 0lands t = 6top (3, 144)h = −16t² + 96t−16t(t − 6) = 0 → t = 0 or t = 6

Interpret every solution: negative times and lengths are thrown out; t = 0 is usually the start, not the answer.

Formula

Formulas you will use

A = lw \qquad A_{\text{square}} = s^2 h = -16t^2 + v_0 t + h_0 \quad (\text{feet, seconds}) x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • A = area, l = length, w = width, s = side.
  • h = height after t seconds, v_0 = starting upward speed, h_0 = starting height.

The area formulas (rectangle/parallelogram, square) and the quadratic formula are on the GED formula sheet. The projectile equation is not on the sheet — but GED questions give it to you when it is needed. Consecutive-integer expressions (n, n + 1, n + 2) must be memorized.

Worked example

Worked example 1 (easy): a rectangular rug

A rug's length is 5 ft more than its width. Its area is 84\text{ ft}^2. Write an equation and find the dimensions.

  1. Let w = width. Then length = w + 5.
  2. Equation: w(w + 5) = 84.
  3. Standard form: w^2 + 5w - 84 = 0.
  4. Factor (multiply to -84, add to 5: 12 and -7): (w + 12)(w - 7) = 0.
  5. w = 7 or w = -12. A width can't be negative, so w = 7.
Area model of the rug, drawn to scale. Width w = 7 ft, length w + 5 = 12 ft. The rug splits into a 7 by 7 square (w squared = 49) and a 7 by 5 strip (5w = 35). Total w(w + 5) = w squared + 5w = 84 square feet.w²7 × 7 = 495w35w = 75w = 7w + 5 = 12 ftRug: length is 5 ft more than the widthw(w + 5) = w² + 5w = 49 + 35 = 84 ft²

Answer: 7 ft by 12 ft.

Check: 7 \times 12 = 84 ✓ and 12 - 7 = 5 ✓

Worked example

Worked example 2: consecutive even integers

The product of two consecutive positive even integers is 168. Find them.

  1. Let n = the smaller even integer. The next even integer is n + 2.
  2. Equation: n(n + 2) = 168.
  3. Standard form: n^2 + 2n - 168 = 0.
  4. Factor (multiply to -168, add to 2: 14 and -12): (n + 14)(n - 12) = 0.
  5. n = 12 or n = -14. The integers are positive, so n = 12.
Area model for n(n + 2) = 168 with n = 12: a 12 by 12 square (n squared = 144) plus a 12 by 2 strip (2n = 24). The rectangle is 12 by 14 and 144 + 24 = 168.n²12 × 12 = 1442n24n = 122n = 12n + 2 = 14Two consecutive even integers: n and n + 2n(n + 2) = n² + 2n = 144 + 24 = 168

Answer: 12 and 14.

Check: 12 \times 14 = 168 ✓, both are even and 2 apart ✓

Worked example

Worked example 3 (real world): a garden path

A 6 ft by 8 ft garden has a path of the same width x all the way around it. The garden and path together cover 120\text{ ft}^2. How wide is the path?

  1. Each dimension grows by 2x: outer size is (6 + 2x) by (8 + 2x).
  2. Equation: (6 + 2x)(8 + 2x) = 120.
  3. Expand: 48 + 12x + 16x + 4x^2 = 120, so 4x^2 + 28x + 48 = 120.
  4. Subtract 120: 4x^2 + 28x - 72 = 0. Divide by 4: x^2 + 7x - 18 = 0.
  5. Factor: (x + 9)(x - 2) = 0, so x = 2 (reject -9).
A 6 ft by 8 ft garden surrounded by a path of width x on every side, drawn to scale with x = 2 ft. The outside measures 6 + 2x = 10 ft by 8 + 2x = 12 ft, and 10 times 12 = 120 square feet.garden8 ft × 6 ft86xxxxpath8 + 2x = 12 ft6 + 2x= 10 ftPath of width x = 2 ft all around (to scale)(6 + 2x)(8 + 2x) = 120 → x = 2 → 10 × 12 = 120

Answer: the path is 2 ft wide.

Check: Outer size 10 \times 12 = 120 ✓

Worked example

Worked example 4 (real world): a thrown ball

A ball is thrown upward from the top of a 64-ft building at 48 ft/s. Its height is h = -16t^2 + 48t + 64. When does it hit the ground?

  1. Ground means h = 0: -16t^2 + 48t + 64 = 0.
  2. Divide every term by -16: t^2 - 3t - 4 = 0.
  3. Factor (multiply to -4, add to -3: -4 and 1): (t - 4)(t + 1) = 0.
  4. t = 4 or t = -1. Time can't be negative, so t = 4.
Graph of h = -16t squared + 48t + 64 for a ball thrown up from a 64-foot roof. It rises to 100 feet at t = 1.5 s and hits the ground (h = 0) at t = 4 s; the solution t = -1 is rejected.t (s)h (ft)0.511.522.533.544.5204060801000roof 64 fttop (1.5, 100)ground t = 4h = −16t² + 48t + 64(t − 4)(t + 1) = 0 → t = 4 (t = −1 rejected)

Answer: after 4 seconds.

Check: -16(16) + 48(4) + 64 = -256 + 192 + 64 = 0 ✓

Worked example

Worked example 5 (GED-hard): a pricing problem

A gym charges $20 per class and averages 100 students. For every $1 increase in price, it loses 4 students. What price(s) would bring in $2,016 per class?

  1. Let x = the number of $1 increases. Price = 20 + x; students = 100 - 4x.
  2. Revenue = price × students: (20 + x)(100 - 4x) = 2016.
  3. Expand: 2000 - 80x + 100x - 4x^2 = 2016, so -4x^2 + 20x + 2000 = 2016.
  4. Subtract 2016: -4x^2 + 20x - 16 = 0. Divide by -4: x^2 - 5x + 4 = 0.
  5. Factor: (x - 1)(x - 4) = 0, so x = 1 or x = 4.
Graph of gym revenue R = (20 + x)(100 - 4x) against x, the number of 1-dollar price increases, with the revenue axis starting at 1,980 dollars. The dashed line at 2,016 dollars meets the curve at x = 1 (price 21 dollars) and x = 4 (price 24 dollars).xrevenue ($)01234561,9801,9902,0002,0102,0202,030$2,016x = 1: $21x = 4: $24R = (20 + x)(100 − 4x)x = number of $1 increases (revenue axis starts at $1,980)

Answer: a price of $21 or $24.

Check: 21 \times 96 = 2{,}016 ✓ and 24 \times 84 = 2{,}016 ✓ (dollars)

Calculator tip

Calculator tip: TI-30XS MultiView

  • Writing the equation is calculator-free — it's about reading carefully. Use the calculator for the solving step.
  • Check a solution fast: substitute it into your original equation. Example: for (6 + 2x)(8 + 2x) = 120 type ( 6 + 2 × 2 ) ( 8 + 2 × 2 ) enter → 120 ✓.
  • Square roots for the quadratic formula: 2nd then x². Negative numbers: the (−) key.
  • Projectile check: for -16t^2 + 48t + 64 at t = 4: (−) 16 × 4 x² + 48 × 4 + 64 enter → 0.

When NOT to use it: don't reach for the calculator to "find" the equation — set it up on paper first. And if the numbers factor nicely, factoring by hand is quicker than the formula.

Common mistake

Common mistakes

  • Using perimeter instead of area. "Area is 40" needs w(w + 3) = 40, not 2w + 2(w + 3) = 40.
  • Adding x instead of 2x for a border. A border on both sides adds 2x to each dimension: (10 + 2x), not (10 + x).
  • Mixing up consecutive types. Consecutive integers: n, n + 1. Consecutive even or odd integers: n, n + 2. Don't use n + 1 for odd numbers.
  • Sum vs. product. "Product" means multiply: n(n + 1) = 56, not n + (n + 1) = 56.
  • Distributing wrong. w(2w - 1) = 2w^2 - w, not 2w^2 - 1. Multiply w by every term.
  • Keeping impossible answers. A width of -8 or a time of -1 s doesn't make sense — reject it. And watch which answer is asked for: the width, the length, or the larger number?
Test strategy

GED test strategy

  • "Which equation…?" questions: test each choice with the true answer if you can find it quickly, or check that the choice uses multiply for area/product and 2x for borders.
  • Plug the answer choices in. On "find the width" questions, try each choice in the story: does length × width give the area?
  • Define the variable in words ("w = width in feet") before writing anything. Most errors come from losing track of what x means.
  • Read the last sentence twice. It tells you what to report: width, length, the larger integer, or the time.
  • Estimate: a square with area 40 has side about 6.3, so a rectangle with area 40 should have sides near that size.
  • Time: if solving is long, remember many questions only ask for the equation.

Check yourself

Write quadratic equations from contextNo calculator

A rectangular patio has a length that is 6 feet more than its width, w. The area of the patio is 55 square feet.

A rectangle with width labeled w and length labeled w + 6. The area is 55 square feet.Area = 55 ft²w + 6w

Which equation can be used to find w?

Answer choices
Hint

The area of a rectangle is length times width. What expression is the length?

Check yourself

Write quadratic equations from contextNo calculatorDrop-down

The product of two consecutive integers is 42. Let n be the smaller integer.

The equation that models this situation is .

Choose an answer from each menu.

Hint

Consecutive integers are 1 apart. "Product" tells you which operation to use.

Summary

  • Area: write one side as w, the other in terms of w, and set length × width = area.
  • Borders: each dimension grows by 2x; for the border alone, subtract the inner area.
  • Consecutive integers: n, n + 1; consecutive even/odd: n, n + 2. "Product" → multiply.
  • Projectiles: h = -16t^2 + v_0 t + h_0; set h to the height asked for (0 for the ground).
  • Put the equation in standard form, solve, then reject answers that don't fit the situation.
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