Quadratic Word Problems: Method with Worked Examples

Quadratic word problems translate a sentence into one quadratic equation, then hand the equation back to the sentence. Every example below follows the same five steps: read the words, name the unknown with its unit, build the equation from the stated relation, solve it, and then reject any root that fails the original words before answering with units. The solving methods themselves (factorisation, completing the square, the quadratic formula) are not re-taught here; they are linked at the end. The class 10 grounding is the CBSE 2026-27 mathematics curriculum, which lists situational problems based on quadratic equations related to day-to-day activities.

The translation method

name the unknown

pick the smallest or most constrained quantity in the question, call it x, and write down its unit

translate the relation

write every other quantity from the words in terms of x, then form the one equation the words state

solve

rearrange to standard form ax^2 + bx + c = 0 and solve by any linked method

context check

test every root against the original words: lengths and speeds are positive, ages are non-negative, counts are whole; reject roots that fail

answer in words

state the final answer in the language of the question, with units, and substitute it back into the original statement

Worked examples

Example 1: area of a rectangle

The length of a rectangle is 5 m more than its width, and its area is 84 m^2. Find the dimensions.

1. Unknown: let the width be w metres; then the length is (w + 5) metres.
2. Relation: w(w + 5) = 84, so w^2 + 5w - 84 = 0.
3. Solve: factorising, (w - 7)(w + 12) = 0, so w = 7 or w = -12.
4. Context check: a width cannot be negative, so -12 is rejected; w = 7 stands.
5. Answer in the original words: width 7 m and length 12 m. Check: 12 = 7 + 5 and 12 x 7 = 84.

The rectangle is 7 m by 12 m.

Example 2: consecutive positive integers

The product of two consecutive positive integers is 462. Find them.

1. Unknown: let the smaller integer be n; the next is n + 1.
2. Relation: n(n + 1) = 462, so n^2 + n - 462 = 0.
3. Solve: discriminant 1 + 1848 = 1849 = 43^2, so n = (-1 + 43)/2 = 21 or n = (-1 - 43)/2 = -22.
4. Context check: the words say positive integers, so -22 is rejected, even though (-22) x (-21) = 462 satisfies the equation. The equation models the words, not the other way round.
5. Answer in the original words: 21 and 22. Check: they are consecutive, positive, and 21 x 22 = 462.

The integers are 21 and 22.

Example 3: speed from a time difference

A train covers 360 km. If its speed were 5 km/h higher, the same journey would take 1 hour less. Find the actual speed.

1. Unknown: let the actual speed be s km/h. Times are 360/s hours and 360/(s + 5) hours.
2. Relation: 360/s - 360/(s + 5) = 1. Multiplying through by s(s + 5): 360(s + 5) - 360s = s(s + 5), so 1800 = s^2 + 5s, that is s^2 + 5s - 1800 = 0.
3. Solve: discriminant 25 + 7200 = 7225 = 85^2, so s = (-5 + 85)/2 = 40 or s = (-5 - 85)/2 = -45.
4. Context check: a speed must be positive, so -45 is rejected; s = 40 stands.
5. Answer in the original words: the actual speed is 40 km/h. Check: 360/40 = 9 hours, 360/45 = 8 hours, and 9 - 8 = 1 hour less, as stated.

The train's actual speed is 40 km/h.

Example 4: ages whose product is given

Arjun is 3 years older than Bina. The product of their ages is 238. Find their ages.

1. Unknown: let Bina's age be b years; Arjun's is (b + 3) years.
2. Relation: b(b + 3) = 238, so b^2 + 3b - 238 = 0.
3. Solve: discriminant 9 + 952 = 961 = 31^2, so b = (-3 + 31)/2 = 14 or b = (-3 - 31)/2 = -17.
4. Context check: an age cannot be negative, so -17 is rejected; b = 14 stands.
5. Answer in the original words: Bina is 14 and Arjun is 17. Check: 17 - 14 = 3 and 14 x 17 = 238.

Bina is 14 years old and Arjun is 17.

Example 5: two numbers with a sum of squares

Two positive numbers differ by 3, and the sum of their squares is 225. Find them.

1. Unknown: let the smaller number be x; the larger is x + 3.
2. Relation: x^2 + (x + 3)^2 = 225, so 2x^2 + 6x - 216 = 0, which divides by 2 to give x^2 + 3x - 108 = 0.
3. Solve: discriminant 9 + 432 = 441 = 21^2, so x = (-3 + 21)/2 = 9 or x = (-3 - 21)/2 = -12.
4. Context check: the words say positive numbers, so -12 is rejected. Note that -12 and -9 do satisfy the sum of squares, which is exactly why the words must be checked last.
5. Answer in the original words: 9 and 12. Check: 12 - 9 = 3 and 9^2 + 12^2 = 81 + 144 = 225.

The numbers are 9 and 12.

Summary

Setup

name one unknown with its unit; express everything else from the words; form one equation

Standard form

rearrange to ax^2 + bx + c = 0 before solving

Solving

use any linked method; word problems change the setup and the checking, not the algebra

Context check

reject roots that fail the words: negative lengths or speeds, negative ages, non-whole counts

Final answer

state it in the original words with units, then substitute it back into the statement

Frequently asked questions

Why reject a root that solves the equation?

A root can satisfy the equation and still fail the words, since lengths and speeds are positive, ages cannot be negative and counts are whole. The equation models the words, so every root must be tested against the original statement.

Do quadratic word problems need a new solving method?

No. Word problems add the setup and the context check. The solving itself uses factorisation, completing the square or the quadratic formula, which the linked chapter page and method guide already teach.

How do I choose what to call x?

Name the smallest or most constrained quantity in the question, with its unit. Then express every other quantity from the words in terms of x before forming the equation.

What if both roots fit the words?

Keep every distinct root that fits the original words, and state each answer with its units. Two different valid roots give two solutions; a repeated root gives one distinct solution. Reject a root only when it breaks a stated condition.

Where are situational quadratic problems examined?

The CBSE Class X 2026-27 mathematics curriculum lists situational problems based on quadratic equations related to day-to-day activities; the official document is linked in the sources section below.

Sources and scope

Method checked against OpenStax, Solve Applications of Quadratic Equations, retrieved 15 November 2026. Class 10 grounding: the CBSE Class X 2026-27 mathematics curriculum lists "situational problems based on quadratic equations related to day-to-day activities", at cbseacademic.nic.in, retrieved 15 November 2026. This page teaches translation into and out of a quadratic; it does not re-teach the solving methods, graphs or the discriminant.

Related reading on this site: quadratic equations (class 10, chapter 4), the step-by-step quadratic equations method guide.