10 min readPathway sheet3-mark or 5-mark question; often a 4-mark case-based question

Word problems on quadratic equations

Let the unknown be x and form the equation. Solve it, then reject roots that make no sense, such as a negative length, age or speed.

Speed problemsTime = distance ÷ speed; 'takes 1 hour less' means time₁ − time₂ = 1

Consecutive numbersx and x + 1 (x and x + 2 for odd or even numbers)

RejectNegative lengths, ages, speeds or numbers of items

AnswerA sentence with units

Method

1
Let the unknown be x, with units. Write other quantities in terms of x.
2
Turn the condition into an equation.
3
Clear fractions and write it as ax² + bx + c = 0.
4
Solve by factorisation or by the quadratic formula.
5★
Reject any root that makes no sense in the situation, and say why.
6
Check with the words of the question and answer with units.

★ key step

Common situations and their equations

Situation	Let	Equation
Area of a rectangle	breadth x, length x + 5	x(x + 5) = area
Speed and time	speed x km/h	d/x − d/(x + k) = time saved
Ages	present age x	(x − 3)(x + 4) = given product
A number and its reciprocal	the number x	x + 1/x = given sum
Cost and quantity	number of items x	price of one item = total cost ÷ x
Right triangle	shorter side x	x² + (x + k)² = hypotenuse²

Example 1: speed of a bus

Problem

A bus travels 300 km at a uniform speed. If the speed were 10 km/h more, the journey would take 1 hour less. Find the speed of the bus.

Working

Let the speed be x km/h. Time = distance ÷ speed.
300/x − 300/(x + 10) = 1
Multiply by x(x + 10): 300(x + 10) − 300x = x(x + 10), so 3000 = x² + 10x.
x² + 10x − 3000 = 0
Product −3000, sum 10: 60 and −50. So (x + 60)(x − 50) = 0.
x = 50 or x = −60. Reject −60, since speed cannot be negative.
Check: 300/50 = 6 hours and 300/60 = 5 hours, which is 1 hour less.

Answer

The speed of the bus is 50 km/h.

Example 2: ages

Problem

The product of Asha's age 4 years ago and her age 6 years from now is 119. Find her present age.

Working

Let her present age be x years.
(x − 4)(x + 6) = 119
x² + 2x − 24 = 119, so x² + 2x − 143 = 0.
(x + 13)(x − 11) = 0, so x = 11 or x = −13.
Reject −13, since age cannot be negative.
Check: (11 − 4)(11 + 6) = 7 × 17 = 119.

Answer

Asha is 11 years old.

Example 3: cost and quantity

Problem

A shopkeeper buys some notebooks for ₹ 600. If each notebook cost ₹ 5 less, he could buy 10 more for the same money. How many notebooks did he buy?

Working

Let the number of notebooks be x. Price of one = ₹ 600/x.
With 10 more notebooks, price of one = ₹ 600/(x + 10).
600/x − 600/(x + 10) = 5
Multiply by x(x + 10): 6000 = 5x(x + 10), so x² + 10x − 1200 = 0.
(x + 40)(x − 30) = 0, so x = 30 (reject −40).
Check: 600 ÷ 30 = ₹ 20 each; 600 ÷ 40 = ₹ 15 each, which is ₹ 5 less.

Answer

He bought 30 notebooks, at ₹ 20 each.

In real life

Bus timetables lead to quadratic equations. So do the price per item in a bulk purchase for a school function and the size of a garden plot. Rejecting the impossible root is everyday common sense.

Exam angle

3 marks: “Two numbers differ by 3 and their product is 504. Find the numbers.” x(x + 3) = 504 gives x² + 3x − 504 = 0. D = 9 + 2016 = 2025, √D = 45. x = 21 or −24. The numbers are 21 and 24, or −24 and −21.

5 marks: “A car travels 240 km at a uniform speed. If the speed were 20 km/h more, it would take 1 hour less. Find the speed of the car.” 240/x − 240/(x + 20) = 1 gives x² + 20x − 4800 = 0, so (x + 80)(x − 60) = 0. Speed = 60 km/h (4 hours at 60, 3 hours at 80).

Remember it

Form, solve, reject, check, say: no negative lengths, ages or speeds.